A Survey on Ordinal Notations Around the Bachmann–Howard Ordinal

Abstract

Various ordinal functions which in the past have been used to describe ordinals not much larger than the Bachmann–Howard ordinal are set into relation. Special efforts are made to reveal the intrinsic connections between Feferman’s $\theta$-functions and the Bachmann hierarchy.

This is a revised and slightly extended version of [8].

Appendix

This appendix is devoted to the proof of Lemmata 2.1, 2.2d.

Lemma A.1

• (a) $\lambda \in Lim\;\Rightarrow \;0<\lambda [0]$.

• (b) $\gamma +\Omega ^\beta<\Omega ^\alpha \; \& \;\eta<\Omega \;\Rightarrow \;\gamma +\Omega ^\beta \eta <\Omega ^\alpha$.

• (c) $\lambda =_\mathrm{NF}\gamma +\Omega ^\beta \eta \not \in \mathrm{ran}(F_0)\; \& \;\Omega ^\alpha <\lambda \;\Rightarrow \;\Omega ^\alpha \le \lambda [0]$.

Proof of (c): From $\Omega ^\alpha <\lambda =\gamma +\Omega ^\beta \eta$ by (b) we get $\Omega ^\alpha \le \gamma +\Omega ^\beta$. If $\eta \in Lim$ then $\lambda [0]=\gamma +\Omega ^\beta$. If $1<\eta =\eta _0{+}1$ then $\gamma +\Omega ^\beta \le \gamma +\Omega ^\beta \eta _0\le \lambda [0]$. If $\eta =1$ then $0<\gamma$ (since $\lambda \not \in \mathrm{ran}(F_0)$) and therefore $\Omega ^{\beta +1}\le \gamma$ which together with $\Omega ^\alpha <\lambda =\gamma +\Omega ^\beta$ yields $\Omega ^\alpha \le \gamma \le \lambda [0]$.

Lemma A.2

$\lambda =_\mathrm{NF}F_\alpha (\beta )\; \& \;0<\beta \;\Rightarrow \;F_\alpha (\beta [n])\le \lambda [n]$.

Proof

1. $\beta \in Lim$: $F_\alpha (\beta [n])=\lambda [n]$.

2. $\beta =\beta _0{+}1$:

2.1. $\alpha =0$: $F_\alpha (\beta [n])=\Omega ^{\beta _0}\le \Omega ^{\beta _0}{\cdot }(1{+}n)=\lambda [n]$.

2.2. $\alpha >0$: $F_\alpha (\beta [n])=F_\alpha (\beta _0)<\lambda ^-\le \lambda [n]$.

Lemma A.3

$F_\zeta (\mu )<\lambda \le F_\zeta (\mu {+}1)\;\Rightarrow \;F_\zeta (\mu )\le \lambda [0]$.

Proof

0. $\lambda =F_\zeta (\mu +1)$:

0.1. $\zeta =0$: $\lambda =\Omega ^{\mu +1}$, $\lambda [\xi ]=\Omega ^\mu (1{+}\xi )$, $\lambda [0]=F_0(\mu )$.

0.2. $\zeta >0$: $F_\zeta (\mu )<\lambda ^-<F_{\zeta [0]}(\lambda ^-)=\lambda [0]$.

1. $\lambda <F_\zeta (\mu +1)$:

1.1. $\lambda =_\mathrm{NF}\gamma +\Omega ^\beta \eta \not \in \mathrm{ran}(F_0)$:

$F_\zeta (\mu )\in \mathrm{ran}(F_0)\; \& \;F_\zeta (\mu )<\lambda \;\buildrel \mathrm{L.A1{c}}\over \Rightarrow \; F_\zeta (\mu )\le \lambda [0]$.

1.2. $\lambda =_\mathrm{NF}F_\alpha (\beta )$: Then $\alpha <\zeta$ and thus $F_\alpha (F_\zeta (\mu ))=F_\zeta (\mu )<F_\alpha (\beta )$. Hence $F_\zeta (\mu )<\beta$ and therefore $F_\zeta (\mu )\buildrel \mathrm{IH}\over \le \beta [0]\le F_\alpha (\beta [0]) \buildrel \mathrm{A2}\over \le \lambda [0]$.

