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Subjective expected utility with a spectral state space

Abstract

An agent faces a decision under uncertainty with the following structure. There is a set \({\mathcal {A}}\) of “acts”; each will yield an unknown real-valued payoff. Linear combinations of acts are feasible; thus, \({\mathcal {A}}\) is a vector space. But there is no pre-specified set of states of nature. Instead, there is a Boolean algebra \({\mathfrak {I}}\) describing information the agent could acquire. For each element of \({\mathfrak {I}}\), she has a conditional preference order on \({\mathcal {A}}\). I show that if these conditional preferences satisfy certain axioms, then there is a unique compact Hausdorff space \({\mathcal {S}}\) such that elements of \({\mathcal {A}}\) correspond to continuous real-valued functions on \({\mathcal {S}}\), elements of \({\mathfrak {I}}\) correspond to regular closed subsets of \({\mathcal {S}}\), and the conditional preferences have a subjective expected utility (SEU) representation given by a Borel probability measure on \({\mathcal {S}}\) and a continuous utility function. I consider two settings; in one, \({\mathcal {A}}\) has a partial order making it a Riesz space or Banach lattice, and \({\mathfrak {I}}\) is the Boolean algebra of bands in \({\mathcal {A}}\). In the other, \({\mathcal {A}}\) has a multiplication operator making it a commutative Banach algebra, and \({\mathfrak {I}}\) is the Boolean algebra of regular ideals in \({\mathcal {A}}\). Finally, given two such vector spaces \({\mathcal {A}}_1\) and \({\mathcal {A}}_2\) with SEU representations on topological spaces \({\mathcal {S}}_1\) and \({\mathcal {S}}_2\), I show that a preference-preserving homomorphism \({\mathcal {A}}_2{{\longrightarrow }}{\mathcal {A}}_1\) corresponds to a probability-preserving continuous function \({\mathcal {S}}_1{{\longrightarrow }}{\mathcal {S}}_2\). I interpret this as a model of changing awareness.

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Fig. 1
Fig. 2
Fig. 3

Notes

  1. 1.

    These results are stated for the Boolean algebra of regular open sets. But algebra of regular closed sets is dual to the algebra of regular open sets, and almost identical proofs work in this case. Johnstone (1982, Proposition 1.13) proves a similar result for the regular elements of an abstract Heyting algebra—it can be applied to both open and closed sets.

  2. 2.

    The integrals on the right side of (2.1) are well defined because the integrands are bounded, because \(\alpha \), \(\beta \), and u are continuous and \({\mathcal {S}}\) is compact.

  3. 3.

    In fact, any Riesz space is automatically a distributive lattice—that is \(\alpha \wedge (\beta \vee \gamma ) = (\alpha \wedge \beta )\vee (\alpha \wedge \gamma )\) and \(\alpha \vee (\beta \wedge \gamma ) = (\alpha \vee \beta )\wedge (\alpha \vee \gamma )\) for all \(\alpha ,\beta ,\gamma \in {\mathcal {A}}\) (Fremlin 2004, 352E(c)).

  4. 4.

    Note that when \(\alpha \ne 0\), we do not require \(n\,\alpha \ge \beta \) for some \(n\in {\mathbb {N}}\)—we just require \(n\,\alpha \not \le \beta \).

  5. 5.

    Naively, it might seem that \({\mathfrak {K}}({\mathcal {T}})\) should be a band for any closed subset \({\mathcal {T}}\) of \({\mathcal {S}}\). To see why this is false, let \({\mathcal {S}}=[-1,1]\) and \({\mathcal {T}}=\{0\}\), and let \(\alpha _n(s)=1-\exp (-n|s|)\) for all \(n\in {\mathbb {N}}\). Then \(\{\alpha _n\}_{n=1}^{\infty }\) is sequence in \({\mathfrak {K}}\{0\}\), but it order-converges to \(\varvec{1}\), which is not in \({\mathfrak {K}}\{0\}\).

  6. 6.

    In fact, \({\mathcal {A}}({\mathcal {T}})={\mathcal {C}}({\mathcal {T}})\), by the Tietze Extension Theorem. But this is not important right now.

  7. 7.

    See “Appendix A” for more information about Stonean spaces.

  8. 8.

    Terminologies vary. Fremlin (2004, Definition 354G) calls these M-spaces. Meyer-Nieberg (1991, Definition 12.1.2) calls them M-spaces with an order unit, while Aliprantis and Border (2006, Sect. 9.5) call them AM-spaces with unit. Meyer-Nieberg (1991) and Aliprantis and Border (2006) reserve the term M-space for the broader class of Banach lattices such that \(\Vert \alpha \vee \beta \Vert =\max \{\Vert \alpha \Vert ,\Vert \beta \Vert \}\) for all \(\alpha ,\beta \ge 0\).

  9. 9.

    Strictly speaking, this is a real commutative algebra, since I have implicitly assumed that \({\mathcal {A}}\) is a vector space over the field \({\mathbb {R}}\) of real numbers. One could also consider complex algebras (where \({\mathcal {A}}\) is a vector space over the field \({\mathbb {C}}\) of complex numbers) or indeed \({\mathbb {F}}\)-algebras for any field \({\mathbb {F}}\). Almost all the literature on Banach algebras deals with complex algebras.

  10. 10.

    More generally, we could replace \(\{2^{-n^2}\}_{n=0}^{\infty }\) with any sequence \(\{w_n\}_{n=0}^{\infty }\) such that \(w_{n+m}\le w_n\cdot w_m\).

  11. 11.

    But \({\mathbb {R}}[x]\)is semisimple under other norms, such as \(\Vert a_0+a_1\,x+\cdots +a_N \, x^N\Vert :=|a_0|+\cdots +|a_N|\).

  12. 12.

    Ingelstam (1964) refers to a logically equivalent condition as R4. See Ingelstam (1964, Sects. 13–14) for conditions under which a Banach algebra is strictly real.

  13. 13.

    A subset is meagre if it is a countable union of nowhere dense subsets. For example, the set of rational numbers is meagre in \({\mathbb {R}}\). So is the Cantor Set.

  14. 14.

    That is, \(a^*_{n,m} = {\overline{a}}_{m,n}\) for all \(n,m\in [1\cdots N]\).

  15. 15.

    This example can be generalized by replacing \({\mathbb {C}}^{N\times N}\) with the algebra of bounded linear operators on a Hilbert space \({\mathcal {H}}\) and letting \({\mathcal {A}}\) be a bounded hermitian operator on \({\mathcal {H}}\).

  16. 16.

    This is not the standard definition of regular found in the literature. But it is equivalent to the standard definition for unitary commutative real Banach algebras (Lemma C2), which suffices for our purposes.

  17. 17.

    \({\mathcal {S}}\) is also called the maximal spectrum, or the structure space, or the Gelfand representation of \({\mathcal {S}}\).

  18. 18.

    To be precise, there is a canonical homeomorphism \(\phi :{\mathcal {K}}{{\longrightarrow }}{\mathcal {S}}\), such that \(\varPhi (\alpha )=\alpha \circ \phi \) for all \(\alpha \in {\mathcal {A}}\).

  19. 19.

    If \({\mathbf { A}}\) is an \(N\times N\) matrix, then \({\mathcal {S}}\) is just a discrete subset of \({\mathbb {C}}\), with at most N points. But the same result is true when \({\mathbf { A}}\) is a bounded hermitian operator on an infinite-dimensional Hilbert space; in this case, its spectrum could be an arbitrary compact subset of \({\mathbb {C}}\) (Conway 1990, Proposition VII.8.10).

  20. 20.

    In this case, \({\mathcal {A}}\) is called natural. For example, suppose that, for any \(\alpha _1,\ldots ,\alpha _N\in {\mathcal {A}}\) such that \(|\alpha _1|+\cdots |\alpha _N|\) is nonzero everywhere, there exist \(\beta _1,\ldots ,\beta _N\in {\mathcal {A}}\) such that \(\alpha _1\beta _1+\cdots +\alpha _N\beta _N=1\). Then \({\mathcal {A}}\) is natural (Dales et al. 2003, Proposition 3.3.2).

  21. 21.

    A (real) B*-algebra is a (real) Banach algebra equipped with an involution (i.e. a self-isomorphism that is its own inverse) denoted *, such that \(\Vert \alpha ^*\cdot \alpha \Vert =\Vert \alpha \Vert ^2\) for all \(\alpha \in {\mathcal {A}}\). They are closely related to C*-algebras, but form a distinct class. See Palmer (1970) or Rosenberg (2015) for more information.

  22. 22.

    Any uniform Banach algebra is commutative and semisimple (Palmer 1994, Propositions 3.1.7 and 3.1.8). So in Theorem 5, we actually only need \({\mathcal {A}}\) to be strictly real and uniform, rather than realistic.

  23. 23.

    Proof.\(\varPsi \) is injective \(\Leftrightarrow \ker (\varPsi )=\{0\}\Leftrightarrow {\mathfrak {K}}({\widetilde{\mathcal {S}}})=\{0\}\Leftrightarrow {\widetilde{\mathcal {S}}}={\mathcal {S}}\), because \({\widetilde{\mathcal {S}}}\) is closed and \({\mathcal {S}}\) is Tychonoff.

  24. 24.

    For “\(\Rightarrow \)”, recall that \({\mathcal {C}}({\mathcal {S}}')\) separates the points of \({\mathcal {S}}'\). For “\(\Leftarrow \)”, use the Tietze Extension Theorem.

  25. 25.

    The proofs are identical to those sketched in Footnotes 20 and 21.

  26. 26.

    Karni and Vierø (2017) use something like the Savage model. But they also allow lotteries over acts.

  27. 27.

    However, Karni (2017, Sect. 2.6) shows how “states as preferences” can be seen as a special case of “states as functions”.

  28. 28.

    See footnote 3 in Sect. 3 for the definition of distributive lattice.

  29. 29.

    This property is also called complete regularity, when Hausdorff is not a background assumption.

  30. 30.

    Thus, \(\{{\mathcal {O}}_1,\ldots ,{\mathcal {O}}_N\}\) is a regular open partition of \({\mathcal {S}}\) if and only if \(\{\mathrm {clos}({\mathcal {O}}_1),\ldots ,\mathrm {clos}({\mathcal {O}}_N)\}\) is a regular closed partition of \({\mathcal {S}}\).

  31. 31.

    Typically, the Gleason cover is defined using the Stone space of the Boolean algebra \({\mathfrak {O}}({\mathcal {S}})\) of regular open sets. But as I have already noted in the proof of Proposition A1, there is a canonical isomorphism \({\mathfrak {R}}({\mathcal {S}})\cong {\mathfrak {O}}({\mathcal {S}})\), so this makes no difference.

  32. 32.

    See Example 5.1 for the definitions of quotient map and embedding.

  33. 33.

    In other words, parts (a) and (c) define a contravariant functor from the category of unitary Archimedean Riesz spaces and uniferent homomorphisms to the category of compact Hausdorff spaces.

  34. 34.

    This can also be proved by adapting the proof of Example 3.6. See Lemma C3 for a similar argument.

  35. 35.

    That is, for any \(\alpha '\in {\mathcal {A}}'\), there is some \({\widetilde{\alpha }}\in {\widetilde{\mathcal {A}}}\) with \({\widetilde{\alpha }}>\alpha '\)—namely, some scalar multiple of \(\varvec{1}'\).

  36. 36.

    For more on regular Banach algebras, see Kaniuth (2009, Sect. 4.2), Dales et al. (2003, Sect. 25.2), or Palmer (1994, Sect. 7.2).

  37. 37.

    In fact, \({\mathcal {I}}\) is countable, because \({\mathbb {R}}\) is separable. But this is not important.

  38. 38.

    See Example 5.1 for the definitions of quotient map and embedding.

  39. 39.

    In other words, the construction in parts (a) and (d) defines a contravariant functor from the category of realistic Banach algebras to the category of compact Hausdorff spaces.

  40. 40.

    A small complication here is that the Gelfand Representation Theorem is normally stated for complex commutative Banach algebras—see, for example, Conway (1990, Theorem VII.8.9), Kaniuth (2009, Theorems 2.2.3 and 2.2.7) or Palmer (1994, Theorem 3.1.2). But Ingelstam (1964) develops the theory for real Banach algebras, which is what we need here.

  41. 41.

    In fact, Claim 1 implies that all elements of \({\mathcal {S}}\) are real-valued. But this is not important.

  42. 42.

    Incidentally, this implies that \(\ker (\varPsi )\) is a band whenever \(\varPsi \) is a regular Riesz homomorphism. An essentially identical argument shows that \(\ker (\varPsi )\) is a regular ideal whenever \(\varPsi \) is a regular algebra homomorphism between realistic Banach algebras.

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Correspondence to Marcus Pivato.

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Supported by Labex MME-DII (ANR11-LBX-0023-01). I would like to thank Dino Borie, Peter Wakker, and the anonymous referee for many helpful comments.

Appendices

Mathematical background

Boolean algebras A Boolean algebra is a set \({\mathcal {B}}\) equipped with a preorder \(\ge \) and binary operations \(\vee \) and \(\wedge \) under which it is a distributive lattice,Footnote 28 along with an order-reversing bijection \(\lnot \) such that \(\lnot (\lnot a)=a\) for all \(a\in {\mathcal {A}}\) (i.e. \(\lnot \) is an involution) and \(\lnot (a \vee b) = (\lnot a)\wedge (\lnot b)\) and \(\lnot (a \wedge b) = (\lnot a)\vee (\lnot b)\) for all \(a,b\in {\mathcal {B}}\) (these are called the De Morgan laws). Finally, we require \({\mathcal {B}}\) to have a unique maximal element 1 and minimal element 0, such that \(\lnot 1=0\) and \(\lnot 0=1\). It follows that \(0\wedge a=0\), \(0\vee a=a=1\wedge a\) and \(1\vee a=1\), for all \(a\in {\mathcal {A}}\). For example, if \({\mathcal {S}}\) is a probability space, then the set of all measurable subsets of \({\mathcal {S}}\) is a Boolean algebra under union, intersection and complementation. As this example suggests, Boolean algebras often represent the information which is available to an agent in some decision environment; that is how I use them in this paper. For more on Boolean algebras, see Givant and Halmos (2008) or Fremlin (2004, Chapter 31).

Let \({\mathcal {B}}\) and \({\widetilde{\mathcal {B}}}\) be Boolean algebras. A homomorphism is an order-preserving function \(\varPhi :{\mathcal {B}}{{\longrightarrow }}{\widetilde{\mathcal {B}}}\) that preserves all lattice operations and preserves negation (i.e. \(\varPhi (a\vee b)=\varPhi (a)\widetilde{\vee }\varPhi (b)\), \(\varPhi (a\wedge b)=\varPhi (a)\widetilde{\wedge }\varPhi (b)\) and \(\varPhi (\lnot b)=\widetilde{\lnot } \varPhi (b)\) for all \(a,b\in {\mathcal {B}}\)) and such that \(\varPhi (1)=\widetilde{1}\) and \(\varPhi (0)=\widetilde{0}\). It is an isomorphism if it is bijective; in this case, the inverse function \(\varPhi ^{-1}:{\widetilde{\mathcal {B}}}{{\longrightarrow }}{\mathcal {B}}\) is also an isomorphism. For example, if \({\mathcal {S}}_1\) and \({\mathcal {S}}_2\) are measure spaces, and \(\phi :{\mathcal {S}}_1{{\longrightarrow }}{\mathcal {S}}_2\) is measurable, then \(\phi ^{-1}\) is a homomorphism from the Boolean algebra of measurable subsets of \({\mathcal {S}}_2\) to the Boolean algebra of measurable subsets of \({\mathcal {S}}_1\).

An anti-homomorphism is an order-reversing function \(\varPhi :{\mathcal {B}}{{\longrightarrow }}{\widetilde{\mathcal {B}}}\) that switches the lattice operations and preserves negation (i.e. \(\varPhi (a\vee b)=\varPhi (a)\widetilde{\wedge }\varPhi (b)\), \(\varPhi (a\wedge b)=\varPhi (a)\widetilde{\vee }\varPhi (b)\) and \(\varPhi (\lnot b)=\widetilde{\lnot } \varPhi (b)\) for all \(a,b\in {\mathcal {B}}\)), such that \(\varPhi (1)=\widetilde{0}\) and \(\varPhi (0)=\widetilde{1}\). It is an anti-isomorphism if it is bijective; in this case, the inverse function \(\varPhi ^{-1}:{\widetilde{\mathcal {B}}}{{\longrightarrow }}{\mathcal {B}}\) is also an anti-isomorphism. For example, the negation operator \(\lnot \) is an anti-isomorphism from \({\mathcal {B}}\) to itself.

Topology Let \({\mathcal {S}}\) be a topological space. It is Hausdorff if, for any \(s_1,s_2\in {\mathcal {S}}\), there exist disjoint open subsets \({\mathcal {O}}_1,{\mathcal {O}}_2\subset {\mathcal {S}}\) with \(s_1\in {\mathcal {O}}_1\) and \(s_2\in {\mathcal {O}}_2\). An open cover of \({\mathcal {S}}\) is a collection \({\mathfrak {O}}\) of open subsets such that \({\mathcal {S}}=\bigcup _{{\mathcal {O}}\in {\mathfrak {O}}} {\mathcal {O}}\). The space \({\mathcal {S}}\) is compact if, for any such open cover \({\mathfrak {O}}\), there is a finite subset \({\mathfrak {Q}}\subseteq {\mathfrak {O}}\) such that \({\mathcal {S}}=\bigcup _{{\mathcal {Q}}\in {\mathfrak {Q}}} {\mathcal {Q}}\). For example, any closed, bounded subset of a Euclidean space \({\mathbb {R}}^N\) is compact and Hausdorff.

A Hausdorff space \({\mathcal {S}}\) is Tychonoff if, for all closed \({\mathcal {R}}\subset {\mathcal {S}}\) and all \(s\in {\mathcal {S}}\setminus {\mathcal {R}}\), there is a continuous function \(\alpha :{\mathcal {S}}{{\longrightarrow }}{\mathbb {R}}\) such that \(\alpha ({\mathcal {R}})=0\) while \(\alpha (s)=1\).Footnote 29 In effect, this means that the topological structure of \({\mathcal {S}}\) is completely “encoded” in the algebraic structure of \({\mathcal {C}}({\mathcal {S}})\). We will repeatedly use the fact that every compact Hausdorff space is Tychonoff (Willard 2004, Theorem 19.3).

A subset \({\mathcal {R}}\subseteq {\mathcal {S}}\) is clopen if \({\mathcal {R}}\) is both closed and open. Let \({\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}})\) denote the collection of all clopen subsets of \({\mathcal {S}}\). Then \({\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}})\) is a Boolean algebra under union, intersection and complementation. A space \({\mathcal {S}}\) is connected if \({\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}})=\{\emptyset ,{\mathcal {S}}\}\). In contrast, \({\mathcal {S}}\) is totally disconnected if the only connected subsets of \({\mathcal {S}}\) are singleton sets. If \({\mathcal {S}}\) is Hausdorff, then it is totally disconnected if and only if every point has a neighbourhood basis of clopen sets. Thus, for such spaces, \({\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}})\) is generally quite large. A Stone space is a totally disconnected, compact Hausdorff space. The Stone Representation Theorem says that for any Boolean algebra \({\mathcal {B}}\), there is a Stone space \({\mathcal {S}}\) (unique up to homeomorphism) such that \({\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}})\) is isomorphic to \({\mathcal {B}}\). This is called the Stone space of \({\mathcal {B}}\) (Fremlin 2004, 311F, 311I).

A topological space is extremely disconnected if the closure of every open set is clopen. Such a space is necessarily totally disconnected. A space \({\mathcal {S}}\) is extremely disconnected if and only if every regular closed set is clopen. In this case, \({\mathfrak {R}}({\mathcal {S}})={\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}})\), and the Boolean algebra operations for regular closed sets are identical to the ordinary set-theoretic operations. A Stonean space is an extremely disconnected, compact Hausdorff space. In particular, if \({\mathcal {S}}\) is any topological space, then the Stone space of the Boolean algebra \({\mathfrak {R}}({\mathcal {S}})\) is a Stonean space (Fremlin 2004, Theorems 314P and 314S); this fact will be important in the proofs of Theorems 1 and 6 and Proposition A2.

SEU representations on topological spaces The proofs of the main results involve two steps: first, I represent the conditional preference structure on a Riesz space or Banach algebra \({\mathcal {A}}\) by a conditional preference structure on the regular closed subsets of a topological space \({\mathcal {S}}\); second, I obtain an SEU representation for the conditional preference structure on \({\mathcal {S}}\). The second step relies on results from two other papers (Pivato and Vergopoulos 2018a, b), but these results must be adapted to the framework of the current paper. The next result is used in the proofs of Theorems 234 and 5.

Proposition A1

Let \({\mathcal {S}}\) be a (nonsingleton) compact Hausdorff space and let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\). Let \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) be a conditional preference structure on \({\mathcal {A}}\) satisfying (Rch\('\)). Then \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfies (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)) if and only if it admits a characteristic SEU representation (2.3), where \(\mu \) is a normal Borel probability measure with full support. Furthermore, \(\mu \) is unique, the elements of \(\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}\) are unique (\(\mu \)-almost everywhere), and u is unique up to positive affine transformation.

Proof

In principle, this is a straightforward consequence of a theorem in Pivato and Vergopoulos (2018a), except for one awkward detail: the results of Pivato and Vergopoulos (2018a) are formulated for conditional preference structures indexed by regular open subsets of \({\mathcal {S}}\). A subset \({\mathcal {O}}\subset {\mathcal {S}}\) is regular open if \({\mathcal {O}}=\mathrm {int}({\mathcal {C}})\) for some closed set \({\mathcal {C}}\). The collection \({\mathfrak {O}}({\mathcal {S}})\) of all regular open subsets is a Boolean algebra under the operations:

$$\begin{aligned} {\mathcal {O}}\wedge {\mathcal {O}}' := {\mathcal {O}}\cap {\mathcal {O}}', \quad {\mathcal {O}}\vee {\mathcal {O}}' := \mathrm {int}[\mathrm {clos}({\mathcal {O}}\cup {\mathcal {O}}')],\quad \text{ and }\quad \lnot {\mathcal {O}}:= \mathrm {int}({\mathcal {O}}^\complement ). \end{aligned}$$

For any \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\), let \({\mathcal {A}}({\mathcal {O}}):=\{\alpha _{\upharpoonleft {\mathcal {O}}}\); \(\alpha \in {\mathcal {A}}\}\). An \({\mathfrak {O}}({\mathcal {S}})\)-indexed conditional preference structure on \({\mathcal {A}}\) is a collection \(\{\succeq _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\), where for all \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\), \(\succeq _{\mathcal {O}}\) is a preference order on \({\mathcal {A}}({\mathcal {O}})\). The structural condition (Rch\('\)) and the axioms (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)) all have the same formulation for an \({\mathfrak {O}}({\mathcal {S}})\)-indexed conditional preference structure \(\{\succeq _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) as they do for an \({\mathfrak {R}}({\mathcal {S}})\)-indexed conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\)—the only difference being that we replace “\(\wedge \)” with “\(\cap \)” and replace “\(\cup \)” with “\(\vee \)” and replace “almost disjoint” with “disjoint”. A collection of disjoint regular open sets \(\{{\mathcal {O}}_1,\ldots ,{\mathcal {O}}_N\}\) is a regular open partition of \({\mathcal {S}}\) if \({\mathcal {O}}_1\vee \cdots \vee {\mathcal {O}}_N={\mathcal {S}}\).Footnote 30

In the framework of Pivato and Vergopoulos (2018a), the acts range over an arbitrary topological space \({\mathcal {X}}\) of outcomes. Furthermore, \({\mathcal {X}}\) is equipped with an “ex post preference order” \(\succeq _{xp}\), and the axiom (Dom\('\)) is formulated with respect to \(\succeq _{xp}\). The framework of Pivato and Vergopoulos (2018a) requires one further axiom, (C), which stipulates that \(\succeq _{xp}\) is continuous relative to the topology of \({\mathcal {X}}\) (i.e. the upper and lower contour sets of \(\succeq _{xp}\) are closed in \({\mathcal {X}}\)). In the present paper, \({\mathcal {X}}={\mathbb {R}}\), and \(\succeq _{xp}\) is just the standard ordering of the real number line, so that axiom (C) is automatically satisfied.

Let \(\mu \) be a Borel probability measure on \({\mathcal {S}}\). A liminal structure on\({\mathfrak {O}}({\mathcal {S}})\)subordinate to\(\mu \) is a collection \(\{\lambda _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\), where, for all \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\), \(\lambda _{\mathcal {O}}\in {\mathbb {L}}^1(\partial {\mathcal {O}},\mu ;[0,1])\), and for any regular open partition \(\{{\mathcal {O}}_1,\ldots ,{\mathcal {O}}_N\}\) of \({\mathcal {S}}\), we have

$$\begin{aligned} \lambda _{{\mathcal {O}}_1}+\cdots +\lambda _{{\mathcal {O}}_N} = 1,\quad \mu \text{-almost } \text{ everywhere } \text{ on }~\partial {\mathcal {O}}_1\cup \cdots \cup \partial {\mathcal {O}}_N. \end{aligned}$$
(A1)

A liminal SEU representation for \({\{\succeq _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}}\) is given by a Borel probability measure \(\mu \) on \({\mathcal {S}}\) and a liminal structure \(\{\lambda _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) subordinate to \(\mu \), along with a continuous utility function \(u:{\mathbb {R}}{{\longrightarrow }}{\mathbb {R}}\), such that, for all \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\) and all \(\alpha ,\beta \in {\mathcal {A}}({\mathcal {O}})\),

$$\begin{aligned} \left( \alpha \succeq _{\mathcal {O}}\beta \right)\iff & {} \left( \displaystyle \int _{{\mathcal {O}}} u\circ \alpha \ \mathrm {d}\mu + \int _{\partial {\mathcal {O}}} (u\circ \alpha ) \, \lambda _{\mathcal {O}}\ \mathrm {d}\mu \right. \nonumber \\\ge & {} \left. \int _{{\mathcal {O}}} u\circ \beta \ \mathrm {d}\mu + \int _{\partial {\mathcal {O}}} (u\circ \beta ) \, \lambda _{\mathcal {O}}\ \mathrm {d}\mu \right) . \end{aligned}$$
(A2)

(Here, I use the fact \(u\circ \alpha \) and \(u\circ \beta \) have unique extensions to \(\partial {\mathcal {O}}\), by continuity.) Theorem 3 of Pivato and Vergopoulos (2018a) says that if \(\{\succeq _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) satisfies (Rch\('\)), then \(\{\succeq _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) satisfies (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)) if and only if it admits a liminal SEU representation (A2). Furthermore, \(\mu \) is unique and each element of \(\{\lambda _{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) is unique \(\mu \)-almost everywhere on its domain.

