An Introduction to General Nonstandard Analysis

Abstract

In this chapter, we develop the general framework of nonstandard analysis and the necessary logic for the transfer principle. We will begin each section with a brief summary for readers who want to postpone the technical details until a later reading. The summary will note any important definitions and results of the section that the reader should know before going on.

Appendix: Nonstandard Models

Horst Osswald

Mathematisches Institut der Universität München

Theresienstr. 39, D–80333 München, Germany

e-mail: osswald@rz.mathematik.uni-muenchen.de

As promised in Sect. 2.9, we shall now show that for each superstructure V of cardinality $\kappa$ there exists a $\kappa ^{+}$-saturated superstructure W and a monomorphism ${}^{*}$ from V into W. ($\kappa ^{+}$ is the smallest cardinality greater than $\kappa$.) By the way, in model theory a monomorphism is often called an elementary embedding.

The proof is elementary. This means that the ambitious existence of countably incomplete $\kappa ^{+}$ good ultrafilter is not used as for example in the proof of Theorem 6.1.8 in the book of Chang and Keisler [2].

The idea of the proof is similar to the proof in algebra that each field k has an algebraically closed extension K. In the algebraic proof one has to find roots of polynomials; here we have to find elements in the intersection of families of internal sets. Roughly speaking, in algebra one constructs a suitable transfinite increasing chain $(k_{\alpha })_{\alpha <\gamma }$ of fields $k_{\alpha }$ with $k_{0}=k$ such that, if $\alpha$ is a successor ordinal, then $k_{\alpha }$ contains exactly one root of each irreducible polynomial over $k_{\alpha -1}$. If $\alpha$ is a limit number, then $k_{\alpha }$ is the union of the preceding fields. The union of all the $k_{\alpha }$ provides an algebraically closed extension of k. Here we construct a transfinite elementary chain; in the successor step we apply the compactness theorem, which says that for each superstructure and each family of internal sets in the superstructure having the finite intersection property there exists a superstructure—with the same formal properties—in which the intersection of the whole family is non-empty. In the limit case we take the so called elementary limit of the preceding structures. In brief, we are going to prove the following

Theorem 2.9.10

Fix a superstructure V(X) over X of cardinality $\kappa$. There exists a superstructure V(Y) over a set Y such that the following two principles hold:  

(T):

(The Transfer Principle) There exists a monomorphism ${}^{*}$ from V(X) into V(Y) with $Y={}^{*}X$.

(PS):

(Polysaturation) $V({}^{*}X)$ is $\kappa ^{+}$ -saturated.

 

The proof of this theorem, essentially adopted from the first edition of this book, slightly modifies Sacks [11] corresponding proof for first order logic.

1.1 Models

In order to work with long chains of superstructures and monomorphisms between them, it is convenient to consider only the internal part of a superstructure, which we will call a model.

Let $\mathbb {N}_{0}$ denote the nonnegative integers. A sequence $V:=(V_{n})_{n\in \mathbb {N}_{0}}$ is called a weak model if $V_{0}\ne \emptyset$ and $V_{n}\subset \mathcal {P} (V_{0}\cup \cdots \cup \ V_{n-1})$ for each $n\ge 1.$ The entities in $V_{0}$ are called the individuals in V, the entities in $V_{n}$ with $n\ge 1$ are called the sets in V.

Since individuals and sets must be treated differently in the proof of Theorem 2.9.10, we think of an individual as an object different from a set, in particular, it is different from the empty set. Moreover, since we are not interested in the elements of an individual, we assume that individuals do not contain any elements. Because of the extensionality axiom, “global” individuals don’t exist in Zermelo-Fraenkel (ZF)-set theory. In many books on poly-saturated models the existence of such “global” individuals is presupposed. Here we suggest to use “local” individuals instead. Local individuals are individuals relative to the model according to equations (M 1) and (M 2) below. These local individuals are sufficient to lay the foundation for the theory of poly-saturated models and they exist in ZF-set theory.

A weak model $V=(V_{n})_{n\in \mathbb {N}_{0}}$ is called a model if  

(M 1):

$V_{0}\cap \bigcup _{1\le n}V_{n}=\emptyset$, which means that individuals in a model are different from sets within the model.

(M 2):

$b\cap \,\bigcup _{n\in \mathbb {N}_{0}}V_{n}=\emptyset$ for each $b\in V_{0}$, which means that individuals in a model are empty relative to the model.

 

A model $V:=(V_{n})_{n\in \mathbb {N}_{0}}$ is called a standard model if $V_{n}=\mathcal {P}(V_{0}\cup \cdots \cup \ V_{n-1})$ for each $n\ge 1.$ Recall that, if $V:=(V_{n})_{n\in \mathbb {N}_{0}}$ is a standard model and $X:=V_{0}$, then $V(X)=\bigcup _{n\in \mathbb {N}_{0}}V_{n}$ has been called a superstructure over X. If V is a standard model, by induction it is easy to see that $V_{n}(X)=X\cup V_{n}$ for each $n\in \mathbb {N}_{0}$, where $V_{n}(X)$ is defined in 2.2.1 letting

$$\begin{aligned} V_{0}(X):=X\; \text {and}\;V_{n+1}(X):=V_{n}(X)\cup \mathcal {P}(V_{n}(X)). \end{aligned}$$

This shows that $V_{n}(X)$ results from $V_{n}$ by adding the set X of individuals – if V is a standard model.

Let $S:=(S_{n})_{n\in \mathbb {N}_{0}}$ be a sequence of sets. We use the following notation:

$$\begin{aligned} S_{<\infty }:=\bigcup _{n\in \mathbb {N}_{0}}S_{n}\;\text {and}\;S_{\le n}:=S_{0}\cup \cdots \cup \ S_{n}. \end{aligned}$$

1.1.1 From Weak Models to Models

As was promised in Remark 2.1.3, we will now show that the individuals of a weak model V can be renamed in such a way that V becomes a model. Let us call a (weak) model $V:=(V_{n})_{n\in \mathbb {N}_{0}}$ a (weak) model over X if $V_{0}=X$.

Proposition 2.9.11

For each set $Y\ne \emptyset$ there exists a bijection c from Y onto X such that each weak model over X is already a model over X.

Proof

Set

$$\begin{aligned} \widetilde{Y}:=\left\{ \left\{ b\right\} \mid b\in Y\right\} . \end{aligned}$$

Then $\mathop {\mathrm{card}}(\widetilde{Y})=\mathop {\mathrm{card}}(Y)$. Define by induction:

$$\begin{aligned} Y_{0}:=\widetilde{Y},\text { }Y_{n+1}:=\mathcal {P}(Y_{n}),\;Y_{\infty }:=\bigcup _{n\in \mathbb {N}_{0}}Y_{n}, \end{aligned}$$

and

$$\begin{aligned} X:=\left\{ \left\{ \left\{ \left\{ y\right\} ,Y_{\infty }\right\} \right\} \mid y\in Y\right\} . \end{aligned}$$

Now define $c(y):=\left\{ \left\{ \left\{ y\right\} ,Y_{\infty }\right\} \right\}$ and rename the elements y of Y by $c(y)\in X$. Notice that

• (1) $Y_{\infty }$ is an infinite set, $\mathop {\mathrm{card}}(a)=1$ and $\mathop {\mathrm{ card}}(\left\{ a,Y_{\infty }\right\} )=2$ for each $a\in \widetilde{Y}\cup X$. It follows that

• (2) the function $i:a\mapsto \left\{ \left\{ a,Y_\infty \right\} \right\}$ defines a bijection from $\widetilde{Y}$ onto X. Moreover, $Y_0=\widetilde{ Y}\notin X$. Since $\emptyset \in Y_n$ for each $n\ge 1$, $Y_n\notin X$ for each $n\in \mathbb {N\;}\cup \;\left\{ \infty \right\} .$

  Let $V:=(V_n)_{n\in \mathbb {N}_{0}}$ be a weak model over X. By induction over $n\in \mathbb {N}_0$, we see that $Y_n\notin V_k$ for each $n\in \mathbb {N}_0$ and each $k\in \left\{ 0,\ldots ,n\right\} .$ It follows that

• (3) $Y_\infty \notin V_k$ for each $k\in \mathbb {N}_0$.

  Now we prove that V is a model over X : 

• (4) We first show that $X\cap \bigcup _{n\ge 1}V_n=\emptyset :$

  Assume that $k\ge 1$ and $\left\{ \left\{ a,Y_\infty \right\} \right\} \in V_k$ for some $a\in \widetilde{Y}$. Then $\left\{ a,Y_\infty \right\} \in V_i$ for some $i<k$. By (1), $\left\{ a,Y_\infty \right\} \notin X=V_0.$ Therefore, there exists a $j<i$ with $Y_\infty \in V_j$, contradicting (3).