Definition. $r(\gamma ):=\left\{ \begin{array}{ll} -1 &{}\hbox {if}\ \gamma \not \in \mathrm{ran}(F_0)\\ \alpha &{}\hbox {if} \ \gamma =_\mathrm{NF}F_\alpha (\beta )\\ \gamma &{}\hbox {if} \ \gamma =\Lambda \end{array}\right.$

Lemma A.4

• (a) $r(F_\alpha (\beta ))=\max \{\alpha ,r(\beta )\}$.

• (b) $\lambda [0]<\delta <\lambda \;\Rightarrow \;r(\delta )\le r(\lambda )$.

• (c) $\lambda =_\mathrm{NF}F_\alpha (\beta )\; \& \;\beta \not \in Lim\; \& \;\lambda ^-<\eta <\lambda \;\Rightarrow \;\lambda ^-\le \eta [1]$.

Proof

(a) 1. $\beta <F_\alpha (\beta )$:

Then $r(F_\alpha (\beta ))=\alpha$ and ($r(\beta )=-1$ or $\beta =_\mathrm{NF}F_{\beta _0}(\beta _1)$ with $\beta _0\le \alpha$).

2. $\beta =F_\alpha (\beta )$: Then $\beta =_\mathrm{NF}F_{\beta _0}(\beta _1)$ with $\alpha <\beta _0=r(\beta )=r(F_\alpha (\beta ))$.

(b) 1. $\lambda =_\mathrm{NF}\gamma +\Omega ^\beta \eta \not \in \mathrm{ran}(F_0)$:

1.1. $\eta \in Lim$: $\gamma +\Omega ^\beta =\lambda [0]<\delta <\gamma +\Omega ^\beta \eta \; \buildrel \mathrm{A1{b}}\over \Rightarrow \;\delta \not \in \mathrm{ran}(F_0)$.

1.2. $\eta =\eta _0{+}1$:

$\gamma +\Omega ^\beta \eta _0<\lambda [0]<\delta <\gamma +\Omega ^\beta (\eta _0{+}1)\not \in \mathrm{ran}(F_0) \; \& \;\Omega ^{\beta +1}|\gamma \;\Rightarrow \;\delta \not \in \mathrm{ran}(F_0)$.

2. $\lambda =_\mathrm{NF}F_\alpha (\beta )$: If $\lambda <F_{\alpha +1}(0)$ then also $\delta <F_{\alpha +1}(0)$ and thus $r(\delta )\le \alpha =r(\lambda )$. Otherwise there exists $\mu$ such that $F_{\alpha {+}1}(\mu )<\lambda <F_{\alpha +1}(\mu {+}1)$. Then by Lemma A3 we get $F_{\alpha {+}1}(\mu )\le \lambda [0]<\delta <F_{\alpha +1}(\mu {+}1)$ and thus $\delta \not \in \mathrm{ran}(F_{\alpha +1})$, i.e. $r(\delta )\le \alpha =r(\lambda )$.

3. $\lambda =\Lambda$: $r(\delta )<\Lambda =r(\Lambda )$.

(c) For $\beta =0\vee \eta =\eta _0{+}1$ the claim is trivial. Assume now $\beta =\beta _0{+}1\; \& \;\eta \in Lim$.

$F_\alpha (\beta _0)<\eta <F_\alpha (\beta _0{+}1)\;\buildrel \mathrm{L.A3}\over \Rightarrow \; \lambda ^-=F_\alpha (\beta _0){+}1\le \eta [0]+1\le \eta [1]$.

Lemma 2.1

$\lambda =_\mathrm{NF}F_\alpha (\beta )\; \& \;\beta \in Lim\; \& \;1\le \xi <\tau _{\beta }\;\Rightarrow \;\lambda [\xi ]=_\mathrm{NF}F_\alpha (\beta [\xi ])$.