There is a Boolean algebra isomorphism from \({\mathfrak {O}}({\mathcal {S}})\) to \({\mathfrak {R}}({\mathcal {S}})\) given by \({\mathcal {O}}\mapsto \mathrm {clos}({\mathcal {O}})\) for all \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\) (with inverse map \({\mathcal {R}}\mapsto \mathrm {int}({\mathcal {R}})\) for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\)). Furthermore, for any \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\), if \({\mathcal {R}}=\mathrm {clos}({\mathcal {O}})\) (so that \({\mathcal {O}}=\mathrm {int}({\mathcal {R}})\)), then there is a bijection

$$\begin{aligned} {\mathcal {A}}({\mathcal {R}})\ni \alpha \mapsto \alpha _{\upharpoonleft {\mathcal {O}}}\in {\mathcal {A}}({\mathcal {O}}), \end{aligned}$$
(A3)

that sends positive functions to positive functions and sends constants to constants. (It is bijective because any element of \({\mathcal {A}}({\mathcal {O}})\) has a unique continuous extension to \(\mathrm {clos}({\mathcal {O}})={\mathcal {R}}\).) Via these bijections, any \({\mathfrak {R}}({\mathcal {S}})\)-indexed conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) can be transformed into an \({\mathfrak {O}}({\mathcal {S}})\)-indexed conditional preference structure \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) as follows: for any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and \(\alpha ,\beta \in {\mathcal {A}}\), if \({\mathcal {O}}=\mathrm {int}({\mathcal {R}})\), then

$$\begin{aligned} \left( \alpha _{\upharpoonleft {\mathcal {R}}} \succeq _{\mathcal {R}}\beta _{\upharpoonleft {\mathcal {R}}}\right) \quad \iff \quad \left( \alpha _{\upharpoonleft {\mathcal {O}}} \succeq '_{\mathcal {O}}\beta _{\upharpoonleft {\mathcal {O}}}\right) . \end{aligned}$$
(A4)

Using the fact that the function \(\mathrm {clos}:{\mathfrak {O}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {R}}({\mathcal {S}})\) is a Boolean algebra isomorphism, and that the bijection (A3) preserves positive functions and constant functions, it is easily verified that \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) satisfies (Rch\('\)), (CEq\('\)), (C\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)) if and only if \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) does.

Now suppose that \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) has a liminal SEU representation (A2). For all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), if \({\mathcal {O}}=\mathrm {int}({\mathcal {R}})\), then \(\partial {\mathcal {R}}=\partial {\mathcal {O}}\). Define the function \(\chi _{\mathcal {R}}:{\mathcal {S}}{{\longrightarrow }}[0,1]\) by

$$\begin{aligned} \chi _{\mathcal {R}}(s):=\left\{ {{\begin{array}{lll} 1 &{}\quad \text{ if } &{}\quad s\in {\mathcal {O}};\\ \lambda _{\mathcal {O}}(s) &{}\quad \text{ if } &{}\quad s\in \partial {\mathcal {R}};\\ 0 &{}\quad \text{ if } &{}\quad s\in {\mathcal {R}}^\complement . \end{array}}}\right. \end{aligned}$$

It follows that \(\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}\) is a characteristic structure on \({\mathfrak {R}}({\mathcal {S}})\), because Eq. (A1) is equivalent to Eq. (2.2). Likewise, the liminal SEU representation (A2) is equivalent to the characteristic SEU representation (2.3) for \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\). \(\square \)

Let \({\mathcal {S}}^*\) be the Stone space of the Boolean algebra \({\mathfrak {R}}({\mathcal {S}})\). Thus, \({\mathcal {S}}^*\) is a Stonean space, and there is a Boolean algebra isomorphism \(\varXi :{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}}^*)={\mathfrak {R}}({\mathcal {S}}^*)\). For any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), let \({\mathcal {R}}^*\) denote the corresponding element of \({\mathfrak {R}}({\mathcal {S}}^*\)). There is a continuous surjection \(\pi :{\mathcal {S}}^*{{\longrightarrow }}{\mathcal {S}}\), such that \(\pi ^{-1}[\mathrm {int}({\mathcal {R}})]\subseteq {\mathcal {R}}^*\subseteq \pi ^{-1}({\mathcal {R}})\) for any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) (Johnstone 1982, Sect. 3.10); the pair \(({\mathcal {S}}^*,\pi )\) is called the Gleason cover of \({\mathcal {S}}\).Footnote 31 For any \(\alpha \in {\mathcal {C}}({\mathcal {S}})\), define \(\alpha ^*:=\alpha \circ \pi \); this is an element of \({\mathcal {C}}({\mathcal {S}}^*)\).

Let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\), and let \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) be a conditional preference structure on \({\mathcal {A}}\). Let \({\mathcal {A}}^*:=\{\alpha ^*\); \(\alpha \in {\mathcal {A}}\}\); then, \({\mathcal {A}}^*\subseteq {\mathcal {C}}({\mathcal {S}}^*)\). A Stonean SEU representation for \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) is given by a conditional preference structure \(\{\succeq ^*_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}}^*)}\) on \({\mathcal {A}}^*\), a normal Borel probability measure \(\mu ^*\) on \({\mathcal {S}}^*\), and a continuous function \(u:{\mathbb {R}}{{\longrightarrow }}{\mathbb {R}}\), such that, for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and all \(\alpha ,\beta \in {\mathcal {A}}\),

$$\begin{aligned} \left( \alpha _{\upharpoonleft {\mathcal {R}}}\succeq _{\mathcal {R}}\beta _{\upharpoonleft {\mathcal {R}}}\right)\iff & {} \left( \alpha ^*_{\upharpoonleft {\mathcal {R}}^*}\succeq _{{\mathcal {R}}^*}\beta ^*_{\upharpoonleft {\mathcal {R}}^*}\right) \nonumber \\\iff & {} \left( \displaystyle \int _{{\mathcal {R}}^*} u\circ \alpha ^* \ \mathrm {d}\mu ^* \ge \ \int _{{\mathcal {R}}^*} u\circ \beta ^* \ \mathrm {d}\mu ^* \right) . \end{aligned}$$
(A5)

The next result is used in the proofs of Theorems 1 and 6.

Proposition A2

Let \({\mathcal {S}}\) be a compact Hausdorff space, and let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\). Let \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) be a conditional preference structure on \({\mathcal {A}}\) which satisfies condition (Rch\('\)). Then, it satisfies (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)), and (TC\('\)) if and only if it admits a Stonean SEU representation (A5), where \(\mu ^*\) has full support on \({\mathcal {S}}^*\). Finally, \(\mu ^*\) is unique, and u is unique up to positive affine transformation.

Proof

In principle, this is a straightforward consequence of a theorem in Pivato and Vergopoulos (2018b). However, as in the proof of Proposition A1, there is a small complication: the results of Pivato and Vergopoulos (2018b) are formulated for conditional preference structures indexed by a Boolean subalgebra \({\mathfrak {B}}\) of the algebra \({\mathfrak {O}}({\mathcal {S}})\) of regular open subsets of \({\mathcal {S}}\). For the purposes of the present paper, we only need the special case when \({\mathfrak {B}}={\mathfrak {O}}({\mathcal {S}})\). Likewise, Pivato and Vergopoulos (2018b) work with a generalization of the Gleason cover to arbitrary Boolean subalgebras of \({\mathfrak {O}}({\mathcal {S}})\), which is developed in Section 8 of Pivato and Vergopoulos (2018c); in the present paper, we only need the classical Gleason cover, defined above. As in the proof of Proposition A1, we will translate an \({\mathfrak {R}}({\mathcal {S}})\)-indexed conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) into an \({\mathfrak {O}}({\mathcal {S}})\)-indexed conditional preference structure \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) via formula (A4). Once again, \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) satisfies (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)) if and only if \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) does.

In the framework of Pivato and Vergopoulos (2018b), there is a topological space \({\mathcal {X}}\) of “outcomes”, along with a Boolean algebra \({\mathfrak {D}}\) of regular open subsets of \({\mathcal {X}}\); the elements of \({\mathcal {A}}\) are continuous functions from \({\mathcal {S}}\) to \({\mathcal {X}}\) which are “comeasurable” with respect to the Boolean algebras \({\mathfrak {B}}\) and \({\mathfrak {D}}\). For the purposes of the present paper, we set \({\mathcal {X}}:={\mathbb {R}}\), and \({\mathfrak {D}}:={\mathfrak {O}}({\mathbb {R}})\). All continuous functions from \({\mathcal {S}}\) to \({\mathbb {R}}\) are “comeasurable” with respect to \({\mathfrak {O}}({\mathcal {S}})\) and \({\mathfrak {O}}({\mathbb {R}})\) (Pivato and Vergopoulos 2018c, Sect. 5). Thus, the comeasurability requirement is automatically satisfied.

In Pivato and Vergopoulos (2018b), \({\mathcal {X}}\) is equipped with an “ex post preference order” \(\succeq _{xp}\), and the axiom (Dom\('\)) is formulated with respect to \(\succeq _{xp}\). The framework of Pivato and Vergopoulos (2018b) requires one further axiom, (M), which stipulates that \(\succeq _{xp}\) must be “measurable” relative to \({\mathfrak {D}}\). When \({\mathfrak {D}}={\mathfrak {O}}({\mathcal {X}})\), axiom (M) reduces to the axiom (C), which says that \(\succeq _{xp}\) must be continuous relative to the topology of \({\mathcal {X}}\) (Pivato and Vergopoulos 2018b, Theorem 3). In the present paper, \({\mathcal {X}}={\mathbb {R}}\), and \(\succeq _{xp}\) is just the standard ordering of the real number line, so that axiom (C) is automatically satisfied.

Finally, in the framework of Pivato and Vergopoulos (2018b), \({\mathcal {S}}\) is assumed to be a locally compact Hausdorff space, and the elements of \({\mathcal {A}}\) are required to be bounded and continuous. In the present paper, \({\mathcal {S}}\) is a compact Hausdorff space, so all continuous functions are automatically bounded.

As already explained above, there is a Boolean algebra isomorphism \({\mathfrak {R}}({\mathcal {S}})\cong {\mathfrak {R}}({\mathcal {S}}^*)\), and for any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), I use \({\mathcal {R}}^*\) to denote its image under this isomorphism. As explained in the proof of Proposition A1, there is also a canonical isomorphism \({\mathfrak {O}}({\mathcal {S}})\cong {\mathfrak {R}}({\mathcal {S}})\) given by \({\mathcal {O}}\mapsto \mathrm {clos}({\mathcal {O}})\). So for any \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\), let \({\mathcal {O}}^*:=\mathrm {clos}({\mathcal {O}})^*\in {\mathfrak {R}}({\mathcal {S}}^*)\); this yields a canonical isomorphism \({\mathfrak {O}}({\mathcal {S}})\cong {\mathfrak {R}}({\mathcal {S}}^*)\).

Let \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) be an \({\mathfrak {O}}({\mathcal {S}})\)-indexed conditional preference structure on \({\mathcal {A}}\). A Stonean SEU representation for \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) is given by a conditional preference structure \(\{\succeq ^*_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}}^*)}\) on \({\mathcal {A}}^*\), a normal Borel probability measure \(\mu ^*\) on \({\mathcal {S}}^*\) and a continuous function \(u:{\mathbb {R}}{{\longrightarrow }}{\mathbb {R}}\), such that, for all \({\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})\) and all \(\alpha ,\beta \in {\mathcal {A}}\),

$$\begin{aligned} \left( \alpha _{\upharpoonleft {\mathcal {O}}}\succeq '_{\mathcal {O}}\beta _{\upharpoonleft {\mathcal {O}}}\right)\iff & {} \left( \alpha ^*_{\upharpoonleft {\mathcal {O}}^*}\succeq ^*_{{\mathcal {O}}^*}\beta ^*_{\upharpoonleft {\mathcal {O}}^*}\right) \nonumber \\\iff & {} \left( \displaystyle \int _{{\mathcal {O}}^*} u\circ \alpha ^* \ \mathrm {d}\mu ^* \ge \int _{{\mathcal {O}}^*} u\circ \beta ^* \ \mathrm {d}\mu ^* \right) . \end{aligned}$$
(A6)

Theorem 2 of Pivato and Vergopoulos (2018b) says that if \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) satisfies (Rch\('\)), then \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) satisfies (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)) if and only if it admits a Stonean SEU representation (A6), where \(\mu ^*\) has full support on \({\mathcal {S}}^*\). Finally, \(\mu ^*\) is unique, and u is unique up to positive affine transformation. But if \(\{\succeq '_{\mathcal {O}}\}_{{\mathcal {O}}\in {\mathfrak {O}}({\mathcal {S}})}\) is obtained from \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) via the translation formula (A4), then clearly, (A6) is equivalent to a Stonean SEU representation (A5) for \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\).

\(\diamond \)

Proofs of results from Section 3

Proof of Example 3.2

It is clear that \({\mathfrak {K}}({\mathcal {R}})\) is solid and a linear subspace. To see that it is order-closed, let \(\{\beta _n\}_{n\in {\mathcal {N}}}\) be a net in \({\mathfrak {K}}({\mathcal {R}})\), and suppose it order-converges to \(\beta \). Thus, there is some decreasing net \(\{\alpha _n\}_{n\in {\mathcal {N}}}\) such that \(\inf \{\alpha _n\}_{n\in {\mathcal {N}}}=0\) and \(|\beta -\beta _n|<\alpha _n\) for all \(n\in {\mathcal {N}}\). Let \({\mathcal {R}}^o:=\mathrm {int}({\mathcal {R}})\); then, \(\beta _n({\mathcal {R}}^o)=0\) for all \(n\in {\mathcal {N}}\). If \(\beta (s)\ne 0\) for some \(s\in {\mathcal {R}}^o\), then there is some \(\epsilon >0\) and some open \({\mathcal {Q}}\subset {\mathcal {R}}^o\) containing s such that \(|\beta (q)|> \epsilon \) for all \(q\in {\mathcal {Q}}\). But then \(\alpha _n(q)\ge |\beta (q)-\beta _n(q)|=|\beta (q)|>\epsilon \) for all \(q\in {\mathcal {Q}}\) and all \(n\in {\mathcal {N}}\). But this contradicts the fact that \(\inf \{\alpha _n\}_{n\in {\mathcal {N}}}=0\). To avoid the contradiction, we must have \(\beta (s)=0\) for all \(s\in {\mathcal {R}}^o\). Thus, \(\beta (r)=0\) for all \(r\in {\mathcal {R}}\), since \({\mathcal {R}}=\mathrm {clos}[{\mathcal {R}}^o]\).

Another proof is to recall from Example 3.4 that \({\mathfrak {K}}({\mathcal {R}})={\mathcal {G}}^\perp \) where \({\mathcal {G}}=\{\gamma \in {\mathcal {C}}({\mathcal {S}})\); \(\mathrm {supp}(\gamma )\subseteq {\mathcal {R}}\}\). Thus, \({\mathfrak {K}}({\mathcal {R}})\) is a band (Fremlin 2004, Fact 352O(c)). \(\square \)

Proof of Example 3.4

It is obvious that \({\mathfrak {K}}({\mathcal {R}})\subseteq {\mathcal {G}}^\perp \). To see that \({\mathcal {G}}^\perp \subseteq {\mathfrak {K}}({\mathcal {R}})\), recall that \({\mathcal {S}}\) is compact Hausdorff and therefore Tychonoff (Willard 2004, Thm. 19.3). Thus, for any \(s\in \mathrm {int}({\mathcal {R}})\), there is some \(\gamma \in {\mathcal {G}}\) with \(\gamma (s)\ne 0\). Thus, \(\alpha (s)=0\) for all \(\alpha \in {\mathcal {G}}^\perp \). It follows that, for any \(\alpha \in {\mathcal {G}}^\perp \), we have \(\alpha (s)=0\) for all \(s\in \mathrm {int}({\mathcal {R}})\) and hence (by continuity) for all \(s\in {\mathcal {R}}\). Thus, \(\alpha \in {\mathfrak {K}}({\mathcal {R}})\). \(\square \)

Proof of Lemma 3.5

A band \({\mathcal {I}}\subset {\mathcal {A}}\) is complemented if \({\mathcal {I}}={\mathcal {G}}^\perp \) for some subset \({\mathcal {G}}\subseteq {\mathcal {A}}\). The set of complemented bands forms a Boolean algebra under the operations in formula (3.1) (Fremlin 2004, Theorem 352Q). However, if \({\mathcal {A}}\) is Archimedean, then every band in \({\mathcal {A}}\) is complemented (Fremlin 2004, Proposition 353B(b)). Thus, for an Archimedean Riesz space, the set of all bands is a Boolean algebra under these operations. \(\square \)

Proof of Example 3.6

We must prove three statements.

  1. (a)

    For all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), \({\mathfrak {K}}(\lnot {\mathcal {R}})={\mathfrak {K}}({\mathcal {R}})^\perp \).

  2. (b)

    For all \({\mathcal {Q}},{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), \({\mathfrak {K}}({\mathcal {Q}}\cup {\mathcal {R}})={\mathfrak {K}}({\mathcal {Q}})\cap {\mathfrak {K}}({\mathcal {R}})\).

  3. (c)

    For all \({\mathcal {Q}},{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), \({\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}})={\mathfrak {K}}({\mathcal {Q}})\Cup {\mathfrak {K}}({\mathcal {R}})\).

Proof of Statement (a). Clearly, \({\mathfrak {K}}({\mathcal {R}})=\{\kappa \in {\mathcal {C}}({\mathcal {S}})\); \(\mathrm {supp}(\kappa )\subseteq {\mathcal {R}}^\complement \}\). Let \(\alpha \in {\mathcal {C}}({\mathcal {S}})\). Then

$$\begin{aligned} \begin{aligned} \left( \alpha \in {\mathfrak {K}}({\mathcal {R}})^\perp \right)&\Longleftrightarrow \left( |\alpha |\wedge |\kappa |=0\quad \hbox {for all}\quad \kappa \in {\mathfrak {K}}({\mathcal {R}})\right) {\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \mathrm {supp}(\alpha )\subseteq {\mathcal {R}}\right) \\&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \mathrm {supp}(\alpha )\subseteq \mathrm {int}({\mathcal {R}})\right) \Longleftrightarrow \left( \alpha (s)=0\quad \hbox {for all}\quad s\in \lnot {\mathcal {R}}\right) \\&\Longleftrightarrow \left( \alpha \in {\mathfrak {K}}(\lnot {\mathcal {R}})\right) . \end{aligned} \end{aligned}$$

Here, \((*)\) is because \(\mathrm {supp}(\alpha )\) is an open set, because \(\alpha \) is a continuous function. The “\({\Longleftarrow }\)” direction of \((\dagger )\) is obvious. To see the “\(\Longrightarrow \)” direction, recall that \({\mathcal {S}}\) is Tychonoff because it is compact Hausdorff. Thus, for any \(s\in {\mathcal {R}}^\complement \), there is some \(\kappa \in {\mathfrak {K}}({\mathcal {R}})\) with \(\kappa (s)\ne 0\). If \(\alpha (s)\ne 0\), then \(|\alpha |\wedge |\kappa |\ne 0\); thus, we must have \(\alpha (s)=0\) for all \(s\in {\mathcal {R}}^\complement \), and hence, \(\mathrm {supp}(\alpha )\subseteq {\mathcal {R}}\).

Proof of Statement (b). Let \(\alpha \in {\mathcal {C}}({\mathcal {S}})\). Then clearly,

$$\begin{aligned} \left( \alpha \in {\mathfrak {K}}({\mathcal {Q}}\cup {\mathcal {R}})\right)\iff & {} \left( \alpha (s)=0\quad \hbox {for all}\quad s\in {\mathcal {Q}}\cup {\mathcal {R}}\right) \\\iff & {} \left( \alpha (q)=0\quad \hbox {for all}\quad q\in {\mathcal {Q}}\quad \hbox {and}\quad \alpha (r)=0\quad \hbox {for all}\quad r\in {\mathcal {R}}\right) \\\iff & {} \left( \alpha \in {\mathfrak {K}}({\mathcal {Q}})\cap {\mathfrak {K}}({\mathcal {R}})\right) . \end{aligned}$$

Proof of Statement (c). \({\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}})\)\({\mathop {=}\limits _{\mathrm {(*)}}}\)\({\mathfrak {K}}\left(\lnot \left[(\lnot {\mathcal {Q}})\cup (\lnot {\mathcal {R}})\right]\right)\)\({\mathop {=}\limits _{\mathrm {(a)}}}\)\(\left({\mathfrak {K}}\left[(\lnot {\mathcal {Q}})\cup (\lnot {\mathcal {R}})\right]\right)^\perp \)\({\mathop {=}\limits _{\mathrm {(b)}}}\)\( \left[{\mathfrak {K}}(\lnot {\mathcal {Q}})\cap {\mathfrak {K}}(\lnot {\mathcal {R}})\right]^\perp \)\({\mathop {=}\limits _{\mathrm {(a)}}}\)\(\left[{\mathfrak {K}}({\mathcal {Q}})^\perp \cap {\mathfrak {K}}({\mathcal {R}})^\perp \right]^\perp \)\({\mathop {=}\limits _{\mathrm {(*)}}}\)\( {\mathfrak {K}}({\mathcal {Q}})\Cup {\mathfrak {K}}({\mathcal {R}})\), as desired. Here (a) and (b) are by Statements (a) and (b), and both \((*)\) are by De Morgan’s laws.

This proves that \({\mathfrak {K}}\) is a Boolean algebra anti-homomorphism. As noted in Example 3.4, \({\mathfrak {K}}\) is surjective. It remains to show that \({\mathfrak {K}}\) is injective. This uses the next observation.

Claim 1

\({\mathfrak {K}}({\mathcal {R}})\ne {\mathcal {A}}\) for any nonempty \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\).

Proof

Let \(\varvec{1}\) be the constant unit function. If \({\mathcal {R}}\ne \emptyset \), then \(\varvec{1}\not \in {\mathfrak {K}}({\mathcal {R}})\); hence, \({\mathfrak {K}}({\mathcal {R}})\ne {\mathcal {A}}\). \(\diamond \)

To see that \({\mathfrak {K}}\) is injective, let \({\mathcal {R}},{\mathcal {Q}}\in {\mathfrak {R}}({\mathcal {S}})\) be distinct, and suppose \({\mathfrak {K}}({\mathcal {Q}})={\mathfrak {K}}({\mathcal {R}})\). I first show that we can assume without loss of generality that \({\mathcal {Q}}\subsetneq {\mathcal {R}}\). If \({\mathcal {R}}\subseteq {\mathcal {Q}}\), then define \({\mathcal {R}}':={\mathcal {Q}}\) and \({\mathcal {Q}}':={\mathcal {R}}\); then, \({\mathcal {Q}}'\subseteq {\mathcal {R}}'\), and in fact \({\mathcal {Q}}'\subsetneq {\mathcal {R}}'\) (because \({\mathcal {R}}\ne {\mathcal {Q}}\)). Otherwise, if \({\mathcal {R}}\not \subseteq {\mathcal {Q}}\), then let \({\mathcal {Q}}':={\mathcal {Q}}\wedge {\mathcal {R}}\). Then \({\mathcal {Q}}'\subseteq {\mathcal {R}}\), and in fact \({\mathcal {Q}}'\subsetneq {\mathcal {R}}\) (because \({\mathcal {R}}\not \subseteq {\mathcal {Q}}\)); furthermore, \({\mathfrak {K}}({\mathcal {Q}}')={\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}}) \ {\mathop {=}\limits _{\mathrm {(*)}}} \ {\mathfrak {K}}({\mathcal {Q}})\Cup {\mathfrak {K}}({\mathcal {R}})={\mathfrak {K}}({\mathcal {R}})\Cup {\mathfrak {K}}({\mathcal {R}})={\mathfrak {K}}({\mathcal {R}})\), where \((*)\) is by Statement (c). In either case, by replacing \({\mathcal {Q}}\) by \({\mathcal {Q}}'\) and \({\mathcal {R}}\) by \({\mathcal {R}}'\), we can assume that \({\mathcal {Q}}\subsetneq {\mathcal {R}}\) while \({\mathfrak {K}}({\mathcal {Q}})={\mathfrak {K}}({\mathcal {R}})\).