• (5) Finally, we prove that b $\cap \ \bigcup _{n\in \mathbb {N} _{0}}V_{n}=\emptyset$ for each $b\in X:$

  Fix $b=\left\{ \left\{ a,Y_{\infty }\right\} \right\} \in X$. Assume that there exist $n\in \mathbb {N}_{0}$ and $x\in V_{n}$ with $x\in b$. Then $x=\left\{ a,Y_{\infty }\right\}$. By (1), $x\notin X=V_{0}$, thus, $n\ge 1$. It follows that there exists a $k<n$ with $Y_{\infty }\in V_{k}$, contradicting (3). $\square$

It depends on the theory we have in mind, which objects we take either as individuals or as sets. If we are not interested in the shape of elements of a mathematical entity, then this entity can be chosen as an individual. For example, in arbitrary Banach spaces the elements may be individuals, but in the Lebesgue $L^{p}$-spaces the elements are functions, thus sets. In general, the real numbers should be individuals, but if we want to study real numbers as equivalence classes of Cauchy-sequences of rationals, then real numbers become sets and now the rationals may be chosen to be individuals.

1.1.2 Languages for Models

Monomorphisms between superstructures are stronger than the usual injective homomorphisms or homeomorphic embeddings. A monomorphism not only preserves the algebraic or topological structure, it preserves every mathematical property. In order to make the phrase “mathematical property” precise, we need a mathematical language strong enough to formalize every mathematical statement.

The introduction of this language and its interpretation is similar to the approach at the beginning of this chapter. However, to keep this section self-contained, we repeat the necessary basic facts in a form suitable for the proof of our main result. Given a model $V:=(V_{n})_{n\in \mathbb {N} _{0}}$, the alphabet of the language $\mathcal {L}_{V}$ has the following symbols:

Logical symbols: $\vee$, $\lnot$, $\exists$, $=$, $\in$.

Variables: A countable number of them will do.

Parameters: The elements of $V_{<\infty }$ are the parameters in $\mathcal {L}_V.$

Auxiliary symbols: Parentheses “$\left( \right.$”,“$\left. \right)$”, point “$\cdot$” and comma “,”.

We assume that all these symbols are pairwise distinct.

A sentence in $\mathcal {L}_{V}$ is built up inductively from these rules:

• (a) If $a,b\in V_{<\infty }$, then $(a\in b)$ and $(a=b)$ are sentences in $\mathcal {L}_{V}.$

Remark 2.9.12

It should be mentioned that in the sentence $a\in b$ the term a may be a set, in particular, a is an n-tuple of elements in $V_{<\infty }$ (see Part (4) of Proposition 2.9.14). See also the proof of Part (10) in that proposition. It would follow that b is an n -placed relation.

• (b) If A and B are sentences in $\mathcal {L}_{V},$ then $(A\vee B)$ and $(\lnot A)$ are sentences in $\mathcal {L}_{V}.$

• (c) Let A be a sentence in $\mathcal {L}_{V}$ and let a, b be parameters in $\mathcal {L}_V$. If x is a variable, not occurring in A, then $(\exists x\in aA_{b}(x))$ is a sentence in $\mathcal {L}_{V}$. Here $A_{b}(x)$ is the string of signs of the alphabet of $\mathcal {L}_{V}$ that results from A by replacing each b, where b occurs in A, with x.

• (d) The set of sentences of $\mathcal {L}_V$ is the smallest set having the properties (a), (b) and (c).

A formula in $\mathcal {L}_{V}$ results from a sentence in $\mathcal { L}_{V}$ by replacing some parameter with a variable x, not occurring in the sentence. We then say that x is free in the formula. We shall write A(x), to indicate that x is free in the formula A.

As usual,we use the following abbreviations:

$(A\rightarrow B)$ for $((\lnot A)\vee B),\;(A\wedge B)$ for $(\lnot ((\lnot A)\vee (\lnot B))),$

$(A\leftrightarrow B)$ for $((A\rightarrow B)\wedge (B\rightarrow A))$,

$(\forall x\in aA_b(x))$ for $(\lnot (\exists x\in a(\lnot A_b(x)))).$

In order to save parentheses we agree that $\lnot ,$ $\exists ,$ $\forall$ bind stronger than $\wedge ,$ $\wedge$ binds stronger than $\vee ,$ $\vee$ binds stronger than $\rightarrow ,$ and $\rightarrow$ binds stronger than $\leftrightarrow .$ The relation “binds stronger” is transitive. Moreover, we will use the following shorthand

$\exists x_1,\ldots ,x_k\;\in \;a\;A$ for $\exists x_1\;\in \;a\ldots \exists x_k\;\in \;a\;A,$

$\forall x_1,\ldots ,x_k\;\in \;a\;A$ for $\forall x_1\;\in \;a\ldots \forall x_k\;\in \;a\;A$.

1.1.3 Interpretation of the Language

Fix a model $V:=(V_n)_{n\in \mathbb {N}_0}$. Here is the truth predicate for sentences in $\mathcal {L}_V:$

• (a) Fix $a,b\in V_{<\infty }$.

  • (i) The sentence $a=b$ is true in V if $a=b$ in the sense of common set theory.

  • (ii) The sentence $a\in b$ is true in V if $a\in b$ in the sense of common set theory.

  By the definition of models, $a\in b$ can never become true in V if b is an individual of V.

• (b) Let A,  B be sentences of $\mathcal {L}_V$.

  • (i) $\lnot A$ is true in V if A is not true in V.

  • (ii) $A\vee B$ is true in V if A is true in V or B is true in V.

  • (iii) $\exists x\in a\;A(x)$ is true in V if there exists a $c \in a$ such that A(x)(c) is true in V. Here the sentence A(x)(c) results from the formula A(x) by replacing x with c.

Following a common practice in model theory, we will write $V\models A$ to denote that A is true in V.

We conclude this section with a remark concerning individuals.

Remark 2.9.13

Recall that if $\emptyset$ is a set in the model V, then a is an individual in V iff

$$\begin{aligned} a\ne \emptyset \; \ \text { and}\;V\models \lnot \exists x(x\in a). \end{aligned}$$

(Here $\exists x(x\in a)$ is a shorthand for the cumbersome formula $\exists x\in a(x\in a)$.) In particular, when we assume that the positive integers are individuals in V, we mean they are coded so that they never appear as sets in the model.

1.1.4 Models Closed Under Definition

We will now study models having nice closure properties: A model $V:=(V_{n})_{n\in \mathbb {N}_{0}}$ is called closed under definition if for each formula A(x) in $\mathcal {L}_{V}$ and each $n\in \mathbb {N}_{0}$

$$\begin{aligned} \left\{ a\in V_{\le n}\mid V\models A(x)(a)\right\} \in V_{n+1}. \end{aligned}$$

Proposition 2.9.14

Suppose that V is closed under definition and the positive integers are individuals in V, coded according to Proposition 2.9.11. Fix $n\ge 1$. Then

• (1) $\ V_{1}\subseteq \cdots \subseteq V_{n}\subseteq \ldots ,$ thus $V_{\le n}=V_{0}\,\cup \,V_{n}.$

• (2) If F is a finite subset of $V_{\le n-1},$ then $F\in V_{n}.$

• (3) If $a,b\in V_{\le n-1},$ then $\left\langle a,b\right\rangle :=\left\{ \left\{ a,b\right\} ,\left\{ b\right\} \right\} \in V_{n+1}$.

• (4) Fix $a_{1},\ldots ,a_{k}\in V_{\le n-1}.$ Then $(a_{1},\ldots ,a_{n}):=\left\{ \left\langle 1,a_{1}\right\rangle ,\ldots ,\left\langle k,a_{k}\right\rangle \right\} \in V_{n+2}$.

• (5) Fix $a_{1},\ldots ,a_{k}\in V_{n}$. Then $a_{1}\cup \cdots \cup \ a_{k}\in V_{n}$ and $a_{1}\cap \cdots \cap \ a_{k}\in V_{n}.$

• (6) If $a,b\in V_{n},$ then $a{\setminus } b\in V_{n}.$

• (7) $\emptyset \in V_{n}.$

• (8) $V_{\le n-1}\in V_{n}.$

• (9) $V_{n-1}\in V_{n}.$

• (10) Fix $a_{1},\ldots ,a_{k}\in V_{n}$. Then

  $$\begin{aligned} a_{1}\times \cdots \times a_{k}=\left\{ (\alpha _{1},\ldots ,\alpha _{k})\mid \alpha _{1}\in a_{1}\wedge \cdots \wedge \ \alpha _{k}\in a_{k}\right\} \in V_{n+3}. \end{aligned}$$

Proof

(1) Let $c\in V_{n}$. Then

$$\begin{aligned} c=\left\{ a\in V_{\le n}\mid V\models a\in c\right\} =\left\{ a\in V_{\le n}\mid V\models \left( x\in c\right) (a)\right\} \in V_{n+1}. \end{aligned}$$

(2) Let $F=\left\{ a_{1},\ldots ,a_{k}\right\}$. Then

$$\begin{aligned} F=\left\{ a\in V_{\le n-1}\mid V\models a=a_{1}\vee \cdots \vee \ a=a_{k}\right\} \in V_{n}. \end{aligned}$$

(3) and (4) These follow from (2).

(5) Since $a_{1},\ldots ,a_{k}\subseteq V_{\le n-1},$

$$\begin{aligned} a_{1}\cup \cdots \cup \ a_{k}=\left\{ a\in V_{\le n-1}\mid V\models a\;\in \;a_{1}\vee \cdots \vee \ a\;\in \;a_{k}\right\} \in V_{n}. \end{aligned}$$

The proof for “$\cap$” is similar.