Proof

We have $\lambda [\xi ]=F_\alpha (\beta [\xi ])\; \& \;\beta [0]<\beta [\xi ]<\beta$. By Lemma A4b this yields $\lambda [\xi ]=F_\alpha (\beta [\xi ])\; \& \;r(\beta [\xi ])\le r(\beta )\le \alpha$, whence $\lambda [\xi ]=_\mathrm{NF}F_\alpha (\beta [\xi ])$.

Lemma 2.2d

$\xi {+}1<\tau _{\lambda }\; \& \;\lambda [\xi ]<\delta \le \lambda [\xi {+}1]\;\Rightarrow \;\lambda [\xi ]\le \delta [1]$.

Proof by induction on $\delta \#\lambda$: If $r(\delta )<r(\lambda [\xi ])$ then, by Lemma A4b, $\lambda [\xi ]\le \delta [0]$.

Assume now that $r(\lambda [\xi ])\le r(\delta )$ ($\dagger$).

1. $\lambda =_\mathrm{NF}\gamma +\Omega ^\beta \eta \not \in \mathrm{ran}(F_0)$.

1.1. $\eta \in Lim$:

$\gamma +\Omega ^\beta (1{+}\xi )=\lambda [\xi ]<\delta <\lambda [\xi {+}1]= \gamma +\Omega ^\beta (1{+}\xi )+\Omega ^\beta \;\Rightarrow \;\lambda [\xi ]\le \delta [0]$.

1.2. $\eta =\eta _0{+}1$: $\gamma +\Omega ^\beta \eta _0+\Omega ^\beta [\xi ]=\lambda [\xi ]<\delta \le \lambda [\xi {+}1]=\gamma +\Omega ^\beta \eta _0+\Omega ^\beta [\xi {+}1]\;\Rightarrow \delta =(\gamma +\Omega ^\beta \eta _0)+\delta _0$ with $\Omega ^\beta [\xi ]<\delta _0\le \Omega ^\beta [\xi {+}1]\;\Rightarrow \delta [0] =\gamma +\Omega ^\beta \eta _0+\delta _0[0]$ with $\Omega ^\beta [\xi ]\buildrel \mathrm{IH}\over \le \delta _0[0]\;\Rightarrow \; \lambda [\xi ]\le \delta [0]$.

2. $\lambda =_\mathrm{NF}F_\alpha (\beta )\; \& \;\beta \in Lim$: Then (1) $\lambda [\xi ]=F_\alpha (\beta [\xi ])$, and (2) $\lambda [\xi ]<\delta <\lambda$.

From $\alpha \buildrel \mathrm{(1)}\over \le r(\lambda [\xi ])\buildrel \mathrm{(\dagger )}\over \le r(\delta )\buildrel \mathrm{(2),L.A4{b}}\over \le r(\lambda )=\alpha$ we get $r(\delta )=\alpha$, i.e. $\delta =_\mathrm{NF}F_\alpha (\eta )$ for some $\eta$. Now from $\lambda [\xi ]<\delta \le \lambda [\xi {+}1]$ we conclude $\beta [\xi ]<\eta \le \beta [\xi {+}1]$ and then, by IH, $\beta [\xi ]\le \eta [0]$. Hence $\lambda [\xi ]\le F_\alpha (\eta [0])\buildrel \mathrm{L.A2}\over \le \delta [0]$.

3. $\lambda =_\mathrm{NF}F_\alpha (\beta )\; \& \;\beta \not \in Lim$:

3.1. $\alpha =0$: Then $\beta =\beta _0{+}1$, and $\lambda [\xi ]=\Omega ^{\beta _0}(1{+}\xi )<\delta \le \Omega ^{\beta _0}(1{+}\xi )+\Omega ^{\beta _0}$ implies $\lambda [\xi ]\le \delta [0]$.

3.2. $\alpha =\alpha _0{+}1$: Then $\lambda [\xi ]=F_{\alpha _0}^{\xi +1}(\lambda ^-)$.