Now, let \({\mathcal {P}}:={\mathcal {R}}\wedge (\lnot {\mathcal {Q}})\); then, \({\mathcal {P}}\wedge {\mathcal {Q}}=\emptyset \) and \({\mathcal {P}}\cup {\mathcal {Q}}={\mathcal {R}}\). Furthermore, \({\mathcal {P}}\ne \emptyset \) because \({\mathcal {Q}}\subsetneq {\mathcal {R}}\). Thus,

$$\begin{aligned} {\mathfrak {K}}({\mathcal {Q}})\quad {\mathop {=}\limits _{\mathrm {(*)}}}\quad {\mathfrak {K}}({\mathcal {R}})\quad {\mathop {=}\limits _{\mathrm {(\dagger )}}}\quad {\mathfrak {K}}({\mathcal {P}}\cup {\mathcal {Q}})\quad {\mathop {=}\limits _{\mathrm {(\diamond )}}}\quad {\mathfrak {K}}({\mathcal {P}})\cap {\mathfrak {K}}({\mathcal {Q}})\subseteq {\mathfrak {K}}({\mathcal {P}}), \end{aligned}$$
(B1)

where \((*)\) is by hypothesis, \((\dagger )\) is because \({\mathcal {R}}={\mathcal {P}}\cup {\mathcal {Q}}\) and \((\diamond )\) is by Statement (b). Thus,

$$\begin{aligned} {\mathfrak {K}}({\mathcal {P}})\quad {\mathop {=}\limits _{\mathrm {(*)}}}\quad {\mathfrak {K}}({\mathcal {P}})\Cup {\mathfrak {K}}({\mathcal {Q}}) \quad {\mathop {=}\limits _{\mathrm {(\dagger )}}}\quad {\mathfrak {K}}({\mathcal {P}}\wedge {\mathcal {Q}})\quad {\mathop {=}\limits _{\mathrm {(\diamond )}}}\quad {\mathfrak {K}}(\emptyset ) ={\mathcal {A}}. \end{aligned}$$

Here, \((*)\) is by Eq. (B1), \((\dagger )\) is by Statement (c) and \((\diamond )\) is because \({\mathcal {P}}\wedge {\mathcal {Q}}=\emptyset \). But \({\mathcal {P}}\ne \emptyset \), so Claim 1 says \({\mathfrak {K}}({\mathcal {P}})\ne {\mathcal {A}}\) Contradiction. To avoid this contradiction, we must have \({\mathfrak {K}}({\mathcal {Q}})\ne {\mathfrak {K}}({\mathcal {R}})\). Thus, \({\mathfrak {K}}\) is injective. Since we already know that \({\mathfrak {K}}\) is a surjective anti-homomorphism, this implies that \({\mathfrak {K}}\) is an anti-isomorphism. \(\square \)

Proof of Example 3.8

Let \({\mathcal {Q}},{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\). Then

$$\begin{aligned} {\mathfrak {K}}({\mathcal {Q}})^\perp \cap {\mathfrak {K}}({\mathcal {R}})^\perp ~&{\mathop {=}\limits _{\mathrm {(a)}}}~ {\mathfrak {K}}(\lnot {\mathcal {Q}})\cap {\mathfrak {K}}(\lnot {\mathcal {R}}) \quad {\mathop {=}\limits _{\mathrm {(b)}}} \quad {\mathfrak {K}}\left[(\lnot {\mathcal {Q}})\cup (\lnot {\mathcal {R}})\right]\\ ~&{\mathop {=}\limits _{\mathrm {(c)}}}~{\mathfrak {K}}\left[\lnot ({\mathcal {Q}}\wedge {\mathcal {R}})\right] \quad {\mathop {=}\limits _{\mathrm {(d)}}} \quad {\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}})^\perp .\nonumber \end{aligned}$$
(B2)

Here, (a), (b) and (d) use the fact that \({\mathfrak {K}}\) is a Boolean algebra anti-isomorphism (by Example 3.6), while (c) is by De Morgan’s Law. Thus,

$$\begin{aligned} \left( {\mathfrak {K}}({\mathcal {Q}})\hbox { and }{\mathfrak {K}}({\mathcal {R}})\hbox { are codisjoint}\right)&\iff \left( {\mathfrak {K}}({\mathcal {Q}})^\perp \cap {\mathfrak {K}}({\mathcal {R}})^\perp =\{0\}\right) \\&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( {\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}})^\perp =\{0\}\right) {\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( {\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}})=\{0\}^\perp = {\mathcal {A}}\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\diamond )}} \left( {\mathcal {Q}}\wedge {\mathcal {R}}=\emptyset \right) \iff \left( {\mathcal {Q}}\hbox { and }{\mathcal {R}}\hbox { are almost disjoint}\right) . \end{aligned}$$

Here, \((*)\) uses Eq. (B2), \((\dagger )\) uses the fact that \({\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra (so \(({\mathcal {I}}^\perp )^\perp ={\mathcal {I}}\) for all \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\)) and \((\diamond )\) uses the fact that \({\mathfrak {K}}({\mathcal {T}})={\mathcal {A}}\) if and only if \({\mathcal {T}}=\emptyset \) (because \({\mathfrak {K}}\) is a Boolean algebra anti-isomorphism). \(\square \)

Let \({\mathcal {A}}\) be a Riesz space, and let \({\mathcal {B}}\subseteq {\mathcal {A}}\) be a a Riesz subspace. We say \({\mathcal {B}}\) is order-dense if, for all \(\alpha \in {\mathcal {A}}\) with \(\alpha >0\), we have \(\alpha =\sup [{\mathcal {B}}(\alpha )]\), where \({\mathcal {B}}(\alpha ):=\{\beta \in {\mathcal {B}}\); \(0\le \beta \le \alpha \}\). In other words, for any \(\gamma \in {\mathcal {A}}\), if \(\gamma \ge \beta \) for all \(\beta \in {\mathcal {B}}(\alpha )\), then \(\gamma \ge \alpha \). For example, \(\ell ^{\infty }({\mathcal {S}})\) is an order-dense Riesz subspace of \({\mathbb {R}}^{\mathcal {S}}\). The next result summarizes some standard facts about Riesz spaces and Banach lattices. It will be important for the proofs of Theorems 123 and 7.

Proposition B1

  1. (a)

    Let \({\mathcal {A}}\) be a unitary Archimedean Riesz space. Then there is a compact Hausdorff space \({\mathcal {S}}\), an order-dense, uniformly dense Riesz subspace \({\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\), a uniferent Riesz isomorphism \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) and a Boolean algebra anti-isomorphism \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) defined by \(\varPhi ^*({\mathcal {R}}):=\varPhi ^{-1}[{\mathfrak {K}}({\mathcal {R}})\cap {\widehat{\mathcal {A}}}]\), for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\). Furthermore, \({\mathcal {S}}\), \({\widehat{\mathcal {A}}}\) and \(\varPhi \) are unique up to homeomorphism.

  2. (b)

    If \({\mathcal {A}}\) is an M-space, then \({\widehat{\mathcal {A}}}={\mathcal {C}}({\mathcal {S}})\) in part (a).

  3. (c)

    Let \({\mathcal {A}}'\) be another unitary Archimedean Riesz space with associated compact Hausdorff space \({\mathcal {S}}'\) and isomorphism \(\varPhi ':{\mathcal {A}}'{{\longrightarrow }}{\widehat{\mathcal {A}}}'\subseteq {\mathcal {C}}({\mathcal {S}}')\). If \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}'\) is a uniferent Riesz homomorphism, then there is a continuous function \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) such that, for all \(\alpha \in {\mathcal {A}}\), \(\varPhi '[\varPsi (\alpha )] = \varPhi (\alpha )\circ \psi \). In other words, if we define \({\widehat{\varPsi }}({\widehat{\alpha }}) :={\widehat{\alpha }}\circ \psi \) for all \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\), then the diagram shown in Fig. 3a commutes.

  4. (d)

    If \(\varPsi \) is injective, then \(\psi \) is a quotient map from \({\mathcal {S}}'\) onto \({\mathcal {S}}\). If \(\varPsi \) is surjective, then \(\psi \) is an embedding of \({\mathcal {S}}'\) into \({\mathcal {S}}\).Footnote 32

  5. (e)

    Let \({\mathcal {A}}''\) be a third unitary Archimedean Riesz space, and suppose there are uniferent Riesz homomorphisms \(\varPsi \), \(\varTheta \) and \(\varXi \) as shown in the left half of diagram (Fig. 3b), with associated continuous functions \(\psi \), \(\theta \) and \(\xi \) from part (c), shown in the right half of Fig. 3b. If the left diagram commutes, then the right diagram also commutes.Footnote 33

Proof

  1. (a)

    Theorem 353M of Fremlin (2004) yields the space \({\mathcal {S}}\), the order-dense, uniformly dense Riesz subspace \({\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and a uniferent Riesz isomorphism \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) (all unique up to homeomorphism). By Example 3.6, \({\mathfrak {K}}:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\) is an anti-isomorphism of Boolean algebras. Since \({\widehat{\mathcal {A}}}\) is order-dense, Proposition 353D of Fremlin (2004) says that the function \({\mathcal {I}}\mapsto {\mathcal {I}}\cap {\widehat{\mathcal {A}}}\) is Boolean algebra isomorphism from \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\) to \({\mathfrak {I}}({\widehat{\mathcal {A}}})\).Footnote 34 Since \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) is a Riesz isomorphism, the function \(\varPhi ^{-1}:{\mathfrak {I}}({\widehat{\mathcal {A}}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is also a Boolean algebra isomorphism. Combining the anti-isomorphism and the two isomorphisms, we get the anti-isomorphism \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) described in the theorem.

  2. (b)

    See Aliprantis and Border (2006, Theorem 9.32), Fremlin (2004, Corollary 354L) or Meyer-Nieberg (1991, Theorem 2.1.3). (This result is variously called Kakutani’s Representation Theorem or the Kakutani–Bohnenblust–Krein Theorem.)

  3. (c)

    Before continuing, it is necessary to stop and explain the construction of the space \({\mathcal {S}}\). Recall that \({\mathbb {R}}\) is a unitary Riesz space (with 1 as the order unit). The elements of \({\mathcal {S}}\) are simply the uniferent Riesz homomorphisms from \({\mathcal {A}}\) to \({\mathbb {R}}\). For any \(\alpha \in {\mathcal {A}}\), the function \({\widehat{\alpha }}=\varPhi (\alpha )\) is simply defined by setting \({\widehat{\alpha }}(s):=s(\alpha )\) for all \(s\in {\mathcal {S}}\). (If we endow \({\mathcal {S}}\) with the topology of pointwise convergence, then it is compact Hausdorff, and \({\widehat{\alpha }}\) is continuous for all \(\alpha \in {\mathcal {A}}\).) Likewise, \({\mathcal {S}}'\) is the space of uniferent Riesz homomomorphisms from \({\mathcal {A}}'\) to \({\mathbb {R}}\) endowed with the topology of pointwise convergence, and for any \(\alpha '\in {\mathcal {A}}'\), the function \({\widehat{\alpha }}'=\varPhi '(\alpha ')\) is simply defined by setting \({\widehat{\alpha }}'(s'):=s'(\alpha ')\) for all \(s'\in {\mathcal {S}}'\). Now, given a uniferent Riesz homomorphism \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}'\), define \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) by setting \(\psi (s'):=s'\circ \varPsi \) for all \(s'\in {\mathcal {S}}'\). (Recall that the composition of two uniferent Riesz homomorphisms is a uniferent Riesz homomorphism.) It is straightforward to prove that \(\psi \) is continuous with respect to the topologies of pointwise convergence on \({\mathcal {S}}\) and \({\mathcal {S}}'\). Finally, for any \(\alpha \in {\mathcal {A}}\), if \({\widehat{\alpha }}:=\varPhi (\alpha )\), \(\alpha ':=\varPsi (\alpha )\) and \({\widehat{\alpha }}':=\varPhi '(\alpha ')\), then for any \(s'\in {\mathcal {S}}'\),

    $$\begin{aligned} {\widehat{\alpha }}\circ \psi (s')= & {} {\widehat{\alpha }}[\psi (s')] \ = \ {\widehat{\alpha }}[s'\circ \varPsi ] \ = \ (s'\circ \varPsi )(\alpha )\\= & {} s'[\varPsi (\alpha )] \ = \ s'(\alpha ') \ = \ {\widehat{\alpha }}'(s') \ = \ \varPhi '[\alpha '](s'). \end{aligned}$$

    This holds for all \(s'\in {\mathcal {S}}'\). Thus, \({\widehat{\alpha }}\circ \psi = \varPhi '(\alpha ')\). But \({\widehat{\alpha }}=\varPhi (\alpha )\) and \(\alpha '=\varPsi (\alpha )\), so this says that \(\varPhi '[\varPsi (\alpha )] = \varPhi (\alpha )\circ \psi \), as claimed

  4. (d)

    First suppose \(\varPsi \) is surjective. Let \(s'_1,s'_2\in {\mathcal {S}}'\), and suppose \(\psi (s'_1)=\psi (s'_2)\). Then \(s'_1\circ \varPsi =s'_2\circ \varPsi \). Since \(\varPsi \) is surjective, this means that \(s'_1(\alpha ')=s'_2(\alpha ')\) for all \(\alpha '\in {\mathcal {A}}'\). Thus, \(s'_1=s'_2\). We deduce that \(\psi \) is injective. Thus, \(\psi :{\mathcal {S}}'{{\longrightarrow }}\psi ({\mathcal {S}}')\) is a homeomorphism, because \({\mathcal {S}}'\) is compact and \({\mathcal {S}}\) is Hausdorff (Willard 2004, Theorem 17.14). Now suppose that \(\varPsi \) is injective. Let \(s\in {\mathcal {S}}\); we must find \(s'\in {\mathcal {S}}'\) such that \(\psi (s')=s\). Let \({\widetilde{\mathcal {A}}}:=\varPsi ({\mathcal {A}})\); then, \({\widetilde{\mathcal {A}}}\) is a Riesz subspace of \({\mathcal {A}}'\), and \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\widetilde{\mathcal {A}}}\) is a uniferent Riesz isomorphism. Define \({\widetilde{s}}:{\widetilde{\mathcal {A}}}{{\longrightarrow }}{\mathbb {R}}\) by \({\widetilde{s}}({\widetilde{\alpha }}):=s\left(\varPsi ^{-1}({\widetilde{\alpha }})\right)\) for all \({\widetilde{\alpha }}\in {\widetilde{\mathcal {A}}}\); then, \({\widetilde{s}}\) is a uniferent Riesz homomorphism. I will extend \({\widetilde{s}}\) to a uniferent Riesz homomorphism on all of \({\mathcal {A}}'\). Recall that \({\mathcal {A}}\) and \({\mathcal {A}}'\) are unitary; let \(\varvec{1}\) and \(\varvec{1}'\) be their order units. Then \(\varPsi (\varvec{1})=\varvec{1}'\), since \(\varPsi \) is uniferent. Thus, \(\varvec{1}'\in {\widetilde{\mathcal {A}}}\), so \({\widetilde{\mathcal {A}}}\) is a majorizing subspace of \({\mathcal {A}}'\).Footnote 35 Since \({\mathbb {R}}\) is a Dedekind-complete Riesz space, a theorem of Luxemburg and Schep (1979) and Lipecki (1980) yields a Riesz homomorphism \(s':{\mathcal {A}}'{{\longrightarrow }}{\mathbb {R}}\) such that \(s'({\widetilde{\alpha }})={\widetilde{s}}({\widetilde{\alpha }})\) for all \({\widetilde{\alpha }}\in {\widetilde{\mathcal {A}}}\) (see also Aliprantis and Burkinshaw (1985, Theorem 2.29)). It is automatically uniferent; thus, \(s'\in {\mathcal {S}}'\). But for any \(\alpha \in {\mathcal {A}}\), \(s'\circ \varPsi (\alpha )={\widetilde{s}}\circ \varPsi (\alpha )=s\circ \varPsi ^{-1}\circ \varPsi (\alpha )=s(\alpha )\). Thus, \(s'\circ \varPsi =s\). In other words, \(\psi (s')=s\). We can perform this construction for any \(s\in {\mathcal {S}}\); thus, \(\psi \) is surjective. Since \({\mathcal {S}}'\) is compact Hausdorff and \(\psi \) is continuous, it follows that \(\psi \) is a closed map, and thus, a quotient map from \({\mathcal {S}}'\) onto \({\mathcal {S}}\) (Willard 2004, Theorems 9.2, 17.5(b) and 17.7).

  5. (e)

    This follows immediately from the construction in part (c). \(\square \)

The next result is a key step in the proofs of Theorems 12 and 3.

Proposition B2

Let \({\mathcal {A}}\) be a unitary Archimedean Riesz space, and let \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) be as in Proposition B1(a). Let \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) be a conditional preference structure on \({\mathcal {A}}\). Then there is a conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) on \({\widehat{\mathcal {A}}}\) satisfying statement (3.3) for all \(\alpha ,\beta \in {\mathcal {A}}\) and \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\).

Furthermore, \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) satisfies (Rch), (CEq), (Dom), (Sep), (CCP), and (TC) if and only if \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfies (Rch\('\)), (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)), and (TC\('\)).

Proof

For any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), let \({\widehat{\mathcal {A}}}({\mathcal {R}}):=\{{\widehat{\alpha }}_{\upharpoonleft {\mathcal {R}}}\); \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\}\); this is a subspace of \({\mathcal {C}}({\mathcal {R}})\). Then define the linear surjection \({\overline{\varPhi }}_{\mathcal {R}}:{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}({\mathcal {R}})\) by setting \({\overline{\varPhi }}_{\mathcal {R}}(\alpha ):=\varPhi (\alpha )_{\upharpoonleft {\mathcal {R}}}\) for all \(\alpha \in {\mathcal {A}}\). Note that \({\overline{\varPhi }}_{\mathcal {R}}\) is order-preserving (i.e. \((\alpha \ge \beta )\Longrightarrow \left({\overline{\varPhi }}_{\mathcal {R}}(\alpha )\ge {\overline{\varPhi }}_{\mathcal {R}}(\beta )\right)\)) and maps scalar multiples of the order unit \(\varvec{1}\) to constant functions. Let \({\mathcal {I}}:=\varPhi ^*({\mathcal {R}})\); then, \({\overline{\varPhi }}_{\mathcal {R}}(\alpha )=0\) if and only if \(\alpha \in {\mathcal {I}}\)—in other words, \({\mathcal {I}}\) is the kernel of \({\overline{\varPhi }}_{\mathcal {R}}\). Thus, \({\overline{\varPhi }}_{\mathcal {R}}\) factors to a well-defined linear bijection \(\varPhi _{\mathcal {R}}:{\mathcal {A}}/{\mathcal {I}}{{\longrightarrow }}{\widehat{\mathcal {A}}}({\mathcal {R}})\), defined by \(\varPhi _{\mathcal {R}}([\alpha ]_{\mathcal {I}}):=\varPhi (\alpha )_{\upharpoonleft {\mathcal {R}}}\) for all \(\alpha \in {\mathcal {A}}\). Let \(\succeq _{\mathcal {R}}\) be the image of \(\succeq _{\mathcal {I}}\) under this bijection. That is, for all \({\widehat{\alpha }},{\widehat{\beta }}\in {\widehat{\mathcal {A}}}({\mathcal {R}})\), define \({\widehat{\alpha }}\succeq _{\mathcal {R}}{\widehat{\beta }}\) if and only if \([\alpha ]_{\mathcal {I}}\succeq _{\mathcal {I}}[\beta ]_{\mathcal {I}}\), where \([\alpha ]_{\mathcal {I}}\) and \([\beta ]_{\mathcal {I}}\) are the (unique) elements of \({\mathcal {A}}/{\mathcal {I}}\) such that \({\widehat{\alpha }}=\varPhi _{\mathcal {R}}([\alpha ]_{\mathcal {I}})\) and \({\widehat{\beta }}=\varPhi _{\mathcal {R}}([\beta ]_{\mathcal {I}})\). Repeating this construction for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), we get a conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfying statement (3.3). It is easily verified that \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) satisfies (Rch), (CEq), (Dom), (Sep), (CCP), and (TC) if and only if \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfies (Rch\('\)), (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)), and (TC\('\)). (Use (3.3), the fact that \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra anti-isomorphism, and the fact that \({\overline{\varPhi }}_{\mathcal {R}}\) is order-preserving and maps multiples of \(\varvec{1}\) to constant functions.) \(\square \)

Proof of Theorem 1

We will combine the representations from Propositions A2 and B2. Let \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) be as in Proposition B1(a), and let \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) be as in Proposition B2. Let \({\mathcal {S}}^*\) be the Stone space of \({\mathcal {S}}\), let \(\varXi :{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {C}}{\mathfrak {l}}{\mathfrak {p}}({\mathcal {S}}^*)={\mathfrak {R}}({\mathcal {S}}^*)\) be the corresponding isomorphism of Boolean algebras, and let \(\pi :{\mathcal {S}}^*{{\longrightarrow }}{\mathcal {S}}\) be the Gleason covering map. Let \({\mathcal {A}}^*:=\{{\widehat{\alpha }}\circ \pi \); \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\}\); then, \({\mathcal {A}}^*\) is a Riesz subspace of \({\mathcal {C}}({\mathcal {S}}^*)\), and the map \({\widehat{\alpha }}\mapsto {\widehat{\alpha }}\circ \pi \) is a Riesz isomorphism from \({\widehat{\mathcal {A}}}\) to \({\mathcal {A}}^*\). For any \(\alpha \in {\mathcal {A}}\), let \({\overline{\varPhi }}(\alpha ):=\varPhi (\alpha )\circ \pi \); this defines a Riesz isomorphism \({\overline{\varPhi }}:{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}^*\). Define \(\varPhi ^\dagger :=\varPhi ^*\circ \varXi ^{-1}:{\mathfrak {R}}({\mathcal {S}}^*){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\); then, \(\varPhi ^\dagger \) is a Boolean algebra isomorphism (because it is a composition of two isomorphisms). Let \(\{\succeq ^*_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}}^*)}\), \(\mu ^*\) and u be the Stonean SEU representation of \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) obtained from Proposition A2.

Given \(\alpha ,\beta \in {\mathcal {A}}\), let \({\widehat{\alpha }}:=\varPhi (\alpha )\) and \({\widehat{\beta }}:=\varPhi (\beta )\), and then, let \({\widetilde{\alpha }}:={\widehat{\alpha }}\circ \pi \) and \({\widetilde{\beta }}:={\widehat{\beta }}\circ \pi \); in other words, \({\widetilde{\alpha }}={\overline{\varPhi }}(\alpha )\) and \({\widetilde{\beta }}={\overline{\varPhi }}(\beta )\). Given \({\mathcal {R}}^*\in {\mathfrak {R}}({\mathcal {S}}^*)\), let \({\mathcal {R}}:=\varXi ^{-1}({\mathcal {R}}^*)\) (so that \({\mathcal {R}}^*:=\varXi ({\mathcal {R}})\)), and let \({\mathcal {I}}:=\varPhi ^*({\mathcal {R}})\); in other words, \({\mathcal {I}}=\varPhi ^\dagger ({\mathcal {R}}^*)\). Then

$$\begin{aligned} \left( [\alpha ]_{\mathcal {I}}\succeq _{\mathcal {I}}[\beta ]_{\mathcal {I}}\right)&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( {\widehat{\alpha }}_{\upharpoonleft {\mathcal {R}}} \succeq _{\mathcal {R}}{\widehat{\beta }}_{\upharpoonleft {\mathcal {R}}}\right) \quad {\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \quad \left( {\widetilde{\alpha }}_{\upharpoonleft {\mathcal {R}}^*}\succeq _{{\mathcal {R}}^*}{\widetilde{\beta }}_{\upharpoonleft {\mathcal {R}}^*}\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\diamond )}} \left( \displaystyle \int _{{\mathcal {R}}^*} u\circ {\widetilde{\alpha }}\ \mathrm {d}\mu ^* \ \ge \ \int _{{\mathcal {R}}^*} u\circ {\widetilde{\beta }}\ \mathrm {d}\mu ^* \right) . \end{aligned}$$

Here, \((*)\) is by statement (3.3) from Proposition B2, while \((\dagger )\) and \((\diamond )\) are by statement (A5) from Proposition A2. (In the notation of (A5), \({\widetilde{\alpha }}=({\widehat{\alpha }})^*\) and \({\widetilde{\beta }}=({\widehat{\beta }})^*\).) Combining \((*)\) and \((\dagger )\) yields the instance of statement (3.2) required for the Stonean SEU representation of \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\). Meanwhile, equivalence \((\diamond )\) yields a Borel SEU representation (2.1) for the conditional preference structure \(\{\succeq ^*_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}}^*)}\). Thus, \(\{\succeq ^*_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}}^*)}\), \(\mu ^*\) and u comprise a Stonean SEU representation for \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\). \(\square \)

Proof of Theorem 2

Combine Propositions B2 and A1. \(\square \)

Proof of Theorem 3

Proposition B1(b) yields a compact Hausdorff space \({\mathcal {S}}\) and a uniferent Riesz isomorphism \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {C}}({\mathcal {S}})\). Thus, in Proposition B2, we have \({\widehat{\mathcal {A}}}={\mathcal {C}}({\mathcal {S}})\). Now apply Proposition A1 to obtain the desired characteristic SEU representation. \(\square \)

Proofs of results from Section 4

We now turn to the proofs of Theorems 45 and 6. First, we need some technical results. Let \({\mathcal {S}}\) be a topological space. A subset \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\) is Tychonoff for \({\mathcal {S}}\) if, for any closed subset \({\mathcal {F}}\subset {\mathcal {S}}\), and any \(s\in {\mathcal {F}}^\complement \), there exists \(\alpha \in {\mathcal {A}}\) such that \(\alpha ({\mathcal {F}})=0\) while \(\alpha (s)\ne 0\). (Thus, the topology of \({\mathcal {S}}\) is Tychonoff if and only if \({\mathcal {C}}({\mathcal {S}})\) itself is Tychonoff for this topology.)