(6) The proof of (6) is similar to the proof of (5).

(7) This is true, because

$$\begin{aligned} \emptyset =\left\{ a\in V_{\le n-1}\mid V\models \lnot a=a\right\} \in V_{n}. \end{aligned}$$

(8) $V_{\le n-1}=\left\{ a\in V_{\le n-1}\mid V\models a=a\right\} \in V_{n}.$

(9) By (8), $V_{0}=V_{\le 0}\in V_{1}$. Let $n>1$. Then, by (1), (6) and (8),

$$\begin{aligned} V_{n-1}=V_{\le n-1}{\setminus } V_{0}\in V_{n}. \end{aligned}$$

(10) Using (4), we obtain:

$$\begin{aligned} a_{1}\times \cdots \times \ a_{k}=\left\{ a\in V_{\le n+2}\mid V\models \exists x_{1}\in a_{1}\ldots \exists x_{k}\in a_{k}(a=\left( x_{1},\ldots ,x_{k}\right) )\right\} \in V_{n+3}, \end{aligned}$$

where we have used the following abbreviations:

$a=\left( x_1,\ldots ,x_k\right)$ for $\forall x\in V_{\le n+1}\left( x\;\in \;a\leftrightarrow x=\left\langle 1,x_1\right\rangle \vee \cdots \vee x=\left\langle k,x_k\right\rangle \right) ,$

$x=\left\langle i,x_i\right\rangle$ for $\forall z\in V_{\le n}(z\;\in \;x\leftrightarrow z=\left\{ i,x_i\right\} \vee z=\left\{ x_i\right\} ),$

$z=\left\{ i,x_i\right\} \;$for $\forall y\in V_{\le n-1}(y\;\in \;z\leftrightarrow y=i\vee y=x_i),$

$z=\left\{ x_{i}\right\}$ for $\forall y\in V_{\le n-1}(y\;\in \;z\leftrightarrow y=x_{i}).\;\square$

1.1.5 Elementary Embeddings

Fix a model $V\,{=}\,(V_{n})_{n\in \mathbb {N}_{0}}$ closed under definition and a second model $W=(W_{n})_{n\in \mathbb {N}_{0}}$. Suppose ${}*$ is a mapping from $V_{<\infty }$ into $W_{<\infty }$, and recall that, if A is a sentence (formula) in $\mathcal {L}_{V}$, then the ${}*$-transform of A , ${}^{*}A$, is the sentence (formula) in $\mathcal {L}_{W}$ obtained by replacing each parameter a in A by ${}*(a)$. We will write $^{*}a$ instead of $*(a)$. A mapping ${}*:V_{<\infty }\rightarrow W_{<\infty }$ is called an elementary embedding from V into W if (E 1) and (E 2) are true:

(E 1) For each $n\in \mathbb {N}_0$, ${}^{*}V_n=W_n$. It follows that $W_n\in W_{<\infty }$.

(E 2) (Transfer Principle) For each sentence A in $\mathcal {L}_V$

$$\begin{aligned} V\models A\Leftrightarrow W\models {}^{*}A. \end{aligned}$$

Proposition 2.9.15

Let ${}*$ be an elementary embedding from V into W and assume that the positive integers are individuals in V, coded according to Proposition 2.9.11. We obtain for each $n\in \mathbb {N}_{0}$:

• (1) The restriction $\ {}*\upharpoonright V_{n}$ maps $V_{n}$ into $W_{n}$, and $W_{n}$ is a set in W.

• (2) The map ${}*$ is injective and a homomorphism, that is, for $a,b\in V_{<\infty }$,

  $$\begin{aligned} a\in b\Leftrightarrow {}^{*}a\in {}^{*}b. \end{aligned}$$

• (3) $*(V_{\le n})=W_{\le n}\in W_{n+1}.$

• (4) W is closed under definition.

• (5) Let A(x) be a formula in $\mathcal {L}_{V}$. Then

  $$\begin{aligned} ^{*}\left\{ a\in V_{\le n}\mid V\models A(x)(a)\right\} =\left\{ b\in W_{\le n}\mid W\models \left( {}^{*}(A(x))\right) (b)\right\} . \end{aligned}$$

• (6) If $\left\{ a_{1},\ldots ,a_{k}\right\}$ is a finite subset of $V_{<\infty }$, then ${}^{*}\left\{ a_{1},\ldots ,a_{k}\right\} =\left\{ {}^{*}a_{1},\ldots ,{}^{*}a_{k}\right\} ,$ in particular, ${}^{*}\emptyset =\emptyset$. It follows that ${}^{*}\left( a_{1},\ldots ,a_{k}\right) =\left( {}^{*}a_{1},\ldots ,{}^{*}a_{k}\right)$ for all $a_{1},\ldots ,a_{k}\in V_{<\infty }$.

• (7) Let $a_{1},\ldots ,a_{k},a$ be sets in V. Let f be a set in V and a mapping from $a_{1}\,\times \, \cdots \,\times \, a_{k}$ into a. Then $^{*}f$ is a mapping from $^{*}a_{1}\times \cdots \times {}^{*}a_{k}$ into $^{*}a$ such that for all $(b_{1},\ldots ,b_{k})\in a_{1}\times \cdots \times a_{k},$

  $$\begin{aligned} ^{*}(f(b_{1},\ldots ,b_{k}))={}^{*}f({}^{*}b_{1},\ldots ,{}^{*}b_{k})). \end{aligned}$$

  If, in addition, f is surjective onto a or injective or bijective, then $^{*}f$ is surjective onto $^{*}a$, injective or bijective, respectively.

Proof

• (1) Since V is closed under definition, $V_{n}\in V_{n+1}$ and $V_{n}\ne \emptyset$ for all $n\in \mathbb {N}_{0}$. Let $a\in V_{n}.$ Since $V\models a\in V_{n},\;W\models {}^{*}a\in W_{n}.$ Therefore, $W_{n}$ is a set in W and $^{*}a\in W_{n}.$

• (2) Let $a,b\in V_{<\infty }.$ Then

  $$\begin{aligned} a=(\in )\;b\Leftrightarrow V\models a=(\in )\;b\Leftrightarrow W\models {}^{*}a=(\in )\;{}^{*}b\Leftrightarrow {}^{*}a=(\in )\;{}^{*}b. \end{aligned}$$

• (3) If $n=0$, the result follows from (E 1). Let $n\ge 1$. Set

  $$\begin{aligned} A:=\forall x\in V_{\le n}(x\in V_0\vee x\in V_n)\wedge \forall x\in V_0(x\in V_{\le n})\wedge \forall x\in V_n(x\in V_{\le n}). \end{aligned}$$

  Since A is true in V, ${}^{*}A$ is true in W. It follows that ${}^{*}V_{\le n}=W_0\cup W_n$. Since $\forall x\in V_k(x\in V_{k+1})$ is true in V for each $k\ge 1$ and therefore the ${}^{*}$-image of this sentence is true in W for all $k\ge 1$, by (E 1), $W_{\le n}=W_0\cup W_n$.

• (4) Let A(x) be a formula in $\mathcal {L}_{W}$. There is an $m\in \mathbb {N}_{0}$ such that the parameters $b_{1},\ldots ,b_{k}$ occurring in A(x) belong to $W_{\le m}$. Since V is closed under definition,

  $$\begin{aligned} V\models \forall x_{1},\ldots ,x_{k}\in V_{\le m}\exists z\in V_{n+1}\forall x\in V_{\le n}(x\in z\leftrightarrow A(x,x_{1},\ldots ,x_{k}))) \end{aligned}$$

  where $A(x,x_{1},\ldots ,x_{k})$ results from A(x) by replacing $b_{i}$ with $x_{i}$, $i=1,\ldots ,k$. We may assume that $x_{1},\ldots ,x_{k}$ do not occur in A(x) and are pairwise different. Since ${}^{*}V_{\le n}=W_{\le n}$ and ${}^{*}V_{n}=W_{n}$, we obtain the fact that

  $$\begin{aligned} W\models \forall x_{1},\ldots ,x_{k}\in W_{\le m}\exists z\in W_{n+1}\forall x\in W_{\le n}(x\in z\leftrightarrow A(x,x_{1},\ldots ,x_{k})). \end{aligned}$$

  It follows that for $b_{1},\ldots ,b_{k}$ there is a $B\in W_{n+1}$ such that for $a\in W_{\le n}$, $a\in B$ iff $W\models A(x)(a)$. Therefore, $B=\left\{ a\in W_{\le n}\mid W\models A(x)(a)\right\} .$

• (5) Define $B:=\left\{ a\in V_{\le n}\mid V\models A(x)(a)\right\}$. Then $B\in V_{n+1}$ and

  $$\begin{aligned} V\models \forall x\in V_{\le n}(x\in B\leftrightarrow A(x)). \end{aligned}$$

  Since $W\models \forall x\in W_{\le n}(x\in {}^{*}B\leftrightarrow {}^{*}A(x))$ and ${}^{*}B\in W_{n+1}$, we obtain

  $$\begin{aligned} ^{*}B=\left\{ b\in W_{\le n}\mid W\models {}^{*}A(x)(b)\right\} . \end{aligned}$$