Hence, by $(\dagger )$, $\delta =_\mathrm{NF}F_\zeta (\eta )$ with $\alpha _0\le \zeta$.

3.2.1. $\alpha _0<\zeta$: $\lambda ^-<F_\zeta (\eta )\;\Rightarrow \;\lambda [\xi {+}1]=F^{\xi +2}_{\alpha _0}(\lambda ^-)<F_\zeta (\eta )$. Contradiction.

3.2.2. $\zeta =\alpha _0$: Then from $F_{\alpha _0}^{\xi +1}(\lambda ^-)=\lambda [\xi ]<\delta =F_{\alpha _0}(\eta )\le \lambda [\xi {+}1]$ we conclude $F_{\alpha _0}^\xi (\lambda ^-)<\eta \le \lambda [\xi ]$. As we will show, this implies $F_{\alpha _0}^\xi (\lambda ^-)\le \eta [1]$, thence $F_{\alpha _0}^{\xi +1}(\lambda ^-)\le F_\zeta (\eta [1])\buildrel \mathrm{L.A2}\over \le \delta [1]$.

Proof of $F_{\alpha _0}^\xi (\lambda ^-)\le \eta [1]$:

(i) $\xi =n{+}1$: Then the claim follows by IH from $\lambda [n]=F^\xi _{\alpha _0}(\lambda ^-)<\eta \le \lambda [n{+}1]$.

(ii) $\xi =0$: $\lambda ^-<\eta <\lambda \;\buildrel \mathrm{L.A4{c}}\over \Rightarrow \;\lambda ^-\le \eta [1]$.

3.3. $\alpha \in Lim$: $\lambda [\xi ]=F_{\alpha [\xi ]}(\lambda ^-)$, and by $(\dagger )$ we have $\delta =_\mathrm{NF}F_\zeta (\eta )$ with $\alpha [\xi ]\le \zeta$.

3.3.1. $\alpha [\xi {+}1]<\zeta$: $\lambda ^-<F_\zeta (\eta )\;\Rightarrow \;F_{\alpha [\xi +1]}(\lambda ^-)<F_\zeta (\eta )=\delta$. Contradiction.

3.3.2. $\alpha [\xi ]<\zeta \le \alpha [\xi {+}1]$:

(i) $\eta \in Lim$: Then $\lambda ^-<\delta [1]=F_\zeta (\eta [1])$ (for $\beta =0$, $\lambda ^-=0$. If $\beta =\beta _0{+}1$, then $F_\alpha (\beta _0)<\delta <F_\alpha (\beta _0{+}1)$ and thus, by Lemma A3, $F_\alpha (\beta _0)\le \delta [0]$).

$\alpha [\xi ]<\zeta \; \& \;\lambda ^-<\delta [1]\;\Rightarrow \;\lambda [\xi ]=F_{\alpha [\xi ]}(\lambda ^-)<\delta [1]$.

(ii) $\eta \not \in Lim$: By IH $\alpha [\xi ]\le \zeta [1]$. Further $\lambda ^-\le \delta ^-$.

From $\alpha [\xi ]\le \zeta [1]$ and $\lambda ^-\le \delta ^-$ we conclude $\lambda [\xi ]=F_{\alpha [\xi ]}(\lambda ^-)\le F_{\zeta [1]}(\delta ^-)\le \delta [1]$.

3.3.3. $\zeta =\alpha [\xi ]$: This case is similar to 3.2.2(ii):

$\lambda [\xi ]=F_\zeta (\lambda ^-)<F_\zeta (\eta )<F_\alpha (\beta )\;\Rightarrow \;\lambda ^-<\eta <F_\alpha (\beta )\;\Rightarrow \lambda [\xi ]=F_\zeta (\lambda ^-)\buildrel \mathrm{L.A4{c}}\over \le F_\zeta (\eta [1]) \buildrel \mathrm{L.A2}\over \le \delta [1]$.

4. $\lambda =\Lambda$: This case is very similar to 3.3, but considerably simpler.