Lemma C1

Let \({\mathcal {S}}\) be a topological space, and let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\) be Tychonoff for the topology of \({\mathcal {S}}\). For any closed \({\mathcal {R}}\subseteq {\mathcal {S}}\), let \({\mathfrak {K}}({\mathcal {R}}):=\{\alpha \in {\mathcal {A}}\); \(\alpha (r)=0\) for all \(r\in {\mathcal {R}}\}\). Then

$$\begin{aligned} \bigcup _{\kappa \in {\mathfrak {K}}({\mathcal {R}})} \mathrm {supp}(\kappa )={\mathcal {R}}^\complement . \end{aligned}$$

Proof

\(\subseteq \)” (by contradiction) If the left side is not contained in \({\mathcal {R}}^\complement \), then there exists \(\alpha \in {\mathcal {A}}\) with \(\mathrm {supp}(\alpha )\not \subseteq {\mathcal {R}}^\complement \), hence \(\mathrm {supp}(\alpha )\cap {\mathcal {R}}\ne \emptyset \). Thus, \(\alpha (r)\ne 0\) for some \(r\in {\mathcal {R}}\).

\(\supseteq \)” Let \(s\in {\mathcal {R}}^\complement \). Since \({\mathcal {A}}\) is Tychonoff, there is some \(\alpha \in {\mathcal {A}}\) such that \(\alpha (s)\ne 0\) while \(\alpha ({\mathcal {R}})=0\). Thus, \(\alpha \in {\mathfrak {K}}({\mathcal {R}})\), and \(s\in \ \mathrm {supp}(\kappa )\). \(\square \)

Throughout the rest of this appendix, I will adopt the standard notational convention that for any \(r\in {\mathbb {R}}\), the symbol r also refers to the constant r-valued function \(r\varvec{1}\). The next lemma establishes that, for strictly real Banach algebras, the definition of “regular” given in Sect. 4 is logically equivalent to the standard definition (Kaniuth 2009, Definition 4.2.1).Footnote 36 Let \({\mathcal {S}}\) be a set, let \({\mathcal {A}}\) be an algebra of real-valued functions on \({\mathcal {S}}\) (i.e. a subalgebra of \({\mathbb {R}}^{\mathcal {S}}\)), and let \(\alpha \in {\mathcal {A}}\). Suppose \(\alpha (s)\ne 0\) for all \(s\in {\mathcal {A}}\). Then \(1/\alpha \) is a well-defined real-valued function on \({\mathcal {S}}\)—however, it might not be an element of \({\mathcal {A}}\). I will say that \({\mathcal {A}}\) is inversion-closed if \(1/\alpha \in {\mathcal {A}}\) whenever \(\alpha \in {\mathcal {A}}\) and \(\alpha (s)\ne 0\) for all \(s\in {\mathcal {A}}\). (It follows that \({\mathcal {A}}\) is unitary.) For example, if \({\mathcal {S}}\) is a topological space, then \({\mathcal {C}}({\mathcal {S}})\) is inversion-closed. Likewise, the algebra \({\mathcal {C}}^n[a,b]\) of n-times differentiable functions is inversion-closed. (However, the algebra of polynomial functions is not inversion-closed.) Crucially for our purposes, the Gelfand representation of a commutative Banach algebra is always inversion-closed (Kaniuth 2009, Theorem 2.2.5; Dales et al. 2003, Theorem 3.2.2(ii)).

Let \({\mathcal {S}}\) be a set, and let \({\mathcal {A}}\) be a collection of real-valued functions on \({\mathcal {S}}\). Recall that the weak topology induced on \({\mathcal {S}}\) by \({\mathcal {A}}\) is the coarsest topology such that every element of \({\mathcal {A}}\) is continuous. Thus, if \({\mathcal {S}}\) has the weak topology, then \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\). Conversely, if \({\mathcal {S}}\) is already a topological space and \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\), then I will say \({\mathcal {A}}\)generates the topology on \({\mathcal {S}}\) (or is generating) if the topology of \({\mathcal {S}}\) is the weak topology induced by \({\mathcal {A}}\).

Lemma C2

Let \({\mathcal {S}}\) be a topological space, and let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\) be an inversion-closed algebra of functions that generates the topology of \({\mathcal {S}}\). If \({\mathcal {A}}\) is regular, then \({\mathcal {A}}\) is Tychonoff for \({\mathcal {S}}\).

Proof

For any \(\alpha \in {\mathcal {A}}\), let \({\mathcal {N}}_\alpha :=\alpha ^{-1}(-{\infty },0]\). Let \({\mathfrak {N}}:=\{{\mathcal {N}}_\alpha \); \(\alpha \in {\mathcal {A}}\}\). These are closed subsets in the weak topology induced by \({\mathcal {A}}\) on \({\mathcal {S}}\).

Claim 1

For any disjoint \({\mathcal {N}},{\mathcal {M}}\in {\mathfrak {N}}\), there exists \(\gamma \in {\mathcal {A}}\) such that \(\gamma ({\mathcal {N}})=0\) while \(\gamma (s)\ne 0\) for all \(s\in {\mathcal {M}}\).

Proof

Find \(\alpha ,\beta \in {\mathcal {A}}\) such that \({\mathcal {N}}={\mathcal {N}}_\alpha \) and \({\mathcal {M}}={\mathcal {N}}_\beta \). Let \(q,r\ge 0\); I will show that \((\alpha +q)^2+(\beta +r)^2\) is invertible. To see this, observe that \((\alpha +q)^2(s)+(\beta +r)^2(s)=0\) if and only if \(s\in \alpha ^{-1}\{-q\}\cap \beta ^{-1}\{-r\} \subseteq {\mathcal {N}}_\alpha \cap {\mathcal {N}}_\beta ={\mathcal {N}}\cap {\mathcal {M}}\). But \({\mathcal {N}}\cap {\mathcal {M}}=\emptyset \) by hypothesis. Thus, \((\alpha +q)^2(s)+(\beta +r)^2(s)\ne 0\) for all \(s\in {\mathcal {S}}\). Thus, \((\alpha +q)^2+(\beta +r)^2\) has a multiplicative inverse in \({\mathcal {A}}\), because \({\mathcal {A}}\) is inversion-closed.

This holds for all \(q,r\ge 0\). Thus, \(\alpha ,\beta \in {\mathcal {A}}\) are never both nonpositive. Since \({\mathcal {A}}\) is regular, there exists \(\gamma \in {\mathcal {A}}\) such that \(\gamma ^2\cdot \alpha \) is positive, while \(\gamma ^2+(\beta +q)^2\) is invertible for all \(q\ge 0\). As explained in Example 4.2, we can assume \(\gamma (s)=0\) for all \(s\in {\mathcal {N}}_\alpha \). Meanwhile, \(\gamma (s)^2+(\beta +q)(s)^2\ne 0\) for all \(s\in {\mathcal {S}}\) (otherwise \(\gamma ^2+(\beta +q)^2\) would not be invertible in \({\mathcal {A}}\), since \({\mathcal {A}}\) is inversion-closed). Thus, \(\gamma (s)\ne 0\) for all \(s\in {\mathcal {N}}_\beta \). \(\diamond \)

Claim 2

For any \({\mathcal {N}}\in {\mathfrak {N}}\) and \(s_0\in {\mathcal {N}}^\complement \), there exists \(\gamma \in {\mathcal {A}}\) such that \(\gamma ({\mathcal {N}})=0\) while \(\gamma (s_0)>0\).

Proof

Suppose \({\mathcal {N}}={\mathcal {N}}_\alpha \) for some \(\alpha \in {\mathcal {A}}\). Then \(\alpha (s_0)=r>0\), since \(s_0\not \in {\mathcal {N}}\). Let \(\beta :=r-\alpha \). Then \(\beta \in {\mathcal {A}}\), and for all \(s\in {\mathcal {S}}\), we have \(\beta (s)\le 0\) if and only if \(\alpha (s)\ge r\). Thus, if \({\mathcal {M}}:={\mathcal {N}}_\beta \), then \(s_0\in {\mathcal {M}}\), \({\mathcal {M}}\in {\mathfrak {N}}\) and \({\mathcal {M}}\) is disjoint from \({\mathcal {N}}\). So Claim 1 yields some \(\gamma \in {\mathcal {A}}\) such that \(\gamma ({\mathcal {N}})=0\) while \(\gamma (s)\ne 0\) for all \(s\in {\mathcal {M}}\). Thus, \(\gamma (s_0)\ne 0\). By replacing \(\gamma \) with \(\gamma ^2\) if necessary, we can ensure that \(\gamma (s_0)> 0\). \(\diamond \)

Let \({\mathfrak {M}}\) be the collection of all sets \({\mathcal {M}}\) of the form \({\mathcal {M}}=\bigcap _{{\mathcal {N}}\in {\mathfrak {N}}'}{\mathcal {N}}\), for any subset \({\mathfrak {N}}'\subseteq {\mathfrak {N}}\).

Claim 3

For any \({\mathcal {M}}\in {\mathfrak {M}}\) and \(s\in {\mathcal {M}}^\complement \), there exists \(\gamma \in {\mathcal {A}}\) such that \(\gamma ({\mathcal {M}})=0\) while \(\gamma (s)>0\).

Proof

Let \({\mathcal {M}}=\bigcap _{{\mathcal {N}}\in {\mathfrak {N}}'}{\mathcal {N}}\) for some \({\mathfrak {N}}'\subseteq {\mathfrak {N}}\). If \(s\not \in {\mathcal {M}}\), then \(s\not \in {\mathcal {N}}\) for some \({\mathcal {N}}\in {\mathfrak {N}}'\). Claim 2 yields \(\gamma \in {\mathcal {A}}\) such that \(\gamma ({\mathcal {N}})=0\) while \(\gamma (s)>0\). As \({\mathcal {M}}\subseteq {\mathcal {N}}\), hence \(\gamma ({\mathcal {M}})=0\). \(\diamond \)

Claim 4

\({\mathfrak {M}}\) is a subbase of closed sets for the topology of \({\mathcal {S}}\).

Proof

Let \({\mathcal {F}}\subseteq {\mathbb {R}}\) be closed. Then \({\mathcal {O}}={\mathbb {R}}\setminus {\mathcal {F}}\) is open. Thus, \({\mathcal {O}}\) is a union of open intervals—in other words, \({\mathcal {O}}=\bigcup _{i\in {\mathcal {I}}} (a_i,b_i)\) for some (possibly infinite) indexing set \({\mathcal {I}}\) and some \(\{a_i\}_{i\in {\mathcal {I}}}\) and \(\{b_i\}_{i\in {\mathcal {I}}}\) with \(a_i<b_i\) for all \(i\in {\mathcal {I}}\).Footnote 37 Thus, \({\mathcal {F}}=\bigcap _{i\in {\mathcal {I}}} \left((-{\infty },a_i]\sqcup [b_i,{\infty })\right)\). Thus, for any \(\alpha \in {\mathcal {A}}\), we have

$$\begin{aligned} \alpha ^{-1}({\mathcal {F}})=\bigcap _{i\in {\mathcal {I}}} \left(\alpha ^{-1}(-{\infty },a_i]\sqcup \alpha ^{-1}[b_i,{\infty })\right). \end{aligned}$$
(C1)

Now, for any \(i\in {\mathcal {I}}\), the function \(\beta _i:=(\alpha -b_i)\cdot (a_i-\alpha )\) is in \({\mathcal {A}}\), and it is easily seen that \({\mathcal {N}}_{\beta _i}=\alpha ^{-1}(-{\infty },a_i]\sqcup \alpha ^{-1}[b_i,{\infty })\). Thus, formula (C1) says that \(\alpha ^{-1}({\mathcal {F}})=\bigcap _{i\in {\mathcal {I}}} {\mathcal {N}}_{\beta _i}\). Thus, \(\alpha ^{-1}({\mathcal {F}})\in {\mathfrak {M}}\) for all \(\alpha \in {\mathcal {A}}\) and closed \({\mathcal {F}}\subseteq {\mathbb {R}}\). These sets form a subbase of closed sets for the weak topology induced by \({\mathcal {A}}\) on \({\mathcal {S}}\). \(\diamond \)

Now, let \({\mathcal {F}}\subseteq {\mathcal {S}}\) be closed. Claim 4 implies that \({\mathcal {F}}=\bigcap _{i\in {\mathcal {I}}} {\mathcal {F}}^i\), where \({\mathcal {I}}\) is some (possibly infinite) indexing set and where, for all \(i\in {\mathcal {I}}\), \({\mathcal {F}}^i\) is a finite union of elements of \({\mathfrak {M}}\)—that is, \({\mathcal {F}}^i=\bigcup _{j=1}^{J_i} {\mathcal {M}}^{i}_{j}\) for some \(J_i\in {\mathbb {N}}\) and \({\mathcal {M}}^{i}_{1},\ldots ,{\mathcal {M}}^{i}_{J_i}\in {\mathfrak {M}}\)

Let \(s\in {\mathcal {F}}^\complement \). Then \(s\not \in {\mathcal {F}}^i\) for some \(i\in {\mathcal {I}}\). Thus, for all \(j\in [1\cdots J_i]\), we have \(s\not \in {\mathcal {M}}^{i}_{j}\), so Claim 3 yields some \(\gamma _j\in {\mathcal {A}}\) such that \(\gamma _j({\mathcal {M}}^{i}_{j})=0\) while \(\gamma _j(s)>0\). Let \(\gamma :=\gamma _1\cdot \gamma _2\cdots \gamma _J\). Then \(\gamma (m)=0\) for all \(m\in {\mathcal {M}}^{i}_{1}\cup \cdots \cup {\mathcal {M}}^{i}_{J_i}\), while \(\gamma (s)>0\). But \({\mathcal {F}}\subseteq {\mathcal {F}}^i={\mathcal {M}}^{i}_{1}\cup \cdots \cup {\mathcal {M}}^{i}_{J_i}\). Thus, \(\gamma ({\mathcal {F}})=0\).

Thus, we have constructed \(\gamma \in {\mathcal {A}}\) such that \(\gamma ({\mathcal {F}})=0\) while \(\gamma (s)\ne 0\). This argument works for any closed \({\mathcal {F}}\subseteq {\mathcal {S}}\) and \(s\in {\mathcal {F}}^\complement \). Thus, \({\mathcal {A}}\) is Tychonoff for \({\mathcal {S}}\). \(\square \)

Lemma C3

Let \({\mathcal {S}}\) be a compact Hausdorff space, and let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\) be a regular, inversion-closed algebra of continuous functions that generates the topology of \({\mathcal {S}}\). Then \({\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra. For any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), let \({\mathfrak {K}}({\mathcal {R}}):=\{\alpha \in {\mathcal {A}}\); \(\alpha (r)=0\) for all \(r\in {\mathcal {R}}\}\). Then \({\mathfrak {K}}:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra anti-isomorphism.

Proof

We will first prove three statements:

  1. (i)

    For all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), \({\mathfrak {K}}(\lnot {\mathcal {R}})={\mathfrak {K}}({\mathcal {R}})^{\perp \!\perp }\).

  2. (ii)

    For all \({\mathcal {Q}},{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), \({\mathfrak {K}}({\mathcal {Q}}\cup {\mathcal {R}})={\mathfrak {K}}({\mathcal {Q}})\cap {\mathfrak {K}}({\mathcal {R}})\).

  3. (iii)

    For all \({\mathcal {Q}},{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), \({\mathfrak {K}}({\mathcal {Q}}\wedge {\mathcal {R}})={\mathfrak {K}}({\mathcal {Q}})\Cup {\mathfrak {K}}({\mathcal {R}})\).

The proof of Statement (ii) is almost identical to that of Statement (b) in the proof of Example 3.6, except that “\({\mathcal {C}}({\mathcal {S}})\)” is replaced with “\({\mathcal {A}}\)”. To prove Statement (i), note that \({\mathfrak {K}}({\mathcal {R}})=\{\kappa \in {\mathcal {A}}\); \(\mathrm {supp}(\kappa )\subseteq {\mathcal {R}}^\complement \}\). Let \(\alpha \in {\mathcal {A}}\). Then

$$\begin{aligned} \begin{aligned} \left( \alpha \in {\mathfrak {K}}({\mathcal {R}})^{\perp \!\perp }\right)&\Longleftrightarrow \left( \alpha \cdot \kappa =0\hbox { for all }\kappa \in {\mathfrak {K}}({\mathcal {R}})\right) {\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \mathrm {supp}(\alpha )\subseteq {\mathcal {R}}\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \mathrm {supp}(\alpha )\subseteq \mathrm {int}({\mathcal {R}})\right) \Longleftrightarrow \left( \alpha (s)=0\hbox { for all }s\in \lnot {\mathcal {R}}\right) \\&\Longleftrightarrow \left( \alpha \in {\mathfrak {K}}(\lnot {\mathcal {R}})\right) . \end{aligned} \end{aligned}$$

Thus, \({\mathfrak {K}}({\mathcal {R}})^{\perp \!\perp }={\mathfrak {K}}(\lnot {\mathcal {R}})\), as desired. Here, \((\dagger )\) is because \(\mathrm {supp}(\alpha )\) is an open set, because \(\alpha \) is a continuous function. The “\({\Longleftarrow }\)” direction of \((*)\) is obvious. To prove the “\(\Longrightarrow \)” direction, note that Lemma C2 says \({\mathcal {A}}\) is Tychonoff for the topology of \({\mathcal {S}}\). If \(\alpha \cdot \kappa =0\) for all \(\kappa \in {\mathfrak {K}}({\mathcal {R}})\), then \(\mathrm {supp}(\alpha )\) must be disjoint from \(\bigcup _{\kappa \in {\mathfrak {K}}({\mathcal {R}})}\mathrm {supp}(\kappa )\). But Lemma C1 says this union is all of \({\mathcal {R}}^\complement \). Thus, \(\mathrm {supp}(\alpha )\) is disjoint from \({\mathcal {R}}^\complement \), so \(\mathrm {supp}(\alpha )\subseteq {\mathcal {R}}\).

It remains to prove Statement (iii). In Example 3.6, the corresponding Statement (c) was derived from Statements (a) and (b) using De Morgan’s law. But we cannot use that approach here because we do not yet know that \({\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra (that is what we are trying to prove). We need the following result.

Claim 1

\([{\mathfrak {K}}({\mathcal {Q}})+{\mathfrak {K}}({\mathcal {R}})]^{\perp \!\perp }= {\mathfrak {K}}\left[(\lnot {\mathcal {R}})\cup (\lnot {\mathcal {Q}})\right]\).

Proof

Let \(\alpha \in {\mathcal {A}}\). Then

$$\begin{aligned} \begin{aligned} \left( \alpha \in [{\mathfrak {K}}({\mathcal {Q}})+{\mathfrak {K}}({\mathcal {R}})]^{\perp \!\perp }\right)&\iff \left( \alpha \cdot \beta =0\hbox { for all }\beta \in {\mathfrak {K}}({\mathcal {Q}})+{\mathfrak {K}}({\mathcal {R}})\right) \\&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \alpha \cdot \beta =0\hbox { for all } \beta \in {\mathfrak {K}}({\mathcal {Q}})\cup {\mathfrak {K}}({\mathcal {R}})\right) \\&\iff \left( \alpha \in [{\mathfrak {K}}({\mathcal {Q}})\cup {\mathfrak {K}}({\mathcal {R}})]^{\perp \!\perp }\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \alpha \in {\mathfrak {K}}({\mathcal {Q}})^{\perp \!\perp }\cap {\mathfrak {K}}({\mathcal {R}})^{\perp \!\perp }\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\diamond )}} \left( \alpha \in {\mathfrak {K}}(\lnot {\mathcal {Q}})\cap {\mathfrak {K}}(\lnot {\mathcal {R}})\right) {\mathop {\Longleftrightarrow }\limits _{(\ddagger )}} \left( \alpha \in {\mathfrak {K}}[(\lnot {\mathcal {R}})\cup (\lnot {\mathcal {Q}})]\right) . \end{aligned} \end{aligned}$$

Here, \((*)\) is because every element of \({\mathfrak {K}}({\mathcal {Q}})+{\mathfrak {K}}({\mathcal {R}})\) is a sum of elements from \({\mathfrak {K}}({\mathcal {Q}})\) and \({\mathfrak {K}}({\mathcal {R}})\), and multiplication distributes over addition. Next \((\dagger )\) is because \(({\mathcal {I}}\cup {\mathcal {J}})^{\perp \!\perp }={\mathcal {I}}^{\perp \!\perp }\cap {\mathcal {J}}^{\perp \!\perp }\) for any \({\mathcal {I}},{\mathcal {J}}\subseteq {\mathcal {C}}({\mathcal {S}})\). Finally, \((\diamond )\) is by Statement (i), while \((\ddagger )\) is by (ii). \(\diamond \)

Now we have

$$\begin{aligned} \begin{aligned} {\mathfrak {K}}({\mathcal {Q}})\Cup {\mathfrak {K}}({\mathcal {R}})=&\, \left(\left[{\mathfrak {K}}({\mathcal {Q}})+{\mathfrak {K}}({\mathcal {R}})\right]^{\perp \!\perp }\right)^{\perp \!\perp }\ \ {\mathop {=}\limits _{\mathrm {(*)}}} \ \ {\mathfrak {K}}\left[(\lnot {\mathcal {R}})\cup (\lnot {\mathcal {Q}})\right]^{\perp \!\perp }\\ {\mathop {=}\limits _{\mathrm {(\dagger )}}}&\, {\mathfrak {K}}\left(\lnot \left[(\lnot {\mathcal {R}})\cup (\lnot {\mathcal {Q}})\right]\right) \ \ {\mathop {=}\limits _{\mathrm {(\diamond )}}} \ \ {\mathfrak {K}}\left[(\lnot \lnot {\mathcal {R}})\wedge (\lnot \lnot {\mathcal {Q}})\right]\\ =&\,{\mathfrak {K}}\left({\mathcal {R}}\wedge {\mathcal {Q}}\right), \end{aligned} \end{aligned}$$

as desired. Here \((*)\) is by Claim 1, \((\dagger )\) is by Statement (i), and \((\diamond )\) is by applying De Morgan’s Law in the Boolean algebra \({\mathfrak {R}}({\mathcal {S}})\). This proves Statement (iii).

Claim 2

\({\mathfrak {K}}:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is surjective.

Proof

Let \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\). Then \({\mathcal {I}}={\mathcal {G}}^{\perp \!\perp }\) for some \({\mathcal {G}}\subseteq {\mathcal {A}}\). Let \({\mathcal {O}}:=\bigcup _{\gamma \in {\mathcal {G}}}\mathrm {supp}(\gamma )\), an open subset of \({\mathcal {S}}\), and let \({\mathcal {R}}:=\mathrm {clos}({\mathcal {O}})\). Then \({\mathcal {R}}\) is a regular closed set. Let \(\alpha \in {\mathcal {A}}\). Then

$$\begin{aligned} \left( \alpha \in {\mathcal {I}}\right)&\iff \left( \alpha \in {\mathcal {G}}^{\perp \!\perp }\right) \iff \left( \alpha \cdot \gamma =0\hbox { for all }\gamma \in {\mathcal {G}}\right) \\&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \alpha (o)=0\hbox { for all }o\in {\mathcal {O}}\right) \iff \left( \mathrm {supp}(\alpha )\subseteq {\mathcal {O}}^\complement \right) \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \mathrm {supp}(\alpha )\subseteq \mathrm {int}({\mathcal {O}}^\complement )={\mathcal {R}}^\complement \right) \iff \left( \alpha \in {\mathfrak {K}}({\mathcal {R}})\right) . \end{aligned}$$

Thus, \({\mathfrak {K}}({\mathcal {R}})={\mathcal {I}}\), as desired. Here the “\({\Longleftarrow }\)” direction of \((*)\) is obvious; for the “\(\Longrightarrow \)” direction, recall that for any \(o\in {\mathcal {O}}\), there is some \(\gamma \in {\mathcal {G}}\) with \(\gamma (o)\ne 0\), by definition of \({\mathcal {O}}\). Meanwhile, \((\dagger )\) is because \(\mathrm {supp}(\alpha )\) is open, because \(\alpha \) is continuous. \(\diamond \)

Claim 2 and Statements (i), (ii) and (iii) imply that for every algebraic identity involving \(\cup \), \(\wedge \) and \(\lnot \) that holds between generic elements of \({\mathfrak {R}}({\mathcal {S}})\), the corresponding algebraic identity involving \(\cap \), \(\Cup \) and \({}^{\perp \!\perp }\) holds between generic elements of \({\mathfrak {I}}({\mathcal {A}})\). Since \({\mathfrak {R}}({\mathcal {S}})\) is a Boolean algebra, it follows that \({\mathfrak {I}}({\mathcal {A}})\) is also a Boolean algebra. The proof that \({\mathfrak {K}}\) is injective is identical to argument in the proof of Example 3.6 (using the fact that \(\varvec{1}\in {\mathcal {A}}\) because \({\mathcal {A}}\) is inversion-closed, hence unitary). At this point, Statements (i), (ii) and (iii) say \({\mathfrak {K}}\) is a Boolean algebra anti-isomorphism from \({\mathfrak {R}}({\mathcal {S}})\) to \({\mathfrak {I}}({\mathcal {A}})\). \(\square \)

The general strategy for proving Theorems 45 and 6 is similar to the strategy used to prove Theorems 12 and 3. The first step is a result that plays the role of Proposition B1 for realistic Banach algebras. Part (a) of this result seems to be a “folk theorem” in the theory of Banach algebras—it is well known by specialists, but it is difficult to find a clear statement of it anywhere in the literature. Thus, I provide a complete statement and proof here, for completeness.

