1
$\begingroup$

In a constructible universe (ZFC + V=L), is there any known upper bound for the constructibility orders of all elements of the continuum, i.e. some separately described ordinal $\alpha$ such that we can prove $\mathcal{P}(\omega)\subset L_\alpha$ ? For example (under some large cardinal axiom), can it be proven that the first inaccessible cardinal is such an upper bound, or can this cardinal still fail at this ? I intuitively suspect undecidabilities in this matter but am no expert in the field. Thanks.

New contributor
user27887 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • 5
    $\begingroup$ Your desired upper bound is $\omega_1$, as follows from the condensation lemma $\endgroup$