• (6) Let $\left\{ a_{1},\ldots ,a_{k}\right\} \subseteq V_{\le n}.$ Then, by (1), $\left\{ {}^{*}a_{1},\ldots ,{}^{*}a_{k}\right\} \subseteq W_{\le n}$ and, by (5),

  $$\begin{aligned} ^{*}\left\{ a_{1},\ldots ,a_{k}\right\} ={}^{*}\left\{ a\in V_{\le n}\mid V\models a=a_{1}\vee \cdots \vee \ a=a_{k}\right\} = \end{aligned}$$
  $$\begin{aligned} \left\{ b\in W_{\le n}\mid W\models b={}^{*}a_{1}\vee \cdots \vee \ b={}^{*}a_{k}\right\} =\left\{ {}^{*}a_{1},\ldots ,{}^{*}a_{k}\right\} . \end{aligned}$$

  Moreover, we obtain for each $n\ge 1$,

  $$\begin{aligned} ^{*}\emptyset ={}^{*}\left\{ a\in V_{\le n-1}\mid V\models \lnot a=a\right\} =\left\{ b\in W_{\le n-1}\mid W\models \lnot b=b\right\} =\emptyset . \end{aligned}$$

• (7) We identify the k-placed functions f with the 2-placed relations, where the second argument is uniquely determined by the first argument, which is a k-tuple. Therefore, if f is a k-placed function, we can use $x_{k+1}=f(x_{1},\ldots ,x_{k})$ as a shorthand for $\left( \left( x_{1},\ldots ,x_{k}\right) ,x_{k+1}\right) \in f.$

  We may assume that there exists an $n\ge 1$ such that $a_{1},\ldots ,a_{k},a,f \in V_{n}.$ Then $V\models A_{1}\wedge A_{2}\wedge A_{3},$ where

  $$\begin{aligned} A_{1}:=\forall x_{1}\in a_{1}\ldots \forall x_{k}\in a_{k}\exists y\in a(\left( \left( x_{1},\ldots ,x_{k}\right) ,y\right) \in f). \end{aligned}$$
  $$\begin{aligned} A_{2}:=\forall x_{1},\ldots ,x_{k},y\in V_{\le n}(\left( (x_{1},\ldots ,x_{k}),y\right) \in f\rightarrow x_{1}\in a_{1}\wedge \cdots \wedge x_{k}\in a_{k}\wedge y\in a). \end{aligned}$$
  $$\begin{aligned} A_{3}:=\forall x_{1},\ldots ,x_{k},y,y^{\prime }\in V_{\le n}(\left( (x_{1},\ldots ,x_{k}),y\right) \in f\wedge \left( (x_{1},\ldots ,x_{k}),y^{\prime }\right) \in f\rightarrow y=y^{\prime }). \end{aligned}$$

  $A_{1}$ means that $\mathop {\mathrm{domain}}(f)\supseteq a_{1}\times \cdots \times a_{k}$; $A_{2}$ means that $\mathop {\mathrm{range}}(f)\subseteq a$ and $\mathop {\mathrm{domain}}(f)\subseteq a_{1}\times \cdots \times \ a_{k}$; $A_{3}$ means that f is a function. Since $V\models A_{1}\wedge A_{2}\wedge A_{3}$ and therefore $W\models {}^{*}A_{1}\wedge {}^{*}A_{2}\wedge {}^{*}A_{3}$, we see that ${}^{*}f$ is a mapping from ${}^{*}a_{1}\times \cdots \times {}^{*}a_{k}$ into ${}^{*}a$. Assume that $f(b_{1},\ldots ,b_{k})=b.$ Then $\left( (b_{1},\ldots ,b_{k}),b\right) \in f$ and ${}\left( (^{*}b_{1},\ldots ,{}^{*}b_{k}),{}^{*}b\right) \in {}^{*}f.$ Therefore, ${}^{*}f({}^{*}b_{1},\ldots ,{}^{*}b_{k})={}^{*}b={}^{*}(f(b_{1},\ldots ,b_{k})).$ The other parts of assertion (7) can be proved in a similar way. $\square$

Since ${}^{*}$ is injective, we may identify each individual a in V with the individual ${}^{*}a$ in W.

1.1.6 Saturated Models

Let us call a non-empty family $\mathcal {B}$ of sets deep if $\mathcal {B}$ has the finite intersection property, which means that the intersection of all finitely many sets in $\mathcal {B}$ is nonempty. Let $\gamma$ be an uncountable cardinal. A model $V=(V_{n})_{n\in \mathbb {N}_{0}}$, which is closed under definition, is called $\gamma$ -saturated if for each $n\ge 1$

$$\begin{aligned} \bigcap \mathcal {B}\ne \emptyset \text { for each deep }\mathcal {B}\subseteq V_{n}\ \text {with}\ \mathop {\mathrm{card}} (\mathcal {B})<\gamma . \end{aligned}$$

It follows that the internal sets in a $\gamma$-saturated model can be treated as though they were $\gamma$-compact (see Sect. 2.9).

1.1.7 From Pre-models to Models

In order to deal with ultrapowers and limits of elementary chains under the same roof, we introduce the notion “pre-model”: A triple $((P_{n})_{n\in \mathbb {N}_{0}},\sim ,E)$ is called a pre-model if the following conditions are true:

(PM 1) $P_0\ne \emptyset$ and $P_0\cap \bigcup _{n\ge 1}P_n=\emptyset$.

(PM 2) $\sim$ is an equivalence relation on $P_{<\infty }$ such that for all $a,b\in P_{<\infty }$ with $a\sim b$ and all $n\in \mathbb {N}_0:$

$$\begin{aligned} a\in P_n\Rightarrow b\in P_n. \end{aligned}$$

(PM 3) The relation E is a subset of $P_{<\infty }\times P_{1\le }$ such that for $a,a^{\prime },b,b^{\prime }\in P_{<\infty }$ with $a^{\prime }\sim a$ and $b^{\prime }\sim b$: 

$$\begin{aligned} a\,E\,b\Rightarrow a^{\prime }\,E\,b^{\prime }. \end{aligned}$$

(PM 4) Transitivity. If $n\ge 1$ and $a\in P_n$, then

$$\begin{aligned} b\,E\,a\Rightarrow b\in P_{\le n-1}. \end{aligned}$$

(PM 5) Extensionality. Fix $a,b\in P_{1\le }.$ Then $a\sim b$ if $(cEa\Leftrightarrow cEb)$ for all $c\in P_{<\infty }.$

Later on we shall be concerned with two important examples of pre-models.

Fix a pre-model $P=((P_{n})_{n\in \mathbb {N}_{0}},\sim ,E)$. Using the Mostowski Collapsing Function (see Definition 2.5.3), we define inductively on $n\in \mathbb {N}_{0}$ sets $V_{n}$ and their elements $\overline{a}^{n}$ such that $(V_{n})$ becomes a model generated by P.

For each $a\in P_{0}$ set $\widetilde{a}^{0}:=\{b\in P_{0}|$ $b\sim a\}$. According to Proposition 2.9.11, we can rename each $\widetilde{a}^{0}$ to $\overline{a}^{0}$ such that the function $\widetilde{a}^{0}\mapsto \overline{a}^{0}$ defines a bijection between $\left\{ \widetilde{a}^{0}\mid a\in P_{0}\right\}$ and $\left\{ \overline{a}^{0}\mid a\in P_{0}\right\} ,$ and such that each weak model over

$$\begin{aligned} V_{0}:=\{\overline{a}^{0}\;|\;a\in P_{0}\} \end{aligned}$$

is a model over $V_{0}$.

Now let $n\ge 1$ and assume that $\overline{a}^{k}$ is already defined for each $k<n$ and each $a\in P_{k}.$ Moreover, assume that $V_{k}:=\{\overline{a }^{k}\;|\;a\in P_{k}\}$ for each $k<n$. If $a\in P_{n},$ set

$$\begin{aligned} \overline{a}^{n}:=\{\overline{c}^{k}\;|\;k<n,c\in P_{k}\;\text {and} \;c\;E\;a\}. \end{aligned}$$

and

$$\begin{aligned} V_{n}:=\{\overline{a}^{n}\;|\;a\in P_{n}\}. \end{aligned}$$

Since $P_{0}\ne \emptyset$, we have $V_{0}\ne \emptyset .$ Since for $n\ge 1$, $\overline{a}^{n}\subset V_{\le n-1}$, we see that $V_{n}\subset \mathcal {P}(V_{\le n-1})$. It follows that $V:=(V_{n})_{n\in \mathbb {N} _{0}}$ is a model over $V_{0}.$ We say that the model V is generated by $((P_{n})_{n\in \mathbb {N}_{0}},\sim ,E).$

Proposition 2.9.16

Fix $n,m\in \mathbb {N}_{0}$ and $a\in P_{n}$ and $b\in P_{m}.$ Then

$$\begin{aligned} \overline{a}^n\;=\;\overline{b}^m\Leftrightarrow a\sim b. \end{aligned}$$

Proof

We prove this result by induction on $\mu _{m,n}:=\max \left\{ n,m\right\}$.