Proposition C4

  1. (a)

    Let \({\mathcal {A}}\) be a realistic Banach algebra. Then there is a compact Hausdorff space \({\mathcal {S}}\), a uniformly dense, inversion-closed subalgebra \({\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) that contains \(\varvec{1}\) and generates the topology of \({\mathcal {S}}\) and a continuous algebra isomorphism \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) with \(\Vert \varPhi \Vert =1\). Furthermore, \({\mathcal {S}}\) and \(\varPhi \) are unique up to homeomorphism.

  2. (b)

    If \({\mathcal {A}}\) is uniform, then \({\widehat{\mathcal {A}}}={\mathcal {C}}({\mathcal {S}})\) and \(\varPhi \) is a Banach algebra isomorphism. If \({\mathcal {A}}\) is perfectly uniform, then \(\varPhi \) is also an isometry.

  3. (c)

    \({\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra, and there is a Boolean algebra anti-isomorphism \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) defined by \(\varPhi ^*({\mathcal {R}}):=\varPhi ^{-1}[{\widehat{\mathcal {A}}}\cap {\mathfrak {K}}({\mathcal {R}})]\), for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\).

  4. (d)

    Let \({\mathcal {A}}'\) be another realistic Banach algebra with associated compact Hausdorff space \({\mathcal {S}}'\) and isomorphism \(\varPhi ':{\mathcal {A}}'{{\longrightarrow }}{\widehat{\mathcal {A}}}'\subseteq {\mathcal {C}}({\mathcal {S}}')\). If \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}'\) is a Banach algebra homomorphism, then there is a continuous function \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) such that, for all \(\alpha \in {\mathcal {A}}\), \(\varPhi '[\varPsi (\alpha )] = \varPhi (\alpha )\circ \psi \). In other words, if we define \({\widehat{\varPsi }}({\widehat{\alpha }}) :={\widehat{\alpha }}\circ \psi \) for all \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\), then the diagram in Fig. 3a commutes.

  5. (e)

    If \(\varPsi \) is injective, then \(\psi \) is a quotient map from \({\mathcal {S}}'\) onto \({\mathcal {S}}\). If \(\varPsi \) is surjective, then \(\psi \) is an embedding of \({\mathcal {S}}'\) into \({\mathcal {S}}\).Footnote 38

  6. (f)

    Let \({\mathcal {A}}''\) be a third realistic Banach algebra, and suppose there are Banach algebra homomorphisms \(\varPsi \), \(\varTheta \) and \(\varXi \) as shown in the left half of diagram (Fig. 3b), with associated continuous functions \(\psi \), \(\theta \) and \(\xi \) from part (c), shown in the right half of Fig. 3b. If the left diagram commutes, then the right diagram also commutes.Footnote 39

Proof

(a) Let \({\mathcal {C}}({\mathcal {S}},{\mathbb {C}})\) be the algebra of all continuous complex-valued functions on \({\mathcal {S}}\)—this is to be distinguished from \({\mathcal {C}}({\mathcal {S}},{\mathbb {R}})\), the algebra of all continuous real-valued functions on \({\mathcal {S}}\) (which I have denoted by \({\mathcal {C}}({\mathcal {S}})\) up until now). The Gelfand Representation Theorem yields a compact Hausdorff space \({\mathcal {S}}\), a subalgebra \({\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}},{\mathbb {C}})\) and a Banach algebra homomorphism \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) with \(\Vert \varPhi \Vert =1\) (Ingelstam 1964, Theorems 4.3 and 4.4).Footnote 40 The kernel of \(\varPhi \) is the Jacobson radical of \({\mathcal {A}}\); since I assume \({\mathcal {A}}\) is semisimple, this kernel is trivial, so \(\varPhi \) is injective, and hence, an algebra isomorphism from \({\mathcal {A}}\) to \({\widehat{\mathcal {A}}}\).

Claim 1

\({\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}},{\mathbb {R}})\).

Proof

Let \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\); then, \({\widehat{\alpha }}=\varPhi (\alpha )\) for some \(\alpha \in {\mathcal {A}}\). Now \({\mathcal {A}}\) is strictly real, so for any \(r>0\), \(\alpha ^2+r\) is invertible; call its inverse \(\beta \), and let \({\widehat{\beta }}:=\varPhi (\beta )\). Then \({\widehat{\beta }}\cdot ({\widehat{\alpha }}^2+r)=\varPhi (\beta )\cdot (\varPhi (\alpha )^2+r)=\varPhi (\beta \cdot (\alpha ^2+r))=\varPhi (\varvec{1})=\widehat{\varvec{1}}\). Thus, \({\widehat{\alpha }}^2+r\) is invertible, for all \(r>0\). Thus, for all \(s\in {\mathcal {S}}\), we must have \({\widehat{\alpha }}^2(s)+r\ne 0\); hence, \({\widehat{\alpha }}^2(s)\ne -r\). Thus, we conclude that \({\widehat{\alpha }}^2\) never takes negative values, for any \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\). Thus, \({\widehat{\alpha }}\) never takes values on the imaginary line \({\mathbb {R}}{\mathbf { i}}\), for any \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\).

Now, to obtain a contradiction, suppose there was some \({\widehat{\beta }}\in {\widehat{\mathcal {A}}}\) such that \({\widehat{\beta }}(s)\not \in {\mathbb {R}}\) for some \(s\in {\mathcal {S}}\). Thus, \({\widehat{\beta }}(s)^2\not \in [0,{\infty })\). By the previous paragraph, \({\widehat{\beta }}(s)\not \in {\mathbb {R}}{\mathbf { i}}\). Thus, \({\widehat{\beta }}(s)^2\) is not a negative number, so \({\widehat{\beta }}(s)^2\not \in {\mathbb {R}}\). Thus, \({\widehat{\beta }}(s)^2+1\not \in {\mathbb {R}}\). Let \(r:=|{\widehat{\beta }}(s)^2+1|\), and define \({\widehat{\gamma }}:=({\widehat{\beta }}^2+\varvec{1})/r\). Then \({\widehat{\gamma }}(s)\not \in {\mathbb {R}}\), and \(|{\widehat{\gamma }}(s)|=1\).

Now, suppose \({\widehat{\beta }}=\varPhi (\beta )\) for some \(\beta \in {\mathcal {A}}\). Thus, if \(\gamma :=(\beta ^2+\varvec{1})/r\), then \({\widehat{\gamma }}=\varPhi (\gamma )\). Since \({\mathcal {A}}\) is strictly real, \(\beta ^2+\varvec{1}\) is invertible. Thus, \(\gamma \) is also invertible. Let \(\gamma ^{-1}\) be its multiplicative inverse, and let \({\widehat{\gamma }}^{-1}:=\varPhi (\gamma ^{-1})\). Then \({\widehat{\gamma }}^{-1}\) is the multiplicative inverse of \({\widehat{\gamma }}\); that is, \({\widehat{\gamma }}\cdot {\widehat{\gamma }}^{-1}=\varvec{1}\). In particular, \({\widehat{\gamma }}(s)\cdot {\widehat{\gamma }}^{-1}(s)=1\). Since \(|{\widehat{\gamma }}(s)|=1\), this means that \({\widehat{\gamma }}^{-1}(s)\) is the complex conjugate of \({\widehat{\gamma }}(s)\).

There exists some \(\theta \in (-\pi ,\pi ]\) such that \({\widehat{\gamma }}(s) = \cos (\theta )+{\mathbf { i}}\sin (\theta )\). Thus, \({\widehat{\gamma }}^{-1}(s) = \cos (\theta )-{\mathbf { i}}\sin (\theta )\). Since \({\widehat{\gamma }}(s)\not \in {\mathbb {R}}\), we know that \(\theta \ne 0\) and \(\theta \ne \pi \). Let \({\widehat{\alpha }}:={\widehat{\gamma }}-{\widehat{\gamma }}^{-1}\). Clearly, \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\). But \({\widehat{\alpha }}(s)-{\widehat{\alpha }}^{-1}(s)=\cos (\theta )+{\mathbf { i}}\sin (\theta )-\cos (\theta )+{\mathbf { i}}\sin (\theta ) =2{\mathbf { i}}\sin (\theta )\), which contradicts our earlier observation that \({\widehat{\alpha }}\) never takes values in \({\mathbb {R}}{\mathbf { i}}\) for any \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\).

To avoid the contradiction, there cannot exist any \({\widehat{\beta }}\in {\widehat{\mathcal {A}}}\) and \(s\in {\mathcal {S}}\) with \({\widehat{\beta }}(s)\not \in {\mathbb {R}}\). Thus, every element of \({\widehat{\mathcal {A}}}\) is real-valued. \(\diamond \)

Before continuing, it is useful to explain the construction of the space \({\mathcal {S}}\). Recall that \({\mathbb {C}}\) is a Banach algebra under the standard operations and the standard norm. The elements of \({\mathcal {S}}\) are actually Banach algebra homomorphisms from \({\mathcal {A}}\) to \({\mathbb {C}}\); indeed, \({\mathcal {S}}\) is the set of all such homomorphisms, equipped with the topology of pointwise convergence.Footnote 41

Claim 2

\({\widehat{\mathcal {A}}}\) is uniformly dense in \({\mathcal {C}}({\mathcal {S}},{\mathbb {R}})\).

Proof

We will first show that \({\widehat{\mathcal {A}}}\) separates the points of \({\mathcal {S}}\). That is, for any distinct \(s_1,s_2\in {\mathcal {S}}\), there is some \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\) such that \({\widehat{\alpha }}(s_1)\ne {\widehat{\alpha }}(s_2)\). To see this, recall that the elements of \({\mathcal {S}}\) are functions from \({\mathcal {A}}\) to \({\mathbb {C}}\); if \({\widehat{\alpha }}=\varPhi (\alpha )\) for some \(\alpha \in {\mathcal {A}}\), then \({\widehat{\alpha }}(s)=s(\alpha )\) for all \(s\in {\mathcal {S}}\). So if \({\widehat{\alpha }}(s_1)={\widehat{\alpha }}(s_2)\) for all \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\), this means that \(s_1(\alpha )=s_2(\alpha )\) for all \(\alpha \in {\mathcal {A}}\), which means that \(s_1\) and \(s_2\) are actually the same function, contradicting the fact that \(s_1\ne s_2\).

Thus, \({\widehat{\mathcal {A}}}\) is a subalgebra of \({\mathcal {C}}({\mathcal {S}},{\mathbb {R}})\) that separates the points of \({\mathcal {S}}\). Thus, Stone–Weierstrass Theorem says that \({\mathcal {A}}\) is uniformly dense in \({\mathcal {C}}({\mathcal {S}},{\mathbb {R}})\) (Aliprantis and Border 2006, Thm. 9.13; Fremlin 2004, Thm. 281E; Conway 1990, Thm. V.8.1). \(\diamond \)

As just noted, for all \(\alpha \in {\mathcal {A}}\), if \({\widehat{\alpha }}=\varPhi (\alpha )\), then \({\widehat{\alpha }}(s)=s(\alpha )\) for all \(s\in {\mathcal {S}}\) (recalling that \({\mathcal {S}}\) is actually a collection of \({\mathbb {C}}\)-valued functions on \({\mathcal {A}}\)). Thus, the topology of pointwise convergence on \({\mathcal {S}}\) (as a subset of \({\mathbb {C}}^{\mathcal {A}}\)) is actually the weak topology induced by \({\widehat{\mathcal {A}}}\). In other words, \({\widehat{\mathcal {A}}}\) generates the topology of \({\mathcal {S}}\). Furthermore, \({\widehat{\mathcal {A}}}\) is inversion-closed (Kaniuth 2009, Theorem 2.2.5; Dales et al. 2003, Theorem 3.2.2(ii)).

Claim 3

\({\widehat{\mathcal {A}}}\) is Tychonoff for the topology on \({\mathcal {S}}\).

Proof

Regularity is a purely algebraic property, so it is preserved by homomorphisms. Thus, since \({\mathcal {A}}\) is regular, \(\varPhi \) is a homomorphism and \({\widehat{\mathcal {A}}}=\varPhi ({\mathcal {A}})\), it follows that \({\widehat{\mathcal {A}}}\) is also regular. As already noted, \({\widehat{\mathcal {A}}}\) is inversion-closed and generates the topology of \({\mathcal {S}}\). Thus, Lemma C2 says that \({\widehat{\mathcal {A}}}\) is Tychonoff for the topology on \({\mathcal {S}}\). \(\diamond \)

Claim 4

\({\mathcal {S}}\) and \(\varPhi \) are unique up to homeomorphism.

Proof

Suppose \({\mathcal {S}}'\) was another compact Hausdorff space, and \(\varPhi ':{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}'\subseteq {\mathcal {C}}({\mathcal {S}}')\) was another Banach algebra isomorphism (for some dense subalgebra \({\widehat{\mathcal {A}}}'\)). For any \(s'\in {\mathcal {S}}'\), define \(\delta _{s'}:{\mathcal {A}}{{\longrightarrow }}{\mathbb {R}}\) by \(\delta _{s'}(\alpha ):={\widehat{\alpha }}'(s')\), where \({\widehat{\alpha }}':=\varPhi '(\alpha )\). It is straightforward to verify that \(\delta _{s'}:{\mathcal {A}}{{\longrightarrow }}{\mathbb {R}}\) is a Banach algebra homomorphism—hence, it is an element of \({\mathcal {S}}\). This yields a function \(\delta :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) defined by \(\delta (s'):=\delta _{s'}\) for all \(s'\in {\mathcal {S}}'\). I will show that \(\delta \) is a homeomorphism.

\(\delta \)is continuous. Let \(s'_1\in {\mathcal {S}}'\); I must show that if \(s'_2\) is “close” to \(s'_1\), then \(\delta (s'_2)\) is close to \(\delta (s'_1)\). Recall that the topology on \({\mathcal {S}}\) is the topology of pointwise convergence on \({\mathbb {R}}^{\mathcal {A}}\). So let \(\alpha \in {\mathcal {A}}\) and let \(\epsilon >0\). Since \({\widehat{\alpha }}':=\varPhi '(\alpha )\) is a continuous function on \({\mathcal {S}}'\), there is some open neighbourhood \({\mathcal {O}}'\) around \(s'_1\) such that \(|{\widehat{\alpha }}'(s'_2)-{\widehat{\alpha }}'(s'_1)|<\epsilon \) for all \(s'_2\in {\mathcal {O}}'\). In other words, \(|\delta (s'_2)(\alpha )-\delta (s'_1)(\alpha )|<\epsilon \) for all \(s'_2\in {\mathcal {O}}'\). We can construct such an \({\mathcal {O}}'\) for any \(\alpha \in {\mathcal {A}}\) and \(\epsilon >0\). Thus, \(\delta \) is continuous at \(s'_1\).

\(\delta \)is injective.\({\mathcal {S}}'\) is compact Hausdorff and therefore Tychonoff, so \({\mathcal {C}}({\mathcal {S}}')\) separates the points of \({\mathcal {S}}\). Thus, for any distinct \(s'_1,s'_2\in {\mathcal {S}}'\), there is some \({\widehat{\beta }}'\in {\mathcal {C}}({\mathcal {S}}')\) such that \({\widehat{\beta }}'(s'_1)\ne {\widehat{\beta }}'(s'_2)\). Let \(\epsilon :=|{\widehat{\beta }}'(s'_1)-{\widehat{\beta }}'(s'_2)|\). \({\widehat{\mathcal {A}}}'\) is dense in \({\mathcal {C}}({\mathcal {S}}')\) in the uniform topology, so there exists \({\widehat{\alpha }}'\in {\widehat{\mathcal {A}}}'\) with \(\Vert {\widehat{\alpha }}'-{\widehat{\beta }}'\Vert _{\infty }<\epsilon /2\). Then \({\widehat{\alpha }}'(s'_1)\ne {\widehat{\alpha }}'(s'_2)\). Thus, if \(\alpha \in {\mathcal {A}}\) is such that \(\varPhi '(\alpha )={\widehat{\alpha }}'\), then \(\delta (s'_2)(\alpha )\ne \delta (s'_1)(\alpha )\). Thus, \(\delta (s'_2)\ne \delta (s'_1)\). This argument works for any distinct \(s'_1,s'_2\in {\mathcal {S}}'\).

\(\delta \)is surjective. (by contradiction) Suppose not. Then \({\mathcal {D}}:=\delta ({\mathcal {S}}')\) is a compact (hence, closed) proper subset of \({\mathcal {S}}\). Let \(s\in {\mathcal {S}}\setminus {\mathcal {D}}\). By Claim 3, there is some \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\) such that \({\widehat{\alpha }}({\mathcal {D}})=0\) but \({\widehat{\alpha }}(s)\ne 0\). Let \(\alpha :=\varPhi ^{-1}({\widehat{\alpha }})\); thus, \(\alpha \in {\mathcal {A}}\), \({\widehat{\alpha }}=\varPhi (\alpha )\), and \(\alpha \ne 0\) because \({\widehat{\alpha }}\ne 0\). Let \({\widehat{\alpha }}':=\varPhi '(\alpha )\). Then for all \(s'\in {\mathcal {S}}'\), we have \({\widehat{\alpha }}'(s')=\delta _{s'}(\alpha )={\widehat{\alpha }}[\delta (s')]=0\) (because \(\delta (s')\in {\mathcal {D}}\)). Thus, \({\widehat{\alpha }}'=0\). But \(\alpha \ne 0\), so this contradicts the fact that \(\varPhi '\) is an isomorphism. To avoid the contradiction, \(\delta \) must be surjective.

At this point, we have established that \(\delta \) is a continuous bijection from \({\mathcal {S}}'\) to \({\mathcal {S}}\). Since \({\mathcal {S}}'\) is compact and \({\mathcal {S}}\) is Hausdorff, this makes \(\delta \) a homeomorphism (Willard 2004, Theorem 17.14). \(\diamond \)

(b) If \({\mathcal {A}}\) is uniform, then \({\widehat{\mathcal {A}}}\) is a closed subset of \({\mathcal {C}}({\mathcal {S}})\) (Palmer 1994, Proposition 3.1.8). But \({\widehat{\mathcal {A}}}\) is a dense subspace of \({\mathcal {C}}({\mathcal {S}})\); thus, we must have \({\widehat{\mathcal {A}}}={\mathcal {C}}({\mathcal {S}})\). Thus, \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {C}}({\mathcal {S}})\) is a continuous algebra isomorphism. The inverse map \(\varPhi ^{-1}:{\mathcal {C}}({\mathcal {S}}){{\longrightarrow }}{\mathcal {A}}\) is a well-defined algebra homomorphism; it is automatically continuous because \({\mathcal {A}}\) is semisimple (Kaniuth 2009, Cor. 2.1.10; Dales et al. 2003, Prop. 5.1.1; Palmer 1994, Thm. 6.1.2). Thus, \(\varPhi \) is a homeomorphism—hence, it is a Banach algebra isomorphism. If \({\mathcal {A}}\) is perfectly uniform, then \(\varPhi \) is also an isometry (Kaniuth 2009, Theorem 2.2.7(iii)).

(c) As explained in the proof of Claim 3, \({\widehat{\mathcal {A}}}\) is regular. Also, \({\widehat{\mathcal {A}}}\) is inversion-closed and generates the topology of \({\mathcal {S}}\). Thus, Lemma C3 says that \({\mathfrak {I}}({\widehat{\mathcal {A}}})\) is a Boolean algebra, and \(\widehat{\mathfrak {K}}:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\widehat{\mathcal {A}}})\) is a Boolean algebra anti-isomorphism. Meanwhile, \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) is a Banach algebra isomorphism; thus, \({\mathfrak {I}}({\mathcal {A}})\) is also a Boolean algebra, and \(\varPhi ^{-1}:{\mathfrak {I}}({\widehat{\mathcal {A}}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra isomorphism. Since \(\varPhi ^*:=\varPhi ^{-1}\circ {\mathfrak {K}}_{\mathcal {A}}\), it follows that \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra anti-isomorphism.

(d,e) The proofs are identical to the proofs of Proposition B1(c, d), except for two changes. First, the term “uniferent Riesz homomorphism” is everywhere replaced with “algebra homomorphism”. Second, in the proof of the statement “If \(\varPsi \) is injective, then \(\psi \) is surjective” in part (e), we have a subalgebra \({\widetilde{\mathcal {A}}}:=\varPsi ({\mathcal {A}})\subseteq {\mathcal {A}}'\), and we must extend a continuous algebra homomorphism \({\widetilde{s}}:{\widetilde{\mathcal {A}}}{{\longrightarrow }}{\mathbb {C}}\) to a continuous algebra homomorphism \(s':{\mathcal {A}}'{{\longrightarrow }}{\mathbb {C}}\). To do this, we use Corollary 4.2.17 of Kaniuth (2009), which is applicable because \({\mathcal {A}}\) is semisimple and regular.

(f) This follows from the construction in part (d). \(\square \)

Proof of Lemma 4.5

This is part of the statement of Proposition C4(c). \(\square \)

Proof of Example 4.6

\({\mathcal {S}}\) is compact Hausdorff, so it is Tychonoff, and thus, \({\mathcal {C}}({\mathcal {S}})\) generates the topology of \({\mathcal {S}}\) (Willard 2004, Theorems 14.12 and 19.3). Meanwhile, \({\mathcal {C}}({\mathcal {S}})\) is regular because \({\mathcal {S}}\) is normal (Willard 2004, Theorem 17.10), and \({\mathcal {C}}({\mathcal {S}})\) is clearly inversion-closed. The claim now follows from Lemma C3 in the case \({\mathcal {A}}={\mathcal {C}}({\mathcal {S}})\). \(\square \)

The next result is analogous to Proposition B2; it is the last step in the proofs of Theorems 45 and 6.

Proposition C5

Let \({\mathcal {A}}\) be a realistic Banach algebra, and let \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) be as in Proposition C4(a, c). Let \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) be a conditional preference structure on \({\mathcal {A}}\). Then there is a conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) on \({\widehat{\mathcal {A}}}\) satisfying statement (3.3) for all \(\alpha ,\beta \in {\mathcal {A}}\) and \({\mathcal {Q}}\in {\mathfrak {R}}({\mathcal {S}})\).

Furthermore, \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) satisfies (Rch), (CEq), (Dom), (Sep), (CCP) and (TC) if and only if \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfies (Rch\('\)), (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)).

Proof

For any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), recall that \({\widehat{\mathcal {A}}}({\mathcal {R}}):=\{{\widehat{\alpha }}_{\upharpoonleft {\mathcal {R}}}\); \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\}\). Define the linear surjection \({\overline{\varPhi }}_{\mathcal {R}}:{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}({\mathcal {R}})\) by setting \({\overline{\varPhi }}_{\mathcal {R}}(\alpha ):=\varPhi (\alpha )_{\upharpoonleft {\mathcal {R}}}\) for all \(\alpha \in {\mathcal {A}}\). Note that \({\overline{\varPhi }}_{\mathcal {R}}\) maps positive elements of \({\mathcal {A}}\) to positive elements of \({\widehat{\mathcal {A}}}({\mathcal {R}})\) and maps scalar multiples of the identity element \(\varvec{1}\) to constant functions. Let \({\mathcal {I}}:=\varPhi ^*({\mathcal {R}})\); then, \({\overline{\varPhi }}_{\mathcal {R}}(\alpha )=0\) if and only if \(\alpha \in {\mathcal {I}}\)—in other words, \({\mathcal {I}}\) is the kernel of \({\overline{\varPhi }}_{\mathcal {R}}\). Thus, \({\overline{\varPhi }}_{\mathcal {R}}\) factors to a well-defined algebra isomorphism \(\varPhi _{\mathcal {R}}:{\mathcal {A}}/{\mathcal {I}}{{\longrightarrow }}{\widehat{\mathcal {A}}}({\mathcal {R}})\), defined by \(\varPhi _{\mathcal {R}}([\alpha ]_{\mathcal {I}}):=\varPhi (\alpha )_{\upharpoonleft {\mathcal {R}}}\) for all \(\alpha \in {\mathcal {A}}\). Let \(\succeq _{\mathcal {R}}\) be the image of \(\succeq _{\mathcal {I}}\) under this isomorphism. That is, for all \({\widehat{\alpha }},{\widehat{\beta }}\in {\widehat{\mathcal {A}}}({\mathcal {R}})\), we define \({\widehat{\alpha }}\succeq _{\mathcal {R}}{\widehat{\beta }}\) if and only if \([\alpha ]_{\mathcal {I}}\succeq _{\mathcal {I}}[\beta ]_{\mathcal {I}}\), where \([\alpha ]_{\mathcal {I}}\) and \([\beta ]_{\mathcal {I}}\) are the (unique) elements of \({\mathcal {A}}/{\mathcal {I}}\) such that \({\widehat{\alpha }}=\varPhi _{\mathcal {R}}([\alpha ]_{\mathcal {I}})\) and \({\widehat{\beta }}=\varPhi _{\mathcal {R}}([\beta ]_{\mathcal {I}})\). Repeating this construction for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), we obtain a conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfying statement (3.3). It is easily verified that \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) satisfies (CEq), (Dom), (Sep), (CCP) and (TC) if and only if \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) satisfies (CEq\('\)), (Dom\('\)), (Sep\('\)), (CCP\('\)) and (TC\('\)). (Use (3.3), the fact that \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is a Boolean algebra anti-isomorphism, and the fact that \({\overline{\varPhi }}_{\mathcal {R}}\) is positivity-preserving and maps multiples of \(\varvec{1}\) to constant functions.) \(\square \)

Proof of Theorem 4

Combine Propositions A1 and C5. \(\square \)

Proof of Theorem 5

Proposition C4(b) yields a compact Hausdorff space \({\mathcal {S}}\), and a Banach algebra isomorphism \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {C}}({\mathcal {S}})\). Thus, in Proposition C5, we have \({\widehat{\mathcal {A}}}={\mathcal {C}}({\mathcal {S}})\). Now apply Proposition A1 to obtain the desired characteristic SEU representation. \(\square \)

Proof of Theorem 6

Combine the representations from Propositions A2 and C5. The details are as in the proof of Theorem 1. \(\square \)

Proofs of results from Section 5

We begin by justifying a claim made in Example 5.1(b).

