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Are there stronger alternative axioms to the axiom of constructibility that additionally tell us an upper bound on how long it can take for things in $V$ to appear in $L$?

For example, let $\kappa \leftarrow \lambda$ denote ordinal exponentiation $\kappa^\lambda$.

$$ \small\textbf{STRAWMAN:}\;\; V_{\alpha} \subset L_{\omega \leftarrow \alpha} $$

I'm curious because I was reading about the differences between $V$ and $L$ and saw a handful of results on Wikipedia relating $V_\alpha$ and $L_\alpha$.

What follows is background, motivation, and my attempt to make sense of the material.

I was reading this answer to this question, which is substantially more advanced than my background, and stumbled on the notion of a Silver indiscernible. I tried to look up what a Silver indiscernible is and ended up on this page describing $0^\#$. That led me to this discussion of the Gödel constructible universe, which is different from the von Neumann universe that I've seen before but don't understand deeply. I tried to understand what the difference between these things is, and noticed there's an alternate axiom, the Axiom of Constructibility that asserts their equality. That got me wondering if we can consistently insist on bounds for when things in $V$ appear in $L$ if we take the axiom of constructibility (and so a set in $V$ always appears in $L$ eventually).

A given "slice" of the von Neumann universe $V_\alpha$ is indexed by an ordinal $\alpha$.

$$ V_0 = \emptyset $$ $$ V_{\beta+1} = P(V_{\beta}) \;\; \text{if $\beta$ is an ordinal number} $$ $$ V_{\lambda} = \bigcup_{\alpha < \lambda} V_{\alpha} \;\; \text{if $\lambda$ is a limit ordinal} $$

And the universe as a whole is the union of all such slices.

Gödel's construction is similar, but in the $\beta+1$ case, it only takes some of the subsets, not all of them.

$$ L_0 = \emptyset $$ $$ L_{\beta+1} = \text{Def}(L_\beta) \;\; \text{if $\beta$ is an ordinal number} $$ $$ L_{\lambda} = \bigcup_{\alpha < \lambda} L_{\alpha} \;\; \text{if $\lambda$ is a limit ordinal} $$

Where $\text{Def}$ is defined as follows.

Let's define a three-argument version of $\text{Def}$ where $\Phi$ is a well-formed formula and $\vec{z}$ is a finite vector of sets.

$$ \text{Def}(X, \Phi, \vec{z}) = 1 \;\;\text{iff}\;\;\text{each element of $\vec{z}$ is in $X$ and $(X, \in)$ is a model of $\Phi(\vec{z})$ } $$

$$ \text{otherwise $\text{Def}(X, \Phi, \vec{z})$ is zero } $$

So, a triple of $X, \Phi, \vec{z}$ gets sent to one if and only if $(X, \in)$ is a model of $\Phi(\vec{z})$ and $\vec{z}$, considered as a set, is a subset of $X$.

We can then define the one-argument version of $\text{Def}$ as follows.

$$ y \in \text{Def}(X) \;\;\text{iff}\;\; \text{there exists a $\Phi$ and $\vec{z}$ so that $\vec{z}_0 = y$ and $\text{Def}(X, \Phi, \vec{z}) = 1$} $$

It's clear that $\text{Def}(X)$ is a subset of $X$ and thus each $L_\alpha$ is clearly a subset of the corresponding $V_\alpha$.

The axiom of constructibility asserts that $V = L$, but doesn't say anything about how long we need to wait in $L$ to get an element that appears in $V_\alpha$.

I'm wondering if there are stronger axioms that do insist on a bound up front.

Greg Nisbet
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1 Answers1

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This is actually already completely determined by $V=L$. In particular, note that $|L_\alpha|=|\alpha|$ for all infinite $\alpha$, while $|V_{\omega+\alpha}|=\beth_\alpha$ for all $\alpha$ (and this is equal to $\aleph_\alpha$ assuming $V=L$). So, $V_{\omega+\alpha}$ cannot be contained in $L_\beta$ for any $\beta<\omega_\alpha$, simply for cardinality reasons. Conversely, I claim that $V=L$ implies $V_{\omega+\alpha}\subseteq L_{\omega_\alpha}$. You can prove this by induction on $\alpha$. Limit steps are trivial; for successor steps, suppose $V_{\omega+\alpha}\subseteq L_{\omega_\alpha}$ and we wish to show that $V_{\omega+\alpha+1}\subseteq L_{\omega_{\alpha+1}}$. Since $V_{\omega+\alpha+1}=P(V_{\omega_\alpha})$, this follows from the fact that all constructible subsets of $L_{\omega_\alpha}$ are contained in $L_{\omega_{\alpha+1}}$ (this is a consequence of Gödel's condensation lemma, by essentially the same argument as the proof that $V=L$ implies GCH).

So, $V=L$ implies that for every $\alpha$, the least $\beta$ such that $V_{\omega+\alpha}\subseteq L_\beta$ is $\beta=\omega_\alpha$.

Eric Wofsey
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