Let $\mu _{m,n}=0$. Then, by the definition of $\overline{a}^0$, $\overline{a}^0=\overline{b}^0\Leftrightarrow \widetilde{a}^0=\widetilde{b} ^0\Leftrightarrow a\sim b$. For the induction step let $\mu _{m,n}>0$.

Assume that $\overline{a}^{n}=\overline{b}^{m}$. Since $V_{0}\cap V_{1\le }=\emptyset$, we have $1\le n,m$. Suppose that $c\,E\,a$ holds. By (PM 4), there exists an $i<n$ with $c\in P_{i}$. It follows that $\overline{c} ^{i}\in \overline{a}^{n}=\overline{b}^{m}$. Therefore, there exists $j<m$ and an element $d\in P_{j}$ with $d\,E\,b$ and $\overline{d}^{j}=\overline{c} ^{i}$. Since $\mu _{i,j}<\mu _{m,n}$, by the induction hypothesis, $d\sim c$. From (PM 3) it follows that $c\,E\,b$ holds. Therefore, $c\,E\,a$ implies $c\,E\,b$ and vice versa. By (PM 5),  $a\sim b$.

Now assume that $a\sim b$. Since $m>0$ or $n>0,$ by (PM 1) and (PM 2), $1\le m,n$. In order to show that $\overline{a}^{n}=\overline{b}^{m}$, let $\overline{c}^{i}\in \overline{a}^{n}$ with $i<n$, $c\in P_{i}$ and $c\,E\,a.$ By (PM 3), $c\,E\,b.$ By (PM 4), there exists a $j<m$ with $c\in P_{j}.$ It follows that $\overline{c}^{j}\in \overline{b}^{m}$. Since $\mu _{i,j}<\mu _{m,n}$, by the induction hypothesis, $\overline{c}^{i}=\overline{c}^{j}\in \overline{b}^{m}$. This proves that $\overline{a}^{n}\subseteq \overline{b} ^{m}$. The proof of $\overline{a}^{n}\supseteq \overline{b}^{m}$ is similar. $\square$

In view of the previous proposition, we may define for each $a\in P_{<\infty },$

$$\begin{aligned} \overline{a}:=\overline{a}^{n}\;\text {if}\;a\in P_{n}. \end{aligned}$$

We obtain the following

Corollary 2.9.17

Fix $a,b\in P_{<\infty }$. Then

• (1) $\overline{a}=\overline{b}\Leftrightarrow a\sim b.$

• (2) $\overline{a}\in \overline{b}\Leftrightarrow a\,E\,b.$

Proof

We only need to prove (2). Assume that $\overline{a}\in \overline{b}$. Since V is a model, $\overline{b}$ is a set in V. Therefore, there exist $i\in \mathbb {N}_{0}$ and $a^{\prime }\in P_{i}$ with $a^{\prime }\,E\,b$ and $\overline{a^{\prime }}=\overline{a}$. By (1), $a^{\prime }\sim a$. By (PM 3), $a\,E\,b.$ Now assume that $a\,E\,b$. Since $b\in P_{n}$ for some $n\ge 1$, by (PM 4), there exists an $i<n$ with $a\in P_{i}$. By the definition of $\overline{b}$, $\overline{a}\in \overline{b}$. $\square$

1.2 Ultrapowers

The aim of this section is the proof of the compactness theorem. We assume that the reader is familiar with filters and ultrafilters (see Sect. 1.2). First we will construct a pre-model generating the ultrapower of a model V:

Fix a non-empty set I, an ultrafilter D on I and a model $V=(V_n)_{n\in \mathbb {N}_0}$, which is closed under definition. We shall use the following abbreviations.

If $(a_{i})_{i\in I}$ is an I-sequence, then we will write $(a_{i})$ instead of $(a_{i})_{i\in I}.$ If $\mathcal {F}(i)$ is an assertion about elements i of I, then we shall write $\mathcal {F}(i)$ a.e. instead of $\left\{ i\in I\mid \mathcal {F}(i)\right\} \in D$. For each $n\in \mathbb {N} _{0}$ set

$$\begin{aligned} F_{n}:=\{(a_{i})\;|\;a_{i}\in V_{n}\;\text {a.e.}\}. \end{aligned}$$

On $F_{<\infty }=\bigcup _{n\in \mathbb {N}_{0}}F_{n}$ we define a relation $\sim$ by setting

$$\begin{aligned} (a_{i})\sim (b_{i}):\Leftrightarrow a_{i}=b_{i}\;\text {a.e.} \end{aligned}$$

and we define a relation $E\subseteq F_{<\infty }\times F_{1\le }$ by setting,

$$\begin{aligned} (a_{i})\;E\;(b_{i})\;:\Leftrightarrow a_{i}\in b_{i}\;\text {a.e.} \end{aligned}$$

Using the ultrafilter properties one can easily prove the following

Lemma 2.9.18

The triple $\left( (F_{n})_{n\in \mathbb {N}_{0}},\sim ,E\right)$ is a pre-model.

The model generated by $((F_n)_{n\in \mathbb {N}_0},\sim ,E)$ is called the D -ultrapower of V and is denoted by $\Pi _D(V)$.

Now we shall show that the D-ultrapower of V has the same formal properties as V. Let A be a (sentence) formula in $\mathcal {L} _{\Pi _{D}(V)}$. By $(A^{i})_{i\in I}$ we denote one of the I-sequences of sentences (formulas) in $\mathcal {L}_{V}$ that result from A by replacing each parameter $\overline{(a_{i})}$ in A by $a_{i}.$ Notice that, if $( \widetilde{A}^{i})_{i\in I}$ is another result, then $\widetilde{A} ^{i}=A^{i}$ a.e.

Theorem 2.9.19

(Theorem of Łŏs) Fix a sentence A in $\mathcal {L}_{\Pi _{D}(V)}$. Then,

$$\begin{aligned} \Pi _{D}(V)\models A\Leftrightarrow V\models A^{i}\, \text {a.e.} \end{aligned}$$

Proof

By induction over the definition of the sentences in $\mathcal {L}_{\Pi _{D}(V)}$.

• (1)

  • (a) Let A be the sentence $(\overline{(a_{i})}\in \overline{(b_{i})})$. Then

    $$\begin{aligned} \Pi _{D}(V)\models \overline{(a_{i})}\in \overline{(b_{i})}\Leftrightarrow \overline{(a_{i})}\in \overline{(b_{i})}\Leftrightarrow a_{i}\in b_{i}\; \text {a.e.}\Leftrightarrow V\models A^{i}\;\text {a.e.} \end{aligned}$$

  • (b) If $A=(\overline{(a_{i})}=\overline{(b_{i})})$, then the proof is similar to the proof under (a).

• (2)

  • (a) Let $A=B\vee C$ be a formula of $\mathcal {L}_{\Pi _D(V)}$. Assume that the assertion is true for B and C.

    Suppose that $\Pi _{D}(V)\models A$. Then $\Pi _{D}(V)\models B$ or $\Pi _{D}(V)\models C.$ By the induction hypothesis, $V\models B^{i}\,$a.e. or $V\models C^{i}\,$a.e. Therefore, $V\models B^{i}\vee C^{i}$ a.e. The result follows, since $A^{i}$ equals $B^{i}\vee C^{i}$ a.e.

    Suppose that $V\models B^{i}\vee C^{i}$ a.e. Since D is an ultrafilter, $V\models B^{i}$ a.e. or $V\models C^{i}$ a.e. By the induction hypothesis, $\Pi _{D}(V)\models B$ or $\Pi _{D}(V)\models C$, thus, $\Pi _{D}(V)\models B\vee C$.

  • (b) Let $A=\lnot B$.

    Assume that $\Pi _{D}(V)\models A$, thus, $\Pi _{D}(V)\nvDash B$. By the induction hypothesis, $\left\{ V\models B^{i}\right\} \notin D$. Since D is an ultrafilter $\left\{ V\nvDash B^{i}\right\} \in D$, thus, $V\models \lnot B^{i}$ a.e.

    Assume that $V\models \lnot B^{i}$ a.e. Since $\emptyset \notin D$, $\left\{ V\vDash B^{i}\right\} \notin D.$ The induction hypothesis implies $\Pi _{D}(V)\models \lnot B$.

  • (c) Let $A=\exists x\in \overline{(a_i)}B(x)$.

    If $\Pi _{D}(V)\models A$, then there is a $\overline{(b_{i})}\in \overline{ (a_{i})}$ with $\Pi _{D}(V)\models B(x)(\overline{(b_{i})})$. By the induction hypothesis, $V\models B(x)^{i}(b_{i})$ a.e. Since $b_{i}\in a_{i}$ a.e., $V\models A^{i}$ a.e.

    Suppose that $V\models A^{i}$ a.e., that is, $Y:=\left\{ i\in I\mid V\models \exists x\in a_{i}(B(x)^{i})\right\} \in D$. We choose an I-sequence $(b_{i})$ with $V\models b_{i}\in a_{i}\wedge B(x)^{i}(b_{i})$ for each $i\in Y$ and set $b_{i}=\emptyset$ if $i\notin Y$. Since $V\models B(x)^{i}(b_{i})$ a.e., by the induction hypothesis, $\Pi _{D}(V)\models B(x)(\overline{(b_{i})})$. Since $b_{i}\in a_{i}$ a.e., $\overline{(b_{i})} \in \overline{(a_{i})}$. It follows that $\Pi _{D}(V)\models A$. $\square$

An important consequence of Łŏs’ theorem is the existence of an elementary embedding from V into the ultrapower $\Pi _{D}(V)$ of V.