Lemma D1

Let \({\mathcal {S}}\) be a topological space, let \({\mathcal {S}}'\) be a compact space, and let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be a regular continuous function. Then \(\psi ({\mathcal {S}}')\) is a regular closed subset of \({\mathcal {S}}\).

Proof

Let \({\widetilde{\mathcal {S}}}:=\psi ({\mathcal {S}}')\). Then \({\widetilde{\mathcal {S}}}\) is compact, because \({\mathcal {S}}'\) is compact. Thus, \({\widetilde{\mathcal {S}}}\) is a closed subset of \({\mathcal {S}}\). Thus, if \({\mathcal {F}}:=\partial {\widetilde{\mathcal {S}}}\), then \({\mathcal {F}}={\widetilde{\mathcal {S}}}\setminus \mathrm {int}({\widetilde{\mathcal {S}}})\). To show that \({\widetilde{\mathcal {S}}}\) is regular, it suffices to show that every point in \({\mathcal {F}}\) is a cluster point of \(\mathrm {int}({\widetilde{\mathcal {S}}})\).

Now, \({\mathcal {F}}\) is closed and has empty interior (because every element of \({\mathcal {F}}\) is a cluster point of \({\widetilde{\mathcal {S}}}^\complement \)). Thus, \({\mathcal {F}}\) is nowhere dense in \({\mathcal {S}}\). Thus, \(\psi ^{-1}({\mathcal {F}})\) is nowhere dense in \({\mathcal {S}}'\), because \(\psi \) is regular. Meanwhile, \(\psi ^{-1}({\mathcal {F}})\) is closed because \({\mathcal {F}}\) is closed and \(\psi \) is continuous. Thus, \(\psi ^{-1}({\mathcal {F}})\) has empty interior.

Let \(s\in {\mathcal {F}}\). Then \(s\in {\widetilde{\mathcal {S}}}\), so \(\psi ^{-1}\{s\}\ne \emptyset \). Let \(s'\in \psi ^{-1}\{s\}\). Then \(s'\in \psi ^{-1}({\mathcal {F}})\). As just noted, \(\mathrm {int}[\psi ^{-1}({\mathcal {F}})]=\emptyset \). So there is a net \(\{s'_i\}_{i\in {\mathcal {I}}}\subseteq {\mathcal {S}}'\setminus \psi ^{-1}({\mathcal {F}})\) (for some indexing set \({\mathcal {I}}\)) that converges to s. For all \(i\in {\mathcal {I}}\), let \(s_i:=\psi (s'_i)\); then, \(s_i\in \psi [{\mathcal {S}}'\setminus \psi ^{-1}({\mathcal {F}})] = \psi ({\mathcal {S}}')\setminus {\mathcal {F}}={\widetilde{\mathcal {S}}}\setminus {\mathcal {F}}=\mathrm {int}({\widetilde{\mathcal {S}}})\). Thus, \(\{s_i\}_{i\in {\mathcal {I}}}\) is a net in \(\mathrm {int}({\widetilde{\mathcal {S}}})\). Furthermore, this net converges to s, because \(\psi \) is continuous. Thus, s is a cluster point of \(\mathrm {int}({\widetilde{\mathcal {S}}})\), desired. \(\square \)

Lemma D2

Let \({\mathcal {S}}\) be a topological space, and let \(\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}\) be a characteristic structure subordinate to a Borel probability measure \(\mu \) on \({\mathcal {S}}\). Let \({\widetilde{\mathcal {S}}}\in {\mathfrak {R}}({\mathcal {S}})\), and let \(({\widetilde{\mu }},\{{\widetilde{\chi }}_{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})})\) be the Bayesian update of \((\mu ,\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})})\), conditional on \({\widetilde{\mathcal {S}}}\).

  1. (a)

    \(\{{\widetilde{\chi }}_{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})}\) is a characteristic structure on \({\mathfrak {R}}({\widetilde{\mathcal {S}}})\), subordinate to \({\widetilde{\mu }}\).

  2. (b)

    Let \(M:={\widetilde{\nu }}_{\widetilde{\mathcal {S}}}[{\widetilde{\mathcal {S}}}]\). For any measurable function \(\gamma :{\mathcal {S}}{{\longrightarrow }}{\mathbb {R}}\) and any \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\),

    $$\begin{aligned} \int _{\widetilde{\mathcal {S}}}\gamma \cdot {\widetilde{\chi }}_{\mathcal {R}}\ \mathrm {d}{\widetilde{\mu }}= \frac{1}{M} \int _{\mathcal {S}}\gamma \cdot \chi _{\mathcal {R}}\ \mathrm {d}\mu . \end{aligned}$$

Proof

  1. (a)

    We must show that (i) \({\widetilde{\chi }}_{\mathcal {R}}(s)=1\) for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and \(\forall _{{\widetilde{\mu }}}\, s\in \mathrm {int}({\mathcal {R}})\); (ii) \({\widetilde{\chi }}_{\mathcal {R}}(s)=0\) for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and \(\forall _{{\widetilde{\mu }}}\, s\in {\widetilde{\mathcal {S}}}\setminus {\mathcal {R}}\); and (iii) \({\widetilde{\chi }}_{{\mathcal {R}}_1}(s)+\cdots +{\widetilde{\chi }}_{{\mathcal {R}}_N}(s) = 1\) for any regular closed partition \(\{{\mathcal {R}}_1,\ldots ,{\mathcal {R}}_N\}\) of \({\widetilde{\mathcal {S}}}\) and \(\forall _{{\widetilde{\mu }}}\)\(s\in {\widetilde{\mathcal {S}}}\).

    1. (i)

      For any measurable \({\mathcal {U}}\subseteq \mathrm {int}({\widetilde{\mathcal {S}}})\), we have \(\nu _{{\widetilde{\mathcal {S}}}}[{\mathcal {U}}]=\mu [{\mathcal {U}}]\), and thus, \({\widetilde{\mu }}[{\mathcal {U}}]=\mu [{\mathcal {U}}]/M\). Furthermore, if \({\mathcal {U}}\subseteq \mathrm {int}({\mathcal {R}})\) for some \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\), then \(\nu _{\mathcal {R}}[{\mathcal {U}}]=\mu [{\mathcal {U}}]\), and thus, \({\widetilde{\nu }}_{\mathcal {R}}[{\mathcal {U}}]=\mu [{\mathcal {U}}]/M\). From this, it follows that \(\mathrm {d}{\widetilde{\nu }}_{\mathcal {R}}/\mathrm {d}{\widetilde{\mu }}(s)=1\) for \(\forall _{{\widetilde{\mu }}}\, s\in \mathrm {int}({\mathcal {R}})\). In other words, \({\widetilde{\chi }}_{\mathcal {R}}(s)=1\) for \(\forall _{{\widetilde{\mu }}}\, s\in \mathrm {int}({\mathcal {R}})\). This proves (i).

    2. (ii)

      If \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\) and \({\mathcal {U}}\subseteq {\widetilde{\mathcal {S}}}\setminus {\mathcal {R}}\), then \(\nu _{\mathcal {R}}[{\mathcal {U}}]=0\), and thus, \({\widetilde{\nu }}_{\mathcal {R}}[{\mathcal {U}}]=0\). Thus, \(\mathrm {d}{\widetilde{\nu }}_{\mathcal {R}}/\mathrm {d}{\widetilde{\mu }}(s)=0\) for \(\forall _{{\widetilde{\mu }}}\, s\in {\widetilde{\mathcal {S}}}\setminus {\mathcal {R}}\). In other words, \({\widetilde{\chi }}_{\mathcal {R}}(s)=0\) for \(\forall _{{\widetilde{\mu }}}\, s\in {\widetilde{\mathcal {S}}}\setminus {\mathcal {R}}\). This proves (ii).

    3. (iii)

      Let \(\{{\mathcal {R}}_1,\ldots ,{\mathcal {R}}_N\}\) be a regular closed partition of \({\widetilde{\mathcal {S}}}\). Then \(\{{\mathcal {R}}_1,\ldots ,{\mathcal {R}}_N,\lnot {\widetilde{\mathcal {S}}}\}\) is a regular closed partition of \({\mathcal {S}}\). Thus, formula (5.2) yields

      $$\begin{aligned} \nu _{{\mathcal {R}}_1}+\cdots +\nu _{{\mathcal {R}}_N}+\nu _{\lnot {\widetilde{\mathcal {S}}}} =\mu . \end{aligned}$$
      (D1)

      On the other hand, \(\{{\widetilde{\mathcal {S}}},\lnot {\widetilde{\mathcal {S}}}\}\) is also a regular closed partition of \({\mathcal {S}}\), so formula (5.2) yields

      $$\begin{aligned} \nu _{{\widetilde{\mathcal {S}}}}+\nu _{\lnot {\widetilde{\mathcal {S}}}} =\mu . \end{aligned}$$
      (D2)

      Subtracting (D2) from (D1) and reorganizing, we obtain

      $$\begin{aligned} \nu _{{\mathcal {R}}_1}+\cdots +\nu _{{\mathcal {R}}_N}=\nu _{{\widetilde{\mathcal {S}}}}. \end{aligned}$$
      (D3)

      Thus, \(\nu _{{\mathcal {R}}_1},\ldots ,\nu _{{\mathcal {R}}_N}\) are absolutely continuous with respect to \(\nu _{{\widetilde{\mathcal {S}}}}\). But \({\widetilde{\mu }}=\frac{1}{M}\,\nu _{{\widetilde{\mathcal {S}}}}\) and \({\widetilde{\nu }}_{{\mathcal {R}}_n}=\frac{1}{M}\,\nu _{{\mathcal {R}}_n}\) for all \(n\in [1\cdots N]\), so this implies that \({\widetilde{\nu }}_{{\mathcal {R}}_1},\ldots ,{\widetilde{\nu }}_{{\mathcal {R}}_N}\) are absolutely continuous with respect to \({\widetilde{\mu }}\). Meanwhile, taking the Radon–Nikodym derivatives of both sides of (D3) relative to \(\nu _{{\widetilde{\mathcal {S}}}}\), we obtain

      $$\begin{aligned} \frac{{ \;\; \mathrm{d}\nu }_{{\mathcal {R}}_1}}{{ \;\; \mathrm{d}\nu }_{{\widetilde{\mathcal {S}}}}}+\cdots +\frac{{ \;\; \mathrm{d}\nu }_{{\mathcal {R}}_N}}{{ \;\; \mathrm{d}\nu }_{{\widetilde{\mathcal {S}}}}}= 1 \quad \text{( }{\widetilde{\mu }}\text{-almost } \text{ everywhere). } \end{aligned}$$
      (D4)

      But \(\nu _{{\widetilde{\mathcal {S}}}}=M\,{\widetilde{\mu }}\) and \(\nu _{{\mathcal {R}}_n}=M\,{\widetilde{\nu }}_{{\mathcal {R}}_n}\) for all \(n\in [1\cdots N]\), so

      $$\begin{aligned} \frac{{ \;\; \mathrm{d}\nu }_{{\mathcal {R}}_n}}{{ \;\; \mathrm{d}\nu }_{{\widetilde{\mathcal {S}}}}} =\frac{M\,\mathrm {d}{\widetilde{\nu }}_{{\mathcal {R}}_n}}{M \ \mathrm {d}{\widetilde{\mu }}} = \frac{\mathrm {d}{\widetilde{\nu }}_{{\mathcal {R}}_n}}{\mathrm {d}{\widetilde{\mu }}} = {\widetilde{\chi }}_{{\mathcal {R}}_n}, \end{aligned}$$
      (D5)

      for all \(n\in [1\cdots N]\). Substituting (D5) into (D4) yields \(\chi _{{\mathcal {R}}_1}(s)+\cdots +\chi _{{\mathcal {R}}_N}(s) = 1\) for \(\forall _{{\widetilde{\mu }}}\)\(s\in {\widetilde{\mathcal {S}}}\). This holds for any regular closed partition \(\{{\mathcal {R}}_1,\ldots ,{\mathcal {R}}_N\}\) of \({\widetilde{\mathcal {S}}}\); this proves (iii).

  2. (b)

    Let \(\gamma :{\mathcal {S}}{{\longrightarrow }}{\mathbb {R}}\) be measurable and let \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\). Then

    $$\begin{aligned} \int _{\widetilde{\mathcal {S}}}\gamma \cdot {\widetilde{\chi }}_{\mathcal {R}}\ \mathrm {d}{\widetilde{\mu }}= & {} \int _{\widetilde{\mathcal {S}}}\gamma \cdot \frac{\mathrm {d}{\widetilde{\nu }}_{\mathcal {R}}}{\mathrm {d}{\widetilde{\mu }}} \ \mathrm {d}{\widetilde{\mu }}\ = \ \int _{\widetilde{\mathcal {S}}}\gamma \ \mathrm {d}{\widetilde{\nu }}_{\mathcal {R}}\ = \ \frac{1}{M} \int _{\widetilde{\mathcal {S}}}\gamma \ \mathrm {d}\nu _{\mathcal {R}}\\= & {} \frac{1}{M} \int _{\widetilde{\mathcal {S}}}\gamma \cdot \frac{\mathrm {d}\nu _{\mathcal {R}}}{\mathrm {d}\mu } \ \mathrm {d}\mu \ = \ \frac{1}{M} \int _{\widetilde{\mathcal {S}}}\gamma \cdot \chi _{\mathcal {R}}\ \mathrm {d}\mu \ \ {\mathop {=}\limits _{\mathrm {(*)}}} \ \ \frac{1}{M} \int _{\mathcal {S}}\gamma \cdot \chi _{\mathcal {R}}\ \mathrm {d}\mu , \end{aligned}$$

    as claimed. Here, \((*)\) is because \(\chi _{\mathcal {R}}(s)=0\) for \(\forall _\mu \, s\in {\widetilde{\mathcal {S}}}^\complement \), because \({\mathcal {R}}\subseteq {\widetilde{\mathcal {S}}}\) and \(\chi _{\mathcal {R}}(s)=0\) for \(\forall _\mu \, s\in {\mathcal {R}}^\complement \). The other steps are straightforward applications of the definitions.

\(\square \)

Lemma D3

Let \({\mathcal {S}}\) and \({\mathcal {S}}'\) be topological spaces, and let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be regular and continuous. Let \(\mu '\) be a Borel probability measure on \({\mathcal {S}}'\), and let \(\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}\) be a characteristic structure subordinate to a Borel measure \(\mu '\). Let \({\overline{\mu }}:=\psi (\mu ')\), and let \(\{{\overline{\chi }}_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}\) be the \(\psi \)-image of \(\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}\).

  1. (a)

    \(\{{\overline{\chi }}_{{\mathcal {R}}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}\) is a characteristic structure subordinate to \({\overline{\mu }}\).

  2. (b)

    For any measurable function \(\gamma :{\mathcal {S}}{{\longrightarrow }}{\mathbb {R}}\) and any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), if \({\mathcal {R}}':=\psi ^\sharp ({\mathcal {R}})\), then

    $$\begin{aligned} \int _{{\widetilde{\mathcal {S}}}} \gamma \cdot {\overline{\chi }}_{\mathcal {R}}\ \mathrm {d}{\overline{\mu }}= \int _{\mathcal {S}}\gamma \cdot {\overline{\chi }}_{\mathcal {R}}\ \mathrm {d}{\overline{\mu }}= \int _{{\mathcal {S}}'} (\gamma \circ \psi )\,\chi '_{{\mathcal {R}}'}\ \mathrm {d}\mu '. \end{aligned}$$

Proof

  1. (a)

    For all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), recall that \({\overline{\chi }}_{\mathcal {R}}:=\mathrm {d}{\overline{\nu }}_{\mathcal {R}}/\mathrm {d}{\overline{\mu }}\), where \({\overline{\nu }}_{\mathcal {R}}:=\psi (\nu '_{{\mathcal {R}}'})\) and \({\mathcal {R}}':=\psi ^\sharp ({\mathcal {R}})\). We must show that (i) \({\overline{\chi }}_{\mathcal {R}}(s)=1\) for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and \(\forall _{\overline{\mu }}\, s\in \mathrm {int}({\mathcal {R}})\); (ii) \({\overline{\chi }}_{\mathcal {R}}(s)=0\) for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and \(\forall _{\overline{\mu }}\, s\in {\mathcal {R}}^\complement \); and (iii) \({\overline{\chi }}_{{\mathcal {R}}_1}(s)+\cdots +{\overline{\chi }}_{{\mathcal {R}}_N}(s) = 1\) for any regular closed partition \(\{{\mathcal {R}}_1,\ldots ,{\mathcal {R}}_N\}\) and \(\forall _{\overline{\mu }}\)\(s\in {\mathcal {S}}\).

    1. (i)

      Let \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\) and let \({\mathcal {R}}':=\psi ^\sharp ({\mathcal {R}})\). Recall that \(\psi ^\sharp ({\mathcal {R}})=\mathrm {clos}(\mathrm {int}[\psi ^{-1}({\mathcal {R}})])\). Thus, \(\mathrm {int}[\psi ^{-1}({\mathcal {R}})]\subseteq {\mathcal {R}}'\). Meanwhile, \(\psi ^{-1}[\mathrm {int}({\mathcal {R}})]\subseteq \mathrm {int}[\psi ^{-1}({\mathcal {R}})]\) because \(\psi \) is continuous. Thus, \(\psi ^{-1}[\mathrm {int}({\mathcal {R}})]\subseteq {\mathcal {R}}'\), and hence, \(\psi ^{-1}[\mathrm {int}({\mathcal {R}})]\subseteq \mathrm {int}({\mathcal {R}}')\), because \(\mathrm {int}({\mathcal {R}}')\) is the largest open subset of \({\mathcal {R}}'\). Thus, for any measurable subset \({\mathcal {U}}\subseteq \mathrm {int}({\mathcal {R}})\), we have

      $$\begin{aligned} \psi ^{-1}({\mathcal {U}})\subseteq \psi ^{-1}[\mathrm {int}({\mathcal {R}})]\subseteq \mathrm {int}({\mathcal {R}}'). \end{aligned}$$
      (D6)

      Thus,

      $$\begin{aligned} {\overline{\nu }}_{\mathcal {R}}[{\mathcal {U}}] \quad {\mathop {=}\limits _{\mathrm {(*)}}}\quad \nu '_{{\mathcal {R}}'}[\psi ^{-1}({\mathcal {U}})] \quad {\mathop {=}\limits _{\mathrm {(\dagger )}}}\quad \mu '[\psi ^{-1}({\mathcal {U}})] \quad {\mathop {=}\limits _{\mathrm {(\diamond )}}}\quad {\overline{\mu }}[{\mathcal {U}}]. \end{aligned}$$
      (D7)

      Here, \((*)\) is because \({\overline{\nu }}_{\mathcal {R}}=\psi (\nu '_{{\mathcal {R}}'})\), while \((\diamond )\) is because \({\overline{\mu }}=\psi (\mu ')\). Finally, \((\dagger )\) is because \(\psi ^{-1}({\mathcal {U}})\subseteq \mathrm {int}({\mathcal {R}}')\) by formula (D6), and \((\mathrm {d}\nu '_{{\mathcal {R}}'}/\mathrm {d}\mu ')(s')=\chi '_{{\mathcal {R}}'}(s')=1\), for \(\forall _{\mu '}\,s'\in \mathrm {int}({\mathcal {R}}')\).

      Equation (D7) holds for all measurable \({\mathcal {U}}\subseteq \mathrm {int}({\mathcal {R}})\). This implies that \((\mathrm {d}{\overline{\nu }}_{\mathcal {R}}/\mathrm {d}{\overline{\mu }})(s)=1\), for \(\forall _{\overline{\mu }}\,s\in \mathrm {int}({\mathcal {R}})\). In other words, \({\overline{\chi }}_{\mathcal {R}}(s)=1\), for \(\forall _{\overline{\mu }}\, s\in \mathrm {int}({\mathcal {R}})\).

    2. (ii)

      If \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), then \({\mathcal {R}}\) is closed, so \({\mathcal {R}}^\complement \) is open, so \(\psi ^{-1}({\mathcal {R}}^\complement )\) is open, because \(\psi \) is continuous. Furthermore, \(\psi ^{-1}({\mathcal {R}}^\complement )\subseteq \mathrm {clos}\left[\psi ^{-1}({\mathcal {R}}^\complement )\right]\), and thus,

      $$\begin{aligned} \psi ^{-1}({\mathcal {R}}^\complement )\subseteq \mathrm {int}\left(\mathrm {clos}\left[\psi ^{-1}({\mathcal {R}}^\complement )\right]\right), \end{aligned}$$
      (D8)

      because \(\mathrm {int}\left(\mathrm {clos}\left[\psi ^{-1}({\mathcal {R}}^\complement )\right]\right)\) is the largest open subset of \(\mathrm {clos}\left[\psi ^{-1}({\mathcal {R}}^\complement )\right]\). Meanwhile,

      $$\begin{aligned} ({\mathcal {R}}')^\complement= & {} \mathrm {clos}\left(\mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]\right)^\complement = \mathrm {int}\left(\mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]^\complement \right)\nonumber \\= & {} \mathrm {int}\left(\mathrm {clos}\left[\psi ^{-1}({\mathcal {R}})^\complement \right]\right) = \mathrm {int}\left(\mathrm {clos}\left[\psi ^{-1}({\mathcal {R}}^\complement )\right]\right). \end{aligned}$$
      (D9)

      Combining (D8) and (D9), we see that \(\psi ^{-1}({\mathcal {R}}^\complement )\subseteq ({\mathcal {R}}')^\complement \). Thus, for any measurable subset \({\mathcal {U}}\subseteq {\mathcal {R}}^\complement \), we have

      $$\begin{aligned} \psi ^{-1}({\mathcal {U}})\subseteq \psi ^{-1}[{\mathcal {R}}^\complement ] \subseteq ({\mathcal {R}}')^\complement . \end{aligned}$$
      (D10)

      Thus,

      $$\begin{aligned} {\overline{\nu }}_{\mathcal {R}}[{\mathcal {U}}] \quad {\mathop {=}\limits _{\mathrm {(*)}}}\quad \nu '_{{\mathcal {R}}'}[\psi ^{-1}({\mathcal {U}})] \quad {\mathop {=}\limits _{\mathrm {(\dagger )}}}\quad 0, \end{aligned}$$
      (D11)

      where \((*)\) is because \({\overline{\nu }}_{\mathcal {R}}=\psi (\nu '_{{\mathcal {R}}'})\), while \((\dagger )\) is because \(\psi ^{-1}({\mathcal {U}})\subseteq ({\mathcal {R}}')^\complement \) by formula (D10), and \((\mathrm {d}\nu '_{{\mathcal {R}}'}/\mathrm {d}\mu ')(s')=\chi '_{{\mathcal {R}}'}(s')=0\), for \(\forall _{\mu '}\,s'\in ({\mathcal {R}}')^\complement \).

      Equation (D11) holds for all measurable \({\mathcal {U}}\subseteq {\mathcal {R}}^\complement \). This implies that \((\mathrm {d}{\overline{\nu }}_{{\mathcal {R}}}/\mathrm {d}{\overline{\mu }})(s)=0\), for \(\forall _{\overline{\mu }}\,s\in {\mathcal {R}}^\complement \). In other words, \({\overline{\chi }}_{\mathcal {R}}(s)=0\) for \(\forall _{\overline{\mu }}\, s\in {\mathcal {R}}^\complement \).

    3. (iii)

      Let \(\{{\mathcal {R}}_1,\ldots ,{\mathcal {R}}_N\}\) be a regular closed partition of \({\mathcal {S}}\). For all \(n\in [1\cdots N]\), let \({\mathcal {R}}'_n:=\psi ^\sharp ({\mathcal {R}}_n)\). Then \(\{{\mathcal {R}}'_1,\ldots ,{\mathcal {R}}'_N\}\) is a regular closed partition of \({\mathcal {S}}'\), because \(\psi \) is regular, and thus, \(\psi ^\sharp :{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {R}}({\mathcal {S}}')\) is a Boolean algebra homomorphism (Fremlin 2004, 314R). Thus,

      $$\begin{aligned} \nu '_{{\mathcal {R}}'_1}+\cdots +\nu '_{{\mathcal {R}}'_N}=\mu ', \end{aligned}$$
      (D12)

      by formula (5.2), because \(\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}\) is a characteristic structure subordinate to \(\mu '\). But \({\overline{\mu }}=\psi (\mu ')\) and \({\overline{\nu }}_{{\mathcal {R}}_n}=\psi (\nu '_{{\mathcal {R}}'_n})\) for all \(n\in [1\cdots N]\), and \(\psi \) acts linearly on measures. Thus, applying \(\psi \) to both sides of Eq. (D12) yields \({\overline{\nu }}_{{\mathcal {R}}_1}+\cdots +{\overline{\nu }}_{{\mathcal {R}}_N}={\overline{\mu }}\). Since these are all nonnegative measures, it follows that \({\overline{\nu }}_{{\mathcal {R}}_1},\ldots ,{\overline{\nu }}_{{\mathcal {R}}_N}\) are absolutely continuous with respect to \({\overline{\mu }}\), and

      $$\begin{aligned} \frac{\ \mathrm {d}{\overline{\nu }}_{{\mathcal {R}}_1}}{\ \mathrm {d}{\overline{\mu }}}+\cdots +\frac{\ \mathrm {d}{\overline{\nu }}_{{\mathcal {R}}_N}}{\ \mathrm {d}{\overline{\mu }}}= 1, \end{aligned}$$
      (D13)

      \({\overline{\mu }}\)-almost everywhere. In other words, \({\overline{\chi }}_{{\mathcal {R}}_1}(s)+\cdots +{\overline{\chi }}_{{\mathcal {R}}_N}(s) = 1\) for \(\forall _{\overline{\mu }}\)\(s\in {\mathcal {S}}\), as desired.

  2. (b)

    Let \(\gamma :{\mathcal {S}}{{\longrightarrow }}{\mathbb {R}}\) be measurable, let \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), and let \({\mathcal {R}}':=\psi ^\sharp ({\mathcal {R}})\). Then

    $$\begin{aligned} \int _{{\widetilde{\mathcal {S}}}} \gamma \cdot {\overline{\chi }}_{\mathcal {R}}\ \mathrm {d}{\overline{\mu }}~&{\mathop {=}\limits _{\mathrm {(*)}}} \ \ \int _{\mathcal {S}}\gamma \cdot {\overline{\chi }}_{\mathcal {R}}\ \mathrm {d}{\overline{\mu }}\ = \ \int _{\mathcal {S}}\gamma \,\frac{\mathrm {d}{\overline{\nu }}_{\mathcal {R}}}{\mathrm {d}{\overline{\mu }}}\ \mathrm {d}{\overline{\mu }}\ = \ \int _{\mathcal {S}}\gamma \ \mathrm {d}{\overline{\nu }}_{\mathcal {R}}\\&{\mathop {=}\limits _{\mathrm {(\dagger )}}} \int _{{\mathcal {S}}'} (\gamma \circ \psi ) { \;\; \mathrm{d}\nu }'_{{\mathcal {R}}'} \ = \ \int _{{\mathcal {S}}'} (\gamma \circ \psi )\,\frac{\mathrm {d}\nu '_{{\mathcal {R}}'}}{\mathrm {d}\mu '}\ \mathrm {d}\mu ' \ = \ \int _{{\mathcal {S}}'} (\gamma \circ \psi )\,\chi '_{{\mathcal {R}}'}\ \mathrm {d}\mu ', \end{aligned}$$

    as claimed. Here, \((*)\) is because \({\overline{\mu }}[{\mathcal {S}}\setminus {\widetilde{\mathcal {S}}}]=0\), and \((\dagger )\) is by the change of variables formula, because \({\overline{\nu }}_{\mathcal {R}}=\psi (\nu '_{{\mathcal {R}}'})\).

\(\square \)

Proof of Proposition 5.2

Let \({\widetilde{\mathcal {S}}}:=\psi ({\mathcal {S}}')\); this is a regular closed subset of \({\mathcal {S}}\), by Lemma D1. Regard \({\widetilde{\mathcal {S}}}\) itself as a state space, and consider the conditional preference structure \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})}}\). We will construct two SEU representations for \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})}}\).

Let \({\overline{\mu }}:=\psi (\mu ')\) and let \(\{{\overline{\chi }}_{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}\) be the \(\psi \)-image of \(\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}\). Let \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\), and let \({\mathcal {R}}':=\psi ^\sharp ({\mathcal {R}})\) (so \({\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')\)). For any \(\alpha ,\beta \in {\mathcal {A}}\),

$$\begin{aligned} \left( \alpha _{\upharpoonleft {\mathcal {R}}}\succeq _{\mathcal {R}}\beta _{\upharpoonleft {\mathcal {R}}}\right)&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \alpha \circ \psi _{\upharpoonleft {\mathcal {R}}'} \succeq '_{{\mathcal {R}}'} \beta \circ \psi _{\upharpoonleft {\mathcal {R}}'}\right) \nonumber \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \displaystyle \int _{{\mathcal {S}}'} (u'\circ \alpha \circ \psi ) \, \chi '_{{\mathcal {R}}'} \ \mathrm {d}\mu ' \ \ge \ \int _{{\mathcal {S}}'} (u'\circ \beta \circ \psi ) \, \chi '_{{\mathcal {R}}} \ \mathrm {d}\mu '\right) \nonumber \\&{\mathop {\Longleftrightarrow }\limits _{(\diamond )}} \left( \displaystyle \int _{{\widetilde{\mathcal {S}}}} (u'\circ \alpha ) \, {\overline{\chi }}_{{\mathcal {R}}} \ \mathrm {d}{\overline{\mu }}\ \ge \ \int _{{\widetilde{\mathcal {S}}}} (u'\circ \beta ) \, {\overline{\chi }}_{{\mathcal {R}}} \ \mathrm {d}{\overline{\mu }}\right) . \end{aligned}$$
(D14)

Here, \((*)\) is by statement (5.1), \((\dagger )\) is by statement (2.3), and \((\diamond )\) is by Lemma D3(b).

Meanwhile, let \(({\widetilde{\mu }},\{{\widetilde{\chi }}_{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})})\) be the Bayesian update of \((\mu ,\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})})\), conditional on \({\widetilde{\mathcal {S}}}\). Then for any \(\alpha ,\beta \in {\mathcal {A}}\) and \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\),

$$\begin{aligned} \left( \alpha _{\upharpoonleft {\mathcal {R}}}\succeq _{\mathcal {R}}\beta _{\upharpoonleft {\mathcal {R}}}\right)&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \displaystyle \int _{{\mathcal {S}}} (u\circ \alpha ) \, \chi _{{\mathcal {R}}} \ \mathrm {d}\mu \ \ge \ \int _{{\mathcal {S}}} (u\circ \beta ) \, \chi _{{\mathcal {R}}} \ \mathrm {d}\mu \right) \nonumber \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \displaystyle \int _{{\widetilde{\mathcal {S}}}} (u\circ \alpha ) \, {\widetilde{\chi }}_{{\mathcal {R}}} \ \mathrm {d}{\widetilde{\mu }}\ \ge \ \int _{{\widetilde{\mathcal {S}}}} (u\circ \beta ) \, {\widetilde{\chi }}_{{\mathcal {R}}} \ \mathrm {d}{\widetilde{\mu }}\right) .\qquad \end{aligned}$$
(D15)

Here \((*)\) is by statement (2.3), and \((\dagger )\) is by Lemma D2(b).

Statements (D14) and (D15) hold for all \(\alpha ,\beta \in {\mathcal {A}}\) and \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\). Thus, \((u',\mu ',\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')})\) and \((u,\mu ,\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})})\) are both characteristic SEU representations for \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})}}\). By the uniqueness of such representations (Proposition A1), we deduce that \(u'\) is a positive affine transform of u, and obtain statement (5.4). Hence, \(\psi \) projects \((\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')},\mu ')\) into \((\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})},\mu )\). \(\square \)

Let \({\mathcal {S}}\) be a topological space. For any \(\alpha \in {\mathcal {C}}_b({\mathcal {S}})\), define \({\mathcal {Z}}_\alpha :=\{s\in {\mathcal {S}}\); \(\alpha (s)=0\}\). More generally, for any subset \({\mathcal {G}}\subseteq {\mathcal {C}}({\mathcal {S}})\), let \({\mathcal {Z}}_{\mathcal {G}}:=\bigcap _{\gamma \in {\mathcal {G}}}{\mathcal {Z}}_\gamma =\{s\in {\mathcal {S}}\); \(\gamma (s)=0\) for all \(\gamma \in {\mathcal {G}}\}\); this is a closed subset of \({\mathcal {S}}\).

Lemma D4

Let \({\mathcal {S}}\) be a set, let \({\mathcal {A}}\subseteq {\mathbb {R}}^{\mathcal {S}}\), and suppose \({\mathcal {A}}\) is Tychonoff for the topology it induces on \({\mathcal {S}}\). For any closed subset \({\mathcal {R}}\subseteq {\mathcal {S}}\), there exists \({\mathcal {G}}\subseteq {\mathcal {A}}\) with \({\mathcal {R}}={\mathcal {Z}}_{\mathcal {G}}\).

Proof

For all \(s\in {\mathcal {S}}\setminus {\mathcal {R}}\), there is some \(\alpha _s\in {\mathcal {A}}\) such that \(\alpha _s(s)\ne 0\) while \(\alpha _s({\mathcal {R}})=0\), because \({\mathcal {A}}\) is Tychonoff. Let \({\mathcal {G}}:=\{\alpha _s\); \(s\in {\mathcal {S}}\setminus {\mathcal {R}}\}\). Then \({\mathcal {Z}}_{\mathcal {G}}={\mathcal {R}}\). \(\square \)

Lemma D5

Let \({\mathcal {S}}\) be a compact Hausdorff space. Let \({\mathcal {A}}\) be a uniformly dense Riesz subspace of \({\mathcal {C}}({\mathcal {S}})\) containing \(\varvec{1}\). Then \({\mathcal {A}}\) is Tychonoff for the topology of \({\mathcal {S}}\).

Proof

Let \({\mathcal {Q}},{\mathcal {R}}\subset {\mathcal {S}}\) be disjoint closed sets. Since \({\mathcal {S}}\) is compact Hausdorff, it is normal (Willard 2004, Theorem 17.10). Thus, Urysohn’s Lemma yields \(\gamma \in {\mathcal {C}}({\mathcal {S}})\) with \(\gamma ({\mathcal {Q}})=-1\) and \(\gamma ({\mathcal {R}})=2\). Since \({\mathcal {A}}\) is uniformly dense in \({\mathcal {C}}({\mathcal {S}})\), there exists \(\alpha \in {\mathcal {A}}\) with \(\Vert \alpha -\gamma \Vert _{\infty }<1\). Thus, \(\alpha (q)<0\) for all \(q\in {\mathcal {Q}}\), and \(\alpha (r)>1\) for all \(r\in {\mathcal {R}}\). Let \(\beta :=0\vee \alpha \wedge \varvec{1}\). Then \(\beta \in {\mathcal {A}}\) (because \({\mathcal {A}}\) is a Riesz subspace containing \(\varvec{1}\)), and \(\beta (q)=0\) for all \(q\in {\mathcal {Q}}\), and \(\beta (r)=1\) for all \(r\in {\mathcal {R}}\). We can perform this construction for any disjoint closed \({\mathcal {Q}},{\mathcal {R}}\subseteq {\mathcal {S}}\)—in particular, when \({\mathcal {R}}=\{s\}\) for some \(s\in {\mathcal {R}}^\complement \). Thus, \({\mathcal {A}}\) is Tychonoff for \({\mathcal {S}}\). \(\square \)

Let \({\mathcal {S}}\) be a compact Hausdorff space; then, \({\mathcal {C}}({\mathcal {S}})\) is both an Archimedean Riesz space and a realistic Banach algebra, and for any \(\alpha ,\gamma \in {\mathcal {C}}({\mathcal {S}})\), we have \(|\alpha |\wedge |\beta |=0\) if and only if \(\alpha \cdot \gamma =0\). Thus, for any subset \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\), and any \({\mathcal {G}}\subseteq {\mathcal {A}}\), we have \(\{\alpha \in {\mathcal {A}}\); \(|\alpha |\wedge |\gamma |=0\) for all \(\gamma \in {\mathcal {G}}\}= \{\alpha \in {\mathcal {A}}\); \(\alpha \cdot \gamma =0\) for all \(\gamma \in {\mathcal {G}}\}\). In other words, \({\mathcal {G}}^\perp ={\mathcal {G}}^{\perp \!\perp }\) in this case, so we can use the notation “\({\mathcal {G}}^\perp \)” to refer to both. Thus, the next result can be applied to both Riesz subspaces and subalgebras of \({\mathcal {C}}({\mathcal {S}})\).

Lemma D6

Let \({\mathcal {S}}\) be a compact Hausdorff space, let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\) a Riesz subspace or subalgebra of \({\mathcal {C}}({\mathcal {S}})\), and let \({\mathcal {G}}\subseteq {\mathcal {A}}\).

  1. (a)

    If \(\mathrm {int}({\mathcal {Z}}_{\mathcal {G}})=\emptyset \), then \({\mathcal {G}}^\perp =\{0\}\).

  2. (b)

    Suppose \({\mathcal {A}}\) is Tychonoff for \({\mathcal {S}}\). Then \({\mathcal {G}}^\perp =\{0\}\) if and only if \(\mathrm {int}({\mathcal {Z}}_{\mathcal {G}})=\emptyset \).

Proof

Let \(\alpha \in {\mathcal {C}}({\mathcal {S}})\); then, \(\alpha \in {\mathcal {G}}^\perp \) if and only if \(\mathrm {supp}(\alpha )\subseteq {\mathcal {Z}}_{\mathcal {G}}\).

  1. (a)

    Suppose \(\mathrm {int}({\mathcal {Z}}_{\mathcal {G}})=\emptyset \). If \(\alpha \in {\mathcal {G}}^\perp \), then \(\mathrm {supp}(\alpha )\subseteq {\mathcal {Z}}_{\mathcal {G}}\). But \(\mathrm {supp}(\alpha )\) is an open set (because \(\alpha \) is continuous); hence, \(\mathrm {supp}(\alpha )=\emptyset \), so \(\alpha =0\). Thus, \({\mathcal {G}}^\perp =\{0\}\).

  2. (b)

    The direction “\({\Longleftarrow }\)” is part (a). I prove “\(\Longrightarrow \)” by contrapositive. Suppose \(\mathrm {int}({\mathcal {Z}}_{\mathcal {G}})\ne \emptyset \). Let \({\mathcal {F}}:=[\mathrm {int}({\mathcal {Z}}_{\mathcal {G}})]^\complement \); then, \({\mathcal {F}}\subsetneq {\mathcal {S}}\) is a closed set. Let \(s\in \mathrm {int}({\mathcal {Z}}_{\mathcal {G}})\). There exists \(\alpha \in {\mathcal {A}}\) such that \(\alpha (s)=1\) while \(\alpha ({\mathcal {F}})=0\) (because \({\mathcal {A}}\) is Tychonoff for \({\mathcal {S}}\)). Thus, \(\mathrm {supp}(\alpha )\subseteq {\mathcal {Z}}_{\mathcal {G}}\), so \(\alpha \in {\mathcal {G}}^\perp \). But \(\alpha \ne 0\). Thus, \({\mathcal {G}}^\perp \ne \{0\}\). \(\square \)

For any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), recall that \({\mathfrak {K}}({\mathcal {R}})=\{\alpha \in {\mathcal {C}}({\mathcal {S}})\); \(\alpha (r)=0\) for all \(r\in {\mathcal {R}}\}\). Let \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\) be the set of bands of \({\mathcal {C}}({\mathcal {S}})\) as an Archimedean Riesz space. This is a Boolean algebra under the operations defined in formula (3.1) prior to Lemma 3.5. Let \({\mathfrak {I}}^*[{\mathcal {C}}({\mathcal {S}})]\) be the set of regular ideals of \({\mathcal {C}}({\mathcal {S}})\) as a realistic Banach algebra. This is a Boolean algebra under the operations defined in formula (4.1) prior to Lemma 4.5. As explained in Example 3.4, \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]=\{{\mathfrak {K}}({\mathcal {R}})\); \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\}\). As explained in Example 4.4, \({\mathfrak {I}}^*[{\mathcal {C}}({\mathcal {S}})]=\{{\mathfrak {K}}({\mathcal {R}})\); \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\}\). Combining these two observations, we see that \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]={\mathfrak {I}}^*[{\mathcal {C}}({\mathcal {S}})]\), so I will refer to both by the notation \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\). Furthermore, by the previous remarks concerning the equivalence of the \({}^\perp \) and \({}^{\perp \!\perp }\) operations on \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\), we see that the operators defined in formulae (3.1) and (4.1) are actually the same, so the Boolean algebra structures described in Lemmas 3.5 and 4.5 are the same; thus, I can simply refer to “the Boolean algebra \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\)”, without specifying whether I regard \({\mathcal {C}}({\mathcal {S}})\) as a Riesz space or a Banach algebra.

Let \({\mathcal {S}}\) and \({\mathcal {S}}'\) be compact Hausdorff spaces; let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be a continuous function. Define the function \(\varPsi :{\mathcal {C}}({\mathcal {S}}){{\longrightarrow }}{\mathcal {C}}({\mathcal {S}}')\) by \(\varPsi (\alpha ):=\alpha \circ \psi \) for all \(\alpha \in {\mathcal {C}}({\mathcal {S}})\). Then \(\varPsi \) is both a Riesz homomorphism and an algebra homomorphism from \({\mathcal {C}}({\mathcal {S}})\) to \({\mathcal {C}}({\mathcal {S}}')\). Since \({\mathcal {G}}^\perp ={\mathcal {G}}^{\perp \!\perp }\) for all \({\mathcal {G}}\subseteq {\mathcal {C}}({\mathcal {S}})\), it follows that \(\varPsi \) is regular as a Riesz homomorphism if and only if \(\varPsi \) is regular as an algebra homomorphism. Thus, the next result applies to both Riesz subspaces and subalgebras of \({\mathcal {C}}({\mathcal {S}})\).

Proposition D7

Let \({\mathcal {S}}\) and \({\mathcal {S}}'\) be compact Hausdorff spaces. Let \({\mathcal {A}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {A}}'\subseteq {\mathcal {C}}({\mathcal {S}}')\) be either uniformly dense Riesz subspaces of \({\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {C}}({\mathcal {S}}')\) containing \(\varvec{1}\), or regular, inversion-closed, topology-generating subalgebras of \({\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {C}}({\mathcal {S}}')\). Let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be continuous, and define \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}'\) by \(\varPsi (\alpha ):=\alpha \circ \psi \) for all \(\alpha \in {\mathcal {A}}\).

  1. (a)

    \(\varPsi \) is regular (as either a Riesz homomorphism or an algebra homomorphism) if and only if \(\psi \) is regular.

  2. (b)

    For any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), define \(\psi ^\sharp ({\mathcal {R}}):=\mathrm {clos}\left(\mathrm {int}[f^{-1}({\mathcal {R}})]\right)\). If \(\psi \) is regular, then \(\psi ^\sharp \) is a Boolean algebra homomorphism from \({\mathfrak {R}}({\mathcal {S}})\) to \({\mathfrak {R}}({\mathcal {S}}')\).

  3. (c)

    For any \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\), define \(\varPsi ^\#({\mathcal {I}}):=\left(\varPsi ({\mathcal {I}})^\perp \right)^\perp \). If \(\varPsi \) is regular, then \(\varPsi ^\#\) is a Boolean algebra homomorphism from \({\mathfrak {I}}({\mathcal {A}})\) to \({\mathfrak {I}}({\mathcal {A}}')\), and this diagram commutes:

    figurea

Proof

First note that \({\mathcal {A}}\) and \({\mathcal {A}}'\) are Tychonoff for the topologies of \({\mathcal {S}}\) and \({\mathcal {S}}'\), respectively. (If \({\mathcal {A}}\) and \({\mathcal {A}}'\) are uniformly dense Riesz subspaces of \({\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {C}}({\mathcal {S}}')\), then this follows from Lemma D5. If \({\mathcal {A}}\) and \({\mathcal {A}}'\) are regular, inversion-closed, topology-generating subalgebras of \({\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {C}}({\mathcal {S}}')\), then it follows from Lemma C2.)

  1. (a)

    \({\Longleftarrow }\)” Let \({\mathcal {G}}\subseteq {\mathcal {A}}\), and suppose \({\mathcal {G}}^\perp =\{0\}\). Thus, \(\mathrm {int}({\mathcal {Z}}_{\mathcal {G}})=\emptyset \), by Lemma D6(b). Since \({\mathcal {Z}}_{\mathcal {G}}\) is closed, this makes it nowhere dense. Thus, \(\psi ^{-1}({\mathcal {Z}}_{\mathcal {G}})\) is nowhere dense, because \(\psi \) is regular. Meanwhile, \(\psi ^{-1}({\mathcal {Z}}_{\mathcal {G}})\) is closed because \({\mathcal {Z}}_{\mathcal {G}}\) is closed and \(\psi \) is continuous. Thus, \(\mathrm {int}\left[\psi ^{-1}({\mathcal {Z}}_{\mathcal {G}})\right]=\emptyset \). But

    $$\begin{aligned} \psi ^{-1}({\mathcal {Z}}_{\mathcal {G}})= & {} \psi ^{-1}\left(\bigcap _{\gamma \in {\mathcal {G}}}{\mathcal {Z}}_\gamma \right) =\bigcap _{\gamma \in {\mathcal {G}}}\psi ^{-1}\left({\mathcal {Z}}_\gamma \right) =\bigcap _{\gamma \in {\mathcal {G}}}{\mathcal {Z}}_{\gamma \circ \psi }\nonumber \\= & {} \bigcap _{\gamma \in {\mathcal {G}}} {\mathcal {Z}}_{\varPsi (\gamma )} =\bigcap _{\gamma '\in \varPsi ({\mathcal {G}})} {\mathcal {Z}}_{\gamma '} = {\mathcal {Z}}_{\varPsi ({\mathcal {G}})}. \end{aligned}$$
    (D17)

    Thus, \(\mathrm {int}({\mathcal {Z}}_{\varPsi ({\mathcal {G}})})=\emptyset \). Thus, \(\varPsi ({\mathcal {G}})^\perp =\{0\}\), by Lemma D6(a).

    \(\Longrightarrow \)” Let \({\mathcal {N}}\subseteq {\mathcal {S}}\) be nowhere dense. Thus, if \({\overline{\mathcal { N}}}:=\mathrm {clos}({\mathcal {N}})\), then \(\mathrm {int}({\overline{\mathcal { N}}})=\emptyset \). Lemma D4 yields some \({\mathcal {G}}\subseteq {\mathcal {A}}\) such that \({\mathcal {Z}}_{\mathcal {G}}={\overline{\mathcal { N}}}\) (because \({\mathcal {A}}\) is Tychonoff for \({\mathcal {S}}\)). Then Lemma D6(a) says that \({\mathcal {G}}^\perp =\{0\}\). Thus, \(\varPsi ({\mathcal {G}})^\perp =\{0\}\), because \(\varPsi \) is regular. Thus, Lemma D6(b) says that \(\mathrm {int}({\mathcal {Z}}_{\varPsi ({\mathcal {G}})})=\emptyset \) (because \({\mathcal {A}}'\) is Tychonoff for \({\mathcal {S}}'\)). But by Eq. (D17), \({\mathcal {Z}}_{\varPsi ({\mathcal {G}})}=\psi ^{-1}({\mathcal {Z}}_{\mathcal {G}})=\psi ^{-1}({\overline{\mathcal { N}}})\). Thus, we deduce that \(\mathrm {int}[\psi ^{-1}({\overline{\mathcal { N}}})]=\emptyset \). Now \(\psi ^{-1}({\mathcal {N}})\subseteq \psi ^{-1}({\overline{\mathcal { N}}})\) and \(\psi ^{-1}({\overline{\mathcal { N}}})\) is closed (because \(\psi \) is continuous); thus, \(\mathrm {clos}[\psi ^{-1}({\mathcal {N}})]\subseteq \psi ^{-1}({\overline{\mathcal { N}}})\). Thus, \(\mathrm {int}\left( \mathrm {clos}[\psi ^{-1}({\mathcal {N}})]\right)=\emptyset \) also. Thus, \(\psi ^{-1}({\mathcal {N}})\) is nowhere dense, as desired.

  2. (b)

    For any \({\mathcal {R}}\subseteq {\mathcal {S}}\), the set \(\psi ^\sharp ({\mathcal {R}}):=\mathrm {clos}\left(\mathrm {int}[\psi ^{-1}({\mathcal {R}})]\right)\) is in \({\mathfrak {R}}({\mathcal {S}}')\) because it is the closure of an open set. Thus, we obtain a function \(\psi ^\sharp :{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {R}}({\mathcal {S}}')\). If \(\psi \) is regular, then \(\psi ^\sharp \) is a Boolean algebra homomorphism from \({\mathfrak {R}}({\mathcal {S}})\) to \({\mathfrak {R}}({\mathcal {S}}')\) (Fremlin 2004, 314R(a)).

  3. (c)

    First note that \(\varPsi ^\#({\mathcal {I}})\in {\mathfrak {I}}({\mathcal {A}}')\), for any \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\), because \(\varPsi ^\#({\mathcal {I}})={\mathcal {G}}^\perp \) for some \({\mathcal {G}}\subseteq {\mathcal {A}}'\)—namely, \({\mathcal {G}}:=\varPsi ({\mathcal {I}})^\perp \). (If \({\mathcal {A}}'\) is a Riesz subspace of \({\mathcal {C}}({\mathcal {S}}')\), then \({\mathcal {G}}^\perp \in {\mathfrak {I}}({\mathcal {A}}')\) by Fremlin (2004, Fact 352O(c)). If \({\mathcal {A}}'\) is a subalgebra of \({\mathcal {C}}({\mathcal {S}}')\), then \({\mathcal {G}}^\perp \in {\mathfrak {I}}({\mathcal {A}}')\) by the definition of “regular ideal”.) Thus, we obtain a function \(\varPsi ^\#:{\mathfrak {I}}({\mathcal {A}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\). I will show that \(\varPsi ^\#\) satisfies the commuting diagram (D16); from this, I will deduce that it is a Boolean algebra homomorphism.

Claim 1

\({\mathfrak {K}}:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) and \({\mathfrak {K}}':{\mathfrak {R}}({\mathcal {S}}'){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\) are Boolean algebra anti-isomorphisms.