For each $b\in V_{<\infty }$ let $\left( b\right)$ be the constant I-sequence $(b_{i})$ with $b_{i}=b$ for each $i\in I$. The mapping $^{*}:V_{<\infty }\rightarrow \Pi _{D}(V)_{<\infty }$ is defined by setting

$$\begin{aligned} ^{*}b:=\overline{(b)}. \end{aligned}$$

Corollary 2.9.20

The function $*$ is an elementary embedding from V into $\Pi _{D}(V).$

Proof

We have to check the properties (E 1) and (E 2).

(E 1) Since $V_{n}\in V_{n+1},\;(V_{n})\in F_{n+1}$, thus $^{*}V_{n}= \overline{\left( V_{n}\right) }\in \Pi _{D}(V)_{<\infty }$. By Corollary 2.9.17 (2), and since $V_{n}$ is a set in V, we obtain

$$\begin{aligned} \overline{(a_{i})}\in \overline{\left( V_{n}\right) }\Leftrightarrow (a_{i})\,\,E\,\,\left( V_{n}\right) \Leftrightarrow \overline{(a_{i})}\in \Pi _{D}(V)_{n}. \end{aligned}$$

This proves that ${}^{*}V_{n}=\Pi _{D}(V)_{n}.$

(E 2) Let A be a sentence in $\mathcal {L}_{V}.$ Then $^{*}A$ is a sentence in $\mathcal {L}_{\Pi _{D}(V)}$ and $\left( ^{*}A\right) ^{i}=A$ a.e. By the theorem of Łŏs, we obtain

$$\begin{aligned} V\models A\Leftrightarrow V\models \,(^{*}A)^{i}\,\text {a.e.} \Leftrightarrow \Pi _{D}(V)\models \,^{*}A.\;\square \end{aligned}$$

Recall from Proposition 2.9.15 (4) that the model $\Pi _{D}(V)$ is closed under definition. Using the ultrapower construction given above, we can prove the Compactness Theorem. The proof is the same as in first order model theory (see [2]).

Proposition 2.9.21

(The Compactness Theorem) Let $V=\left( V_{n}\right) _{n\in \mathbb {N}_{0}}$ be a model which is closed under definition, let $n\ge 1$ and let $\mathcal {B}$ be a deep subset of $V_{n}$. Then there exists a model $W=\left( W_{n}\right) _{n\in \mathbb {N}_{0}}$ and an elementary embedding $^{*}$ from V into W, such that $\bigcap {}^{*}[\mathcal {B} ]\ne \emptyset$. (Notice that $\bigcap {}^{*}[\mathcal {B}]=\bigcap \left\{ {}^{*}A\mid A\in \mathcal {B}\right\}$.)

Proof

We may assume that $\mathcal {B}$ is closed under finite intersections. Let I be the set of finite non-empty subsets of $\mathcal {B}.$ Since $\mathcal { B}$ is deep, for each $i\in I$ there exists an $a_{i}\in V_{\le n-1}$ such that $a_{i}\in \bigcap i\in \mathcal {B}.$ For each set $A\in \mathcal {B}$ define

$$\begin{aligned} \widetilde{A}:=\{i\in I\,\,|\,\,a_{i}\in A\}\subset I, \end{aligned}$$

and

$$\begin{aligned} G:=\{\widetilde{A}\,\,|\,\,A\in \mathcal {B}\}\subseteq \mathcal {P}\left( I\right) . \end{aligned}$$

Then $\emptyset \notin G,$ because $\left\{ A\right\} \in \widetilde{A}$ for each $A\in \mathcal {B}$. Since $\mathcal {B}\ne \emptyset$, $G\ne \emptyset .$ Now let $A,B\in \mathcal {B}$. Then $\widetilde{A\cap B}\in G$ and it is easy to see that $\widetilde{A\cap B}\subset \widetilde{A}\cap \widetilde{B}.$ It follows that $F(G):=\left\{ B\subseteq I\mid \exists C\in G(C\subseteq B)\right\}$ is a filter. Fix an ultrafilter $D\supseteq F(G)$ on I. Let $\Pi _{D}(V)$ be the D-ultrapower of V and let $*:V_{<\infty }\rightarrow \Pi _{D}(V)_{<\infty }$ be the elementary embedding introduced previously. Since $a_{i}\in V_{\le n-1}$ and D is an ultrafilter, $(a_{i})\in F_{\le n-1}$, thus $\overline{(a_{i})}\in \Pi _{D}(V)_{\le n-1}$. In order to show that $\overline{(a_{i})}\in \bigcap {}^{*}\left[ \mathcal {B}\right]$, fix $A\in \mathcal {B}$. Then

$$\begin{aligned} \left\{ i\in I\mid V\vDash a_{i}\in A\right\} =\left\{ i\in I\mid a_{i}\in A\right\} =\widetilde{A}\in G\subset F(G)\subset D. \end{aligned}$$

By the theorem of Łŏs, $\Pi _{D}(V)\vDash \overline{(a_{i})}\in \overline{(A)}={}^{*}A$. $\square$

1.3 Elementary Chains and Their Elementary Limits

Here we extend the notions “elementary chain” and “elementary limit” to our notion of a model. Let $\lambda$ be an ordinal number different from 0 and let $\left( V^\alpha \right) _{\alpha <\lambda }$ be a $\lambda$-sequence of models $V^\alpha =(V_n^\alpha )_{n\in \mathbb {N}_0}$, which are closed under definition. Then the pair $\left( \left( V^\alpha \right) _{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda }\right)$ is called an elementary chain if for each $\alpha \le \beta \le \gamma <\lambda$

   (EC 1) $*_{\alpha \beta }$ is an elementary embedding from $V^\alpha$ into $V^\beta .$

   (EC 2) $*_{\alpha \alpha }(a)=a$ for each $a\in V_{<\infty }^\alpha .$

   (EC 3) $*_{\beta \gamma }\circ \,*_{\alpha \beta }=\,*_{\alpha \gamma }$.

Starting from an elementary chain C we define a pre-model such that the generated model becomes the elementary limit of C. To this end fix an elementary chain $\left( \left( V^{\alpha }\right) _{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda }\right)$. We define for each $n\in \mathbb {N}_{0}$,

$$\begin{aligned} P_{n}:=\{(a,\alpha )\,\,|\,\,\,\alpha <\lambda \, \text {and }a\in V_{n}^{\alpha }\}. \end{aligned}$$

Two elements $(a,\alpha ),(b,\beta )\in P_{<\infty }=\bigcup _{n\in \mathbb {N}_{0}}P_{n}$ are called equivalent if there exists a $\gamma <\lambda$ with $\alpha ,\beta \le \gamma$ such that $*_{\alpha \gamma }(a)=*_{\beta \gamma }(b)$, in which case we shall write $(a,\alpha )\sim (b,\beta )$.

For $(a,\alpha )\in P_{<\infty }$ and $(b,\beta )\in P_{1\le }$ we set $(a,\alpha )\;E\;(b,\beta )$ if there exists a $\gamma <\lambda$ with $\alpha ,\beta \le \gamma$ such that $*_{\alpha \gamma }(a)\in *_{\beta \gamma }(b).$ The simple proof of the next result is left to the reader :

Lemma 2.9.22

$((P_{n})_{n\in \mathbb {N}_{0}},\sim ,E)$ is a pre-model.

The model, generated by the pre-model $((P_{n})_{n\in \mathbb {N}_{0}},\sim ,E),$ is called the elementary limit of the elementary chain $((V^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda })$ and is denoted by $V^{\lambda }=(V_{n}^{\lambda })_{n\in \mathbb {N}_{0}}$.

We shall now collect some immediate consequences of the construction of the elementary limit (see Corollary 2.9.17 (1) and (2)).

Proposition 2.9.23

Fix an elementary chain $((V^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda })$ and let $V^{\lambda }$ be its elementary limit. Fix $n\in \mathbb {N}_{0}$. Then

• (1) $V_{n}^{\lambda }=\{ \overline{(a,\alpha )}\,\,|\,\,(a,\alpha )\in P_{n}\}=\{\overline{(a,\alpha ) }\,\,|\,\alpha <\lambda$ and $\,a\in V_{n}^{\alpha }\}$.

  Fix $\alpha ,\beta <\lambda ,a\in V_{<\infty }^{\alpha },b\in V_{<\infty }^{\beta }$.

• (2) $\overline{(a,\alpha )}=\overline{(b,\beta )}$ iff there is a $\gamma$ with $\alpha ,\beta \le \gamma <\lambda$ and $*_{\alpha \gamma }(a)=*_{\beta \gamma }(b).$ It follows that $\overline{(a,\alpha )}= \overline{(*_{\alpha \beta }(a),\beta )}$ if $\;\alpha \le \beta <\lambda .$

• (3) $\overline{(a,\alpha )}\in \overline{(b,\beta )}\,$ iff there is a $\gamma <\lambda$ with $\alpha ,\beta \le \gamma$ and $*_{\alpha \gamma }(a)\in *_{\beta \gamma }(b).$

• (4) $\overline{(V_{n}^{\alpha },\alpha )}=\overline{(V_{n}^{\beta },\beta )}$, because for each $\gamma$ with $\alpha ,\beta \le \gamma <\lambda$

  $$\begin{aligned} *_{\alpha \gamma }(V_{n}^{\alpha })=V_{n}^{\gamma }=*_{\beta \gamma }(V_{n}^{\beta }). \end{aligned}$$

Definition 2.9.24

Fix an elementary chain $((V^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda })$ and its elementary limit $V^{\lambda }.$ For each $\alpha \le \lambda$ define a mapping $*_{\alpha \lambda }:V_{<\infty }^{\alpha }\rightarrow V_{<\infty }^{\lambda }$ by setting

$$\begin{aligned} *_{\alpha \lambda }(a):=\overline{(a,\alpha )}\;\text {for}\;\alpha <\lambda \;\text {and}\;*_{\lambda \lambda }\left( \overline{(a,\alpha )} \right) :=\overline{(a,\alpha )}. \end{aligned}$$

Proposition 2.9.25

Fix $\alpha <\lambda .$ Then $*_{\alpha \lambda }$ is an elementary embedding from $V^{\alpha }$ into $V^{\lambda }.$ It follows that, $((V^{\alpha })_{\alpha <\lambda +1},(*_{\alpha \beta })_{\alpha \le \beta <\lambda +1})$ is an elementary chain.

Proof

To prove (E 1), fix $n\in \mathbb {N}_{0}.$ We have to show that $V_{n}^{\lambda }=*_{\alpha \lambda }(V_{n}^{\alpha }):$

“$\subseteq$” Fix $x\in V_{n}^{\lambda }$. Then there exist $\beta <\lambda$ and $b\in V_{n}^{\beta }$ with $x=\overline{(b,\beta )}.$ By (3) and (4),

$$\begin{aligned} x=\overline{(b,\beta )}\in \overline{(V_{n}^{\beta },\beta )}=\overline{ (V_{n}^{\alpha },\alpha )}=*_{\alpha \lambda }V_{n}^{\alpha }. \end{aligned}$$

“$\supseteq$” Fix $x\in *_{\alpha \lambda }(V_{n}^{\alpha })= \overline{(V_{n}^{\alpha },\alpha )}.$ Since $V_{n}^{\alpha }$ is a set in $V^{\alpha }$, $\overline{\left( V_{n}^{\alpha },\alpha \right) }$ is a set in $V^{\lambda }$. Therefore, there exist $\beta <\lambda$ and $b\in V_{<\infty }^{\beta }$ with $x=\overline{(b,\beta )}.$ By Corollary 2.9.17 (2), $(b,\beta )\,E\,\left( V_{n}^{\alpha },\alpha \right)$, thus there exists a $\gamma \ge \alpha ,\beta$ with $*_{\beta \gamma }(b)\in *_{\alpha \gamma }(V_{n}^{\alpha })=V_{n}^{\gamma }.$ We obtain

$$\begin{aligned} x=\overline{(b,\beta )}=\overline{(*_{\beta \gamma }(b),\gamma )}\in V_{n}^{\lambda }. \end{aligned}$$

This proves (E 1). To prove (E 2), we will prove by induction on the definition of the sentences A in $\bigcup _{\alpha <\lambda }\mathcal {L} _{V^{\alpha }}$ that if A is in $\mathcal {L}_{V^{\alpha }}$ then

$$\begin{aligned} V^{\alpha }\models A\Leftrightarrow V^{\lambda }\models \;^{*_{\alpha \lambda }}A. \end{aligned}$$

• (1)

  • (a) Let $A=(a\in b).$ Then

    $$\begin{aligned} V^{\alpha }\models A\Leftrightarrow a\in b\Leftrightarrow (a,\alpha )\,E\,(b,\alpha )\Leftrightarrow \overline{(a,\alpha )}\,\in \,\overline{ (b,\alpha )} \end{aligned}$$
    $$\begin{aligned} \Leftrightarrow *_{\alpha \lambda }(a)\in *_{\alpha \lambda }(b)\Leftrightarrow V^{\lambda }\models {}^{*_{\alpha \lambda }}A. \end{aligned}$$

    The second “$\Leftarrow$ ” of the previous computation can be seen as follows: If $(a,\alpha )\,E\,(b,\alpha )$, then $*_{\alpha \gamma }(a)\in *_{\alpha \gamma }(b)$ for some $\gamma \ge \alpha$, thus $a\in b$.

  • (b) If $A=(a=b)$, then the proof is similar to the proof of (1) (a).

• (2) If $A=(B\vee C)$ or $A=\lnot B$, then the assertion follows immediately from the induction hypothesis.

Let $A=\exists x\in aB(x)$.

Assume that $V^\alpha \models A.$ Then there is a $b\in a$ with $V^\alpha \models B(x)(b)$. By the induction hypothesis, $V^{\lambda }\models \, ^{*_{\alpha \lambda }}B(x) \left( \overline{(b,\alpha )}\right)$. Since $\overline{(b,\alpha )}\in \overline{(a,\alpha )}$, $V^{\lambda } \models \, ^{*_{\alpha \lambda }}A$.

Assume that $V^{\lambda }\models \,^{*_{\alpha \lambda }}A.$ Then $V^{\lambda }\models \,^{*_{\alpha \lambda }}B(x)\left( \overline{ (b,\delta )}\right)$ for some $\overline{(b,\delta )}\in \overline{ (a,\alpha )}$. Set $\gamma :=\max \{\alpha ,\delta \}$. Since $\overline{ (*_{\delta \gamma }(b),\gamma )}=\overline{(b,\delta )}$, $V^{\lambda }\models \,^{*_{\gamma \lambda }}(^{*_{\alpha \gamma }}B(x)\left( *_{\delta \gamma }(b)\right) )$. By the induction hypothesis, $V^{\gamma }\models \,^{*_{\alpha \gamma }}B(x)\left( *_{\delta \gamma }(b)\right)$. Since $*_{\delta \gamma }(b)\in *_{\alpha \gamma }(a)$, we obtain the fact that $V^{\gamma }\models \exists x\in *_{\alpha \gamma }(a)\,^{*_{\alpha \gamma }}B(x)$, that is, $V^{\gamma }\models \,^{*_{\alpha \gamma }}\exists x\in aB(x)$. Since the parameters of $\exists x\in aB(x)$ belong to $V_{<\infty }^{\alpha }$, we obtain $V^{\alpha }\models \exists x\in aB(x)$. $\square$

1.4 Existence of Polysaturated Nonstandard Models

In order to finish the proof of Theorem 2.9.10, we need one more lemma.

Lemma 2.9.26

Fix a model $W:=(W_{n})_{n\in \mathbb {N}_{0}}$ that is closed under definition. Then there exist a model U and an elementary embedding $*$ from W into U such that for each $n\ge 1$ and each deep $\mathcal {B}\subseteq W_{n}$, $\bigcap {}^{*}[\mathcal {B}]\ne \emptyset$.

Proof

There exists a cardinal number $\theta$ and a listing $(\mathcal {B}_{\alpha })_{\alpha <\theta }$ of all deep subsets $\mathcal {B}_{\alpha }\subseteq W_{n}$ for some $n\ge 1$. By transfinite recursion, we define an elementary chain $((W^{\alpha })_{\alpha <\theta },(*_{\alpha \beta })_{\alpha \le \beta <\theta })$ in the following way: Set

$$\begin{aligned} W^{0}:=W,\,*_{00}:=\mathop {\mathrm{identity}}\upharpoonright W_{<\infty }. \end{aligned}$$

Assume that $\lambda <\theta$ and that $W^{\alpha },$ $*_{\alpha \beta }$ are already defined for each $\alpha \le \beta <\lambda$ such that the following conditions $(1,\lambda )$ and $(2,\lambda )$ hold:

$$\begin{aligned} (1,\lambda )\;((W^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda })\;\ \text {is an elementary chain}, \end{aligned}$$

$$\begin{aligned} (2,\lambda )\;\bigcap {}^{*_{0\alpha }}\left[ \mathcal {B}_{\beta }\right] \ne \emptyset \;\ \text {for each}\;\beta <\alpha <\lambda . \end{aligned}$$

In order to construct $W^{\lambda }$ and $*_{\alpha \lambda }$ for each $\alpha <\lambda$, we have to consider two cases:

Case 1: $\lambda =\gamma +1$ is a successor ordinal.

Since $\mathcal {B}_{\gamma }$ is deep and $*_{0\gamma }$ is an elementary embedding from $W^{0}$ into $W^{\gamma }$, $^{*_{0\gamma }} \left[ \mathcal {B}_{\gamma }\right]$ is deep. By the compactness theorem, there exist a model $W^{\lambda }$ and an elementary embedding $*_{\gamma \lambda }$ from $W^{\gamma }$ into $W^{\lambda },$ such that $\bigcap ^{*_{\gamma \lambda }\circ *_{0\gamma }}\left[ \mathcal {B} _{\gamma }\right] \ne \emptyset$. We now define for each $\alpha \le \gamma$:

$$\begin{aligned} *_{\alpha \lambda }:=*_{\gamma \lambda }\circ *_{\alpha \gamma }\;\text { and}\;*_{\lambda \lambda }:=\mathop {\mathrm{identity}}\upharpoonright W_{<\infty }^{\lambda }. \end{aligned}$$

Notice that $(1,\lambda +1)$ and $(2,\lambda +1)$ are true.

Case 2: $\lambda$ is a limit ordinal.

Let $W^{\lambda }$ be the elementary limit of $((W^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda }).$ For each $\alpha <\lambda$, the elementary embedding $*_{\alpha \lambda }$ from $W^{\alpha }$ into $W^{\lambda }$ is defined in Definition 2.9.24. Set $*_{\lambda \lambda }:=\mathop {\mathrm{identity}}\upharpoonright W_{<\infty }^{\lambda }$. Notice that $(1,\lambda +1)$ and $(2,\lambda +1)$ are true.

We thus obtain an elementary chain $((W^{\lambda })_{\lambda <\theta },(*_{\alpha \lambda })_{\alpha \le \lambda <\theta }).$ Let $U:=W^{\theta }$ be its elementary limit. Set $^{*}:=*_{0\theta }.$ By Proposition 2.9.25, $*$ is an elementary embedding from W into U. It is easy to check that $\bigcap {}^{*}\left[ \mathcal {B}\right] \ne \emptyset$ if $\mathcal {B}$ is a deep subset of $W_{n}$ with $n\ge 1$. $\square$

Now we are able to finish the proof of Theorem 2.9.10: fix a superstructure V(X) of cardinality $\kappa$. Let $V=(V_{n})_{n\in \mathbb { N}_{0}}$ be the standard model over X. Then $V(X)=V_{<\infty }$. Let $\kappa ^{+}$ be the smallest cardinal number greater than $\kappa .$ Then $\kappa ^{+}$ is a regular cardinal, is, for each $\rho <\kappa ^{+}$ and each $\rho$-sequence $(\alpha _{\beta })_{\beta <\rho }$ in $\kappa ^{+}$, $\sup _{\beta <\rho }\alpha _{\beta }<\kappa ^{+}$. By transfinite recursion, we construct again an elementary chain $((V^{\alpha })_{\alpha <\kappa ^{+}},(*_{\alpha \beta })_{\alpha \le \beta <\kappa ^{+}})$: Set

$$\begin{aligned} V^{0}:=V\,\,\ \text {and}\;*_{00}:=\mathop {\mathrm{identity}}\upharpoonright V_{<\infty }. \end{aligned}$$

Fix an ordinal $\lambda$ strictly between 0 and $\kappa ^{+}$ and assume that there already exists an elementary chain $((V^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda })$ such that

We now define $V^{\lambda }$ and $*_{\alpha \lambda }$ for each $\alpha \le \lambda$:

First assume that $\lambda =\gamma +1$ is a successor ordinal.

Then, by Lemma 2.9.26, there exists a model $V^{\lambda }$ and an elementary embedding $*_{\gamma \lambda }$ from $V^{\gamma }$ into $V^{\lambda }$, such that for each $n\ge 1$ and each deep $\mathcal {B} \subseteq V_{n}^{\gamma }$, $\bigcap ^{*_{\gamma \lambda }}\left[ \mathcal {B}\right] \ne \emptyset .$ For each $\alpha <\lambda$ set

$$\begin{aligned} *_{\alpha \lambda }:=*_{\gamma \lambda }\circ *_{\alpha \gamma }\;\ \text {and}\;*_{\lambda \lambda }:=\mathop {\mathrm{identity}}\upharpoonright V_{<\infty }^{\lambda }. \end{aligned}$$

Note that $((V^{\alpha })_{\alpha <\lambda +1},(*_{\alpha \beta })_{\alpha \le \beta <\lambda +1})$ is an elementary chain such that $\diamondsuit \left( \lambda +1\right)$ is true.

Now assume that $\lambda$ is a limit number.

Let $V^{\lambda }$ be the elementary limit of $((V^{\alpha })_{\alpha <\lambda },(*_{\alpha \beta })_{\alpha \le \beta <\lambda })$. For $\alpha \le \lambda$ let $*_{\alpha \lambda }$ be the elementary embedding from $V^{\alpha }$ into $V^{\lambda }$, defined in Definition 2.9.24. Notice that $((V^{\alpha })_{\alpha <\lambda +1},(*_{\alpha \beta })_{\alpha \le \beta <\lambda +1})$ is an elementary chain and $\diamondsuit \left( \lambda +1\right)$ is true.

Let $W:=V^{\kappa ^{+}}$ be the elementary limit of $((V^{\alpha })_{\alpha <\kappa ^{+}},(*_{\alpha \beta })_{\alpha \le \beta <\kappa ^{+}}).$ Set ${}^{*}:=$ $*_{0\kappa ^{+}}$ Then ${}^{*}$ is an elementary embedding from V into W. To prove that W is $\kappa ^{+}$ -saturated, fix $n\ge 1$ and a deep set $\mathcal {B}\subseteq W_{n}$ with $\mathop {\mathrm{card}}(\mathcal {B})<\kappa ^{+}$. Since $\kappa ^{+}$ is a regular cardinal, there exists a $\delta <\kappa ^{+}$ such that

$$\begin{aligned} \mathcal {B}\subseteq \left\{ \overline{(b,\beta )}\mid \beta <\delta \;\ \text {and}\;b\in V_{n}^{\beta }\right\} . \end{aligned}$$

Set $\mathcal {B}^{\prime }:=\left\{ ^{*_{\beta \delta }}(b)\mid \beta <\delta ,\;b\in V_{n}^{\beta }\;\ \text {and}\;\overline{(b,\beta )}\in \mathcal {B}\right\} .$ Notice that $^{*_{\delta \kappa ^{+}}}[\mathcal {B} ^{\prime }]=\mathcal {B}$ and that $\mathcal {B}^{\prime }$ is deep. From the construction of $((V^{\alpha })_{\alpha <\kappa ^{+}},(*_{\alpha \beta })_{\alpha \le \beta <\kappa ^{+}})$ it follows that $\bigcap ^{*_{\delta (\delta +1)}}\left[ \mathcal {B}^{\prime }\right] \ne \emptyset$. Since $*_{(\delta +1)\kappa ^{+}}$ is an elementary embedding from $V^{\delta +1}$ into $V^{\kappa ^{+}}=W,$ by (EC 3),

$$\begin{aligned} \bigcap \mathcal {B}=\bigcap {}^{*_{\delta \kappa ^{+}}}\left[ \mathcal {B} ^{\prime }\right] =\bigcap {}^{*_{(\delta +1)\kappa ^{+}}\circ \;*_{\delta (\delta +1)}}\left[ \mathcal {B}^{\prime }\right] \ne \emptyset . \end{aligned}$$

Now set $Y:=W_{0}.$ Then $Y={}^{*}V_{0}={}^{*}X$, and V(Y) is a superstructure over Y with $W_{<\infty }\subseteq V(Y)$.

Finally, we will prove that the internal entities in V(Y) coincide with the elements in $W_{<\infty }$: Let $B\in V(Y)$ be internal. Then there exists an $A\in V(X){\setminus } X=V_{1\le }$ with $B\in {}^{*}A$. Since ${}^{*}A\in W_{1\le }$, we obtain the fact that $B\in W_{<\infty }$. Now assume that $B\in W_{<\infty }$. Then $B\in W_{n}={}^{*}V_{n}$ for some $n\in \mathbb {N}_{0}$. This proves that B is internal. $\square$

We end this appendix with an important application of Theorem 2.9.10.

Theorem 2.9.27

Fix an internal set A in $V(^{*}X)$, a set I in V(X) and a mapping $f:I\rightarrow A$. Then there exists an internal mapping $F:\ ^{*}I\rightarrow A$ with $f(i)=F(^{*}i)$ for all $i\in I$. Since the mapping $*$ is injective, we may identify $i\in I$ with $^{*}i\in \ ^{*}I$, thus F is an internal extension of f.

Proof

If I is standard finite, then f is internal, thus f is its own internal extension (see Proposition 2.9.14 Part (2)). Therefore, we may assume that I is infinite, thus $A\ne \emptyset$. Let $\mathcal {E}\left( I\right)$ denote the set of all finite subsets of I. For all $E\in \mathcal {E}\left( I\right)$ define,

$$\begin{aligned} \mathcal {B}_{E}:=\left\{ F:\ ^{*}I\rightarrow A\mid F\text { is internal and }\forall i\in E\left( f(i)=F(^{*}i)\right) \right\} . \end{aligned}$$

Note that $\mathcal {B}_{E}$ is internal for each $E\in \mathcal {E}\left( I\right)$ and that there exists an internal $F:\ ^{*}I\rightarrow A$ such that $f(i)=F(^{*}i)$ for all $i\in E$. Therefore, $\left\{ \mathcal { B}_{E}\mid E\in \mathcal {E}\left( I\right) \right\}$ has the finite intersection property. Since its cardinality is smaller than $\kappa ^{+}$, there exists an $F\in \mathcal {B}_{E}$ for all $E\in \mathcal {E}\left( I\right)$. This F is an internal extension of f. $\square$