Proof

If \({\mathcal {A}}\) and \({\mathcal {A}}'\) are regular, inversion-closed, topology-generating subalgebras of \({\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {C}}({\mathcal {S}}')\), then the claim follows immediately from Lemma C3. On the other hand, suppose \({\mathcal {A}}\) and \({\mathcal {A}}'\) are uniformly dense (hence, order-dense) Riesz subspaces of \({\mathcal {C}}({\mathcal {S}})\) and \({\mathcal {C}}({\mathcal {S}}')\). Recall from Example 3.6 that the functions \({\mathfrak {K}}_0:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\) and \({\mathfrak {K}}'_0:{\mathfrak {R}}({\mathcal {S}}'){{\longrightarrow }}{\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}}')]\) are Boolean algebra anti-isomorphisms. If \({\mathcal {A}}\) and \({\mathcal {A}}'\) are order-dense, then the functions \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}})]\ni {\mathcal {I}}{{\longrightarrow }}{\mathcal {I}}\cap {\mathcal {A}}\in {\mathfrak {I}}({\mathcal {A}})\) and \({\mathfrak {I}}[{\mathcal {C}}({\mathcal {S}}')]\ni {\mathcal {I}}'{{\longrightarrow }}{\mathcal {I}}'\cap {\mathcal {A}}'\in {\mathfrak {I}}({\mathcal {A}}')\) are Boolean algebra isomorphisms (Fremlin 2004, Proposition 353D). Thus, if we define \({\mathfrak {K}}({\mathcal {R}}):={\mathfrak {K}}_0({\mathcal {R}})\cap {\mathcal {A}}\) for any \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), and \({\mathfrak {K}}'({\mathcal {R}}'):={\mathfrak {K}}'_0({\mathcal {R}}')\cap {\mathcal {A}}'\) for any \({\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')\), then we get Boolean algebra anti-isomorphisms \({\mathfrak {K}}:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) and \({\mathfrak {K}}':{\mathfrak {R}}({\mathcal {S}}'){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\). \(\diamond \)

Claim 1 implies that

$$\begin{aligned} {\mathfrak {K}}'(\lnot {\mathcal {R}}')={\mathfrak {K}}'({\mathcal {R}}')^\perp ,\quad \text{ for } \text{ all }\quad {\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}'). \end{aligned}$$
(D18)

Now let \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\), and let \({\mathcal {I}}:={\mathfrak {K}}({\mathcal {R}})\); then, \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\). Since \({\mathcal {A}}\) is Tychonoff for the topology of \({\mathcal {S}}\), Lemma C1 says that

$$\begin{aligned} \bigcup _{\alpha \in {\mathcal {I}}} \mathrm {supp}(\alpha ) = {\mathcal {R}}^\complement . \end{aligned}$$
(D19)

For any \(\alpha \in {\mathcal {I}}\), we have

$$\begin{aligned} \mathrm {supp}[\varPsi (\alpha )] = \mathrm {supp}[\alpha \circ \psi ] = \psi ^{-1}[\mathrm {supp}(\alpha )]. \end{aligned}$$
(D20)

Thus,

$$\begin{aligned} \bigcup _{\alpha '\in \varPsi ({\mathcal {I}})} \mathrm {supp}(\alpha ')~&{\mathop {=}\limits _{\mathrm {(*)}}} \bigcup _{\alpha \in {\mathcal {I}}} \psi ^{-1}[\mathrm {supp}(\alpha )]\nonumber \\&= \psi ^{-1}\left[ \bigcup _{\alpha \in {\mathcal {I}}} \mathrm {supp}(\alpha )\right] \ \ {\mathop {=}\limits _{\mathrm {(\dagger )}}} \ \ \psi ^{-1}({\mathcal {R}}^\complement ), \end{aligned}$$
(D21)

where \((*)\) is by (D20) and \((\dagger )\) is by (D19). Thus,

$$\begin{aligned} \left( \bigcup _{\alpha '\in \varPsi ({\mathcal {I}})} \mathrm {supp}(\alpha ')\right)^\complement \quad {\mathop {=}\limits _{\mathrm {(*)}}}\quad \psi ^{-1}({\mathcal {R}}^\complement )^\complement = \psi ^{-1}({\mathcal {R}}), \end{aligned}$$
(D22)

where \((*)\) is by (D21). Thus, for any \(\gamma '\in {\mathcal {A}}'\),

$$\begin{aligned} \left( \gamma '\in \varPsi ({\mathcal {I}})^\perp \right)&\iff \left( \displaystyle \mathrm {supp}(\gamma ')\subseteq \left( \bigcup _{\alpha '\in \varPsi ({\mathcal {I}})} \mathrm {supp}(\alpha ')\right)^\complement \right) \nonumber \\&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( \mathrm {supp}(\gamma ')\subseteq \psi ^{-1}({\mathcal {R}})\right) \nonumber \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}}\left( \mathrm {supp}(\gamma ')\subseteq \mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]\right) . \end{aligned}$$
(D23)

where \((*)\) is by (D22), and \((\dagger )\) is because \(\mathrm {supp}(\gamma )\) is open because \(\gamma \) is continuous. Furthermore, \({\mathcal {A}}'\) is Tychonoff for \({\mathcal {S}}'\), so (D23) implies that

$$\begin{aligned} \bigcup _{\gamma '\in \varPsi ({\mathcal {I}})^\perp } \mathrm {supp}(\gamma ') = \mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]. \end{aligned}$$
(D24)

Thus, for any \(\gamma '\in {\mathcal {A}}'\),

$$\begin{aligned} \left( \gamma '\in \varPsi ^\#({\mathcal {I}})\right)&\iff \left( \gamma '\in (\varPsi ({\mathcal {I}})^\perp )^\perp \right) \ {\mathop {\Longleftrightarrow }\limits _{(*)}} \ \left( \mathrm {supp}(\gamma ')\subseteq \left(\mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]\right)^\complement \right) \nonumber \\&{\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( \mathrm {supp}(\gamma ')\subseteq \mathrm {int}\left[\left(\mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]\right)^\complement \right]\right) \end{aligned}$$
(D25)

where \((*)\) is by (D24) and \((\dagger )\) is again because \(\mathrm {supp}(\gamma ')\) is open. But

$$\begin{aligned} \mathrm {int}\left[\left(\mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]\right)^\complement \right] \ \ = \ \ \left[\mathrm {clos}\left(\mathrm {int}\left[\psi ^{-1}({\mathcal {R}})\right]\right)\right]^\complement \ \ = \ \ \psi ^\sharp ({\mathcal {R}})^\complement . \end{aligned}$$
(D26)

Thus, for any \(\gamma '\in {\mathcal {A}}'\),

$$\begin{aligned} \left( \gamma '\in \varPsi ^\#({\mathcal {I}})\right) \ \ {\mathop {\Longleftrightarrow }\limits _{(*)}} \ \ \left( \mathrm {supp}(\gamma ')\subseteq \psi ^\sharp ({\mathcal {R}})^\complement \right) \iff \left( \gamma '\in {\mathfrak {K}}'\left[\psi ^\sharp ({\mathcal {R}})\right]\right) , \end{aligned}$$

where \((*)\) is by substituting Eq. (D26) into statement (D25). It follows that

$$\begin{aligned} \varPsi ^\#({\mathcal {I}})= {\mathfrak {K}}'\left[\psi ^\sharp ({\mathcal {R}})\right]. \end{aligned}$$
(D27)

Recall that \({\mathcal {I}}={\mathfrak {K}}({\mathcal {R}})\). Thus, Eq. (D27) becomes

$$\begin{aligned} {\mathfrak {K}}'\left[\psi ^\sharp ({\mathcal {R}})\right] = \varPsi ^\#\left[{\mathfrak {K}}({\mathcal {R}})\right]. \end{aligned}$$
(D28)

Equation (D28) holds for all \({\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})\). Thus, the diagram (D16) commutes. Since \({\mathfrak {K}}\) is an anti-isomorphism, it is invertible, and \({\mathfrak {K}}^{-1}\) is also an anti-isomorphism. Thus, \(\varPsi ^\#:= {\mathfrak {K}}'\circ \psi ^\sharp \circ {\mathfrak {K}}^{-1}\) is a composition of two Boolean algebra anti-isomorphisms and a homomorphism, so it is also a homomorphism. \(\square \)

Proposition D8

Let \({\mathcal {A}}\) and \({\mathcal {A}}'\) be realistic Banach algebras, with Gelfand representations \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and \(\varPhi ':{\mathcal {A}}'{{\longrightarrow }}{\widehat{\mathcal {A}}}'\subseteq {\mathcal {C}}({\mathcal {S}}')\) for some compact Hausdorff spaces \({\mathcal {S}}\) and \({\mathcal {S}}'\) and uniformly dense subalgebras \({\widehat{\mathcal {A}}}\) and \({\widehat{\mathcal {A}}}'\), along with Boolean algebra anti-homomorphisms \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) and \({\varPhi '}^*:{\mathfrak {R}}({\mathcal {S}}'){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\) as in Proposition C4(a, c).

Let \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}'\) be a regular Banach algebra homomorphism, and let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be the associated continuous function from Proposition C4(d). Then \(\psi \) is regular. For all \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\), define \(\varPsi ^\#({\mathcal {I}}):=\varPsi ({\mathcal {I}})^{{\perp \!\perp }{\perp \!\perp }}\). Then \(\varPsi ^\#:{\mathfrak {I}}({\mathcal {A}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\) is a Boolean algebra homomorphism, and we have a commuting diagram:

figureb

Proof

Let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be the continuous function from Proposition C4(d).

Claim 1

\(\psi \) is regular.

Proof

Define \({\widehat{\varPsi }}:{\widehat{\mathcal {A}}}{{\longrightarrow }}{\widehat{\mathcal {A}}}'\) by setting \({\widehat{\varPsi }}({\widehat{\alpha }}):={\widehat{\alpha }}\circ \psi \) for all \({\widehat{\alpha }}\in {\widehat{\mathcal {A}}}\). Then Proposition C4(d) says that the following diagram commutes:

figurec

Now, \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\) and \(\varPhi ':{\mathcal {A}}'{{\longrightarrow }}{\widehat{\mathcal {A}}}'\) are algebra isomorphisms (by Proposition C4(a)). As \({\mathcal {A}}\) and \({\mathcal {A}}'\) are regular algebras, hence \({\widehat{\mathcal {A}}}\) and \({\widehat{\mathcal {A}}}'\) are regular (because regularity is a purely algebraic property, hence preserved by isomorphisms). Furthermore, since \(\varPhi \) is bijective, the inverse function \(\varPhi ^{-1}:{\widehat{\mathcal {A}}}{{\longrightarrow }}{\mathcal {A}}\) is well defined and is also an algebra isomorphism; thus, diagram (D30) says that \({\widehat{\varPsi }}:=\varPhi '\circ \varPsi \circ \varPhi ^{-1}\). Now \(\varPsi \) is a regular homomorphism (by hypothesis), and composition by algebra isomorphisms preserves regularity of morphisms (since it is a purely algebraic property); thus, \({\widehat{\varPsi }}\) is also a regular homomorphism. Thus, Proposition D7(a) says that \(\psi \) is regular, because \({\widehat{\varPsi }}\) is regular, and the subalgebras \({\widehat{\mathcal {A}}}\) and \({\widehat{\mathcal {A}}}'\) are inversion-closed and topology generating (by Proposition C4(a)) and regular (as already noted above). \(\diamond \)

For any \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\), define \(\varPhi _*({\mathcal {I}}):=\varPhi ({\mathcal {I}})\); then, \(\varPhi _*:{\mathfrak {I}}({\mathcal {A}}){{\longrightarrow }}{\mathfrak {I}}({\widehat{\mathcal {A}}})\) is a Boolean algebra isomorphism, because \(\varPhi \) is an algebra isomorphism. For any \({\widehat{\mathcal {I}}}'\in {\mathfrak {I}}({\widehat{\mathcal {A}}}')\), define \((\varPhi ')^{-1}_*({\widehat{\mathcal {I}}}'):=\varPhi ^{-1}({\widehat{\mathcal {I}}}')\); then, \((\varPhi ')^{-1}_*:{\mathfrak {I}}({\widehat{\mathcal {A}}}'){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\) is a Boolean algebra isomorphism, because \((\varPhi ')^{-1}\) is an algebra isomorphism.

Finally, for any \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\), define \(\varPsi ^\#({\mathcal {I}}):=\varPsi ({\mathcal {I}})^{{\perp \!\perp }{\perp \!\perp }}\). Then

$$\begin{aligned} \varPsi ^\#({\mathcal {I}})~&{\mathop {=}\limits _{\mathrm {(*)}}}~(\varPhi ')^{-1} \left( {\widehat{\varPsi }}\left[ \varPhi ({\mathcal {I}})\right]\right)^{{\perp \!\perp }{\perp \!\perp }} \quad {\mathop {=}\limits _{\mathrm {(\dagger )}}}\quad (\varPhi ')^{-1} \left( {\widehat{\varPsi }}\left[ \varPhi ({\mathcal {I}})\right]^{{\perp \!\perp }{\perp \!\perp }} \right)\nonumber \\&{\mathop {=}\limits _{\mathrm {(\diamond )}}}~(\varPhi ')^{-1} \left( {\widehat{\varPsi }}^\# \left[ \varPhi ({\mathcal {I}})\right] \right) \quad {\mathop {=}\limits _{\mathrm {(\ddagger )}}}\quad (\varPhi ')^{-1}_*\circ {\widehat{\varPsi }}^\#\circ \varPhi _*({\mathcal {I}}). \end{aligned}$$
(D31)

\((*)\) is because \(\varPsi =(\varPhi ')^{-1} \circ {\widehat{\varPsi }}\circ \varPhi \) by diagram (D30). Next, \((\dagger )\) is because \((\varPhi ')^{-1}({\widehat{\mathcal {I}}}^{\prime {\perp \!\perp }{\perp \!\perp }})=(\varPhi ')^{-1}({\widehat{\mathcal {I}}}')^{{\perp \!\perp }{\perp \!\perp }}\) for any \({\widehat{\mathcal {I}}}'\in {\mathfrak {I}}({\widehat{\mathcal {A}}}')\), because \((\varPhi ')^{-1}\) is an algebra isomorphism. Next, \((\diamond )\) is by the definition of \({\widehat{\varPsi }}^\#\). Finally, \((\dagger )\) is by the definitions of \(\varPhi _*\) and \((\varPhi ')^{-1}_*\).

Equation (D31) holds for all \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\); thus, \(\varPsi ^\#=(\varPhi ')^{-1}_*\circ {\widehat{\varPsi }}^\#\circ \varPhi _*\). Thus, \(\varPsi ^\#\) is a Boolean algebra homomorphism from \({\mathfrak {I}}({\mathcal {A}})\) to \({\mathfrak {I}}({\mathcal {A}}')\) (because it is a composition of two isomorphisms and a homomorphism), and we have the commuting diagram (a):

figured

Since \(\varPhi ^{-1}_*:{\mathfrak {I}}({\widehat{\mathcal {A}}}) {{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) is also a Boolean algebra isomorphism, we also have the commuting diagram (b). Finally, diagram (D16) from Proposition D7(c) yields diagram (c) (after appropriate translation of notation). Now, combine (b) and (c) to obtain diagram (D29), by recalling that \(\varPhi ^*=\varPhi ^{-1}_*\circ {\mathfrak {K}}\) and \(\varPhi ^{\prime *}=(\varPhi '_*)^{-1}\circ {\mathfrak {K}}'\) by definition. \(\square \)

Proposition D9

Let \({\mathcal {A}}\) and \({\mathcal {A}}'\) be Archimedean unitary Riesz spaces, with representations \(\varPhi :{\mathcal {A}}{{\longrightarrow }}{\widehat{\mathcal {A}}}\subseteq {\mathcal {C}}({\mathcal {S}})\) and \(\varPhi :{\mathcal {A}}'{{\longrightarrow }}{\widehat{\mathcal {A}}}'\subseteq {\mathcal {C}}({\mathcal {S}}')\) for some compact Hausdorff spaces \({\mathcal {S}}\) and \({\mathcal {S}}'\) and dense Riesz subspaces \({\widehat{\mathcal {A}}}\) and \({\widehat{\mathcal {A}}}'\), along with Boolean algebra anti-homomorphisms \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) and \({\varPhi '}^*:{\mathfrak {R}}({\mathcal {S}}'){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\) as described in Proposition B1(a).

Let \(\varPsi :{\mathcal {A}}{{\longrightarrow }}{\mathcal {A}}'\) be a regular Riesz homomorphism, and let \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) be the associated continuous function from Proposition B1(c). Then \(\psi \) is regular. For all \({\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})\), define \(\varPsi ^\#({\mathcal {I}}):=\varPsi ({\mathcal {I}})^{\perp \perp }\). Then \(\varPsi ^\#:{\mathfrak {I}}({\mathcal {A}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}}')\) is a Boolean algebra homomorphism, and we have the commuting diagram (D29).

Proof

The proof is essentially identical to the proof of Proposition D8, except that the words “Banach algebra”, “subalgebra” and “algebra homomorphism” are everywhere changed to “Riesz space”, “Riesz subspace” and “Riesz homomorphism”, and \({\mathfrak {I}}(\_)\) refers to the Boolean algebra of bands of a Riesz space, rather than regular ideals of an algebra. Likewise, all references to Proposition C4 are changed to Proposition B1. \(\square \)

Proof of Theorem 7

(a) Use Proposition B1(c) to obtain a continuous function \(\psi :{\mathcal {S}}'{{\longrightarrow }}{\mathcal {S}}\) satisfying the commuting diagram in Fig. 3a. Now, \(\varPsi \) is regular (because it is a preference homomorphism); thus, Proposition D9 says that \(\psi \) is regular. Thus, Proposition D7(b) says \(\psi ^\sharp :{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {R}}({\mathcal {S}}')\) is a Boolean algebra homomorphism, and Proposition D9 yields the commuting diagram (D29). Let \({\widetilde{\mathcal {S}}}:=\psi ({\mathcal {S}}')\); then, \({\widetilde{\mathcal {S}}}\) is a regular closed subset of \({\mathcal {S}}\) by Lemma D1. Define \(\varPhi ^*:{\mathfrak {R}}({\mathcal {S}}){{\longrightarrow }}{\mathfrak {I}}({\mathcal {A}})\) as in Proposition B1(a).

Claim 1

\(\ker (\varPsi )=\varPhi ^*({\widetilde{\mathcal {S}}})\).Footnote 42

Proof

Let \(\alpha \in {\mathcal {A}}\), and let \({\widehat{\alpha }}:=\varPhi (\alpha )\in {\widehat{\mathcal {A}}}\).

$$\begin{aligned} \begin{aligned} \left( \varPsi (\alpha )=0\right)&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( (\varPhi ')^{-1}\circ {\widehat{\varPsi }}({\widehat{\alpha }})=0\right) \ {\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \ \left( {\widehat{\varPsi }}({\widehat{\alpha }})=0\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\diamond )}}\left( {\widehat{\alpha }}\circ \psi =0\right) \ {\mathop {\Longleftrightarrow }\limits _{(\ddagger )}} \ \left( {\widehat{\alpha }}({\widetilde{s}})=0\hbox { for all } {\widetilde{s}}\in {\widetilde{S}}\right) \ \iff \ \left( {\widehat{\alpha }}\in {\mathfrak {K}}({\widetilde{S}})\right) \\&\iff \left( \alpha \in \varPhi ^{-1}[{\mathfrak {K}}({\widetilde{S}})]\right) \ {\mathop {\Longleftrightarrow }\limits _{(@)}} \ \left( \alpha \in \varPhi ^*({\widetilde{\mathcal {S}}})\right) , \end{aligned} \end{aligned}$$

as claimed. Here, \((*)\) is by the commuting diagram in Fig. 3a, \((\dagger )\) is because \((\varPsi ')^{-1}\) is injective, and \((\diamond )\) is by the definition of the function \({\widehat{\varPsi }}\). Next, \((\ddagger )\) is because \({\widetilde{\mathcal {S}}}:=\psi ({\mathcal {S}}')\), and (@) is because \(\varPhi ^*({\widetilde{\mathcal {S}}})=\varPhi ^{-1}[ {\mathfrak {K}}({\widetilde{\mathcal {S}}})]\) by the definition of \(\varPhi ^*\). \(\diamond \)

Let \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) and \({\{\succeq '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}}\) be the conditional preference structures on \({\widehat{\mathcal {A}}}\) and \({\widehat{\mathcal {A}}}'\) obtained from \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) and \(\{\succeq '_{{\mathcal {I}}'}\}_{{\mathcal {I}}'\in {\mathfrak {I}}({\mathcal {A}}')}\) via Proposition B2.

Claim 2

\(\psi \) is a preference homomorphism from \({\{\succeq '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}}\) to \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\).

Proof

Let \({\widehat{\alpha }},{\widehat{\beta }}\in {\widehat{\mathcal {A}}}\) and \({\mathcal {R}}\in {\mathfrak {R}}({\widetilde{\mathcal {S}}})\). We must show that statement (5.1) holds. In other words, if \({\mathcal {R}}':=\psi ^\sharp ({\mathcal {R}})\), \({\widehat{\alpha }}':={\widehat{\alpha }}\circ \psi = {\widehat{\varPsi }}({\widehat{\alpha }})\) and \({\widehat{\beta }}':={\widehat{\beta }}\circ \psi ={\widehat{\varPsi }}({\widehat{\beta }})\), then we want:

$$\begin{aligned} \left( {\widehat{\alpha }}_{\upharpoonleft {\mathcal {R}}} \succeq _{\mathcal {R}}{\widehat{\beta }}_{\upharpoonleft {\mathcal {R}}}\right) \iff \left( {\widehat{\alpha }}'_{\upharpoonleft {\mathcal {R}}'} \succeq '_{{\mathcal {R}}'} {\widehat{\beta }}'_{\upharpoonleft {\mathcal {R}}'}\right) . \end{aligned}$$
(D32)

To see this, let \({\mathcal {I}}:=\varPhi ^*({\mathcal {R}})\) and \({\mathcal {I}}':=\varPhi ^{\prime *}({\mathcal {R}}')\); then, \({\mathcal {I}}'=\varPsi ^\#({\mathcal {I}})\), by commuting diagram (D29). Furthermore, \(\ker (\varPsi )\subseteq {\mathcal {I}}\), because \(\ker (\varPsi )=\varPhi ^*({\widetilde{\mathcal {S}}})\) (by Claim 1), \({\widetilde{\mathcal {S}}}\supseteq {\mathcal {R}}\), and \(\varPhi ^*\) is a Boolean algebra anti-isomorphism by Proposition B1(a). Let \(\alpha :=\varPhi ^{-1}({\widehat{\alpha }})\) and \(\beta :=\varPhi ^{-1}({\widehat{\beta }})\). Then let \(\alpha ':=\varPsi (\alpha )\) and \(\beta ':=\varPsi (\beta )\); then, \({\widehat{\alpha }}':=\varPhi '(\alpha ')\) and \({\widehat{\beta }}':=\varPhi '(\beta ')\), by the commuting diagram in Fig. 3a. Thus,

$$\begin{aligned} \left( {\widehat{\alpha }}_{\upharpoonleft {\mathcal {R}}} \succeq _{\mathcal {R}}{\widehat{\beta }}_{\upharpoonleft {\mathcal {R}}}\right)&{\mathop {\Longleftrightarrow }\limits _{(*)}} \left( [\alpha ]_{\mathcal {I}}\succeq _{\mathcal {I}}[\beta ]_{\mathcal {I}}\right) {\mathop {\Longleftrightarrow }\limits _{(\dagger )}} \left( [\alpha ']_{{\mathcal {I}}'} \succeq '_{{\mathcal {I}}'} [\beta ']_{\mathcal {I}}'\right) \\&{\mathop {\Longleftrightarrow }\limits _{(\diamond )}} \left( {\widehat{\alpha }}'_{\upharpoonleft {\mathcal {R}}'} \succeq '_{{\mathcal {R}}'} {\widehat{\beta }}'_{\upharpoonleft {\mathcal {R}}'}\right) , \end{aligned}$$

which proves (D32). Here, \((*)\) is by statement (3.2), because \({\{\succeq _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})}}\) is derived from \(\{\succeq _{\mathcal {I}}\}_{{\mathcal {I}}\in {\mathfrak {I}}({\mathcal {A}})}\) via Proposition B2. Likewise, \((\diamond )\) is by statement (3.2), because \({\{\succeq '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')}}\) is derived from \(\{\succeq '_{{\mathcal {I}}'}\}_{{\mathcal {I}}'\in {\mathfrak {I}}({\mathcal {A}}')}\) via Proposition B2. Finally, \((\dagger )\) is by formula (5.7), because \(\varPsi \) is a Riesz preference homomorphism and \(\ker (\varPsi )\subseteq {\mathcal {I}}\). \(\diamond \)

Proposition 5.2 and Claim 2 imply that \(u'\) is a positive affine transform of u, and \(\psi \) projects \((\{\chi '_{{\mathcal {R}}'}\}_{{\mathcal {R}}'\in {\mathfrak {R}}({\mathcal {S}}')},\mu ')\) onto \((\{\chi _{\mathcal {R}}\}_{{\mathcal {R}}\in {\mathfrak {R}}({\mathcal {S}})},\mu )\). This proves Part (a). Part (b) follows from Proposition B1(e). \(\square \)

Proof of Theorem 8

The proof is almost identical to the proof of Theorem 7, except that all references of “Riesz space”, “Riesz subspace” and “Riesz homomorphism” are replaced by “Banach algebra”, “subalgebra” and “algebra homomorphism”. Likewise, the proof uses Propositions C4C5 and D8 instead of Propositions B1B2 and D9. \(\square \)

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Pivato, M. Subjective expected utility with a spectral state space. Econ Theory 69, 249–313 (2020). https://doi.org/10.1007/s00199-018-01173-5

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Keywords

  • Subjective expected utility
  • Awareness
  • Subjective state space
  • Riesz space
  • Banach lattice
  • Commutative Banach algebra

JEL Classification

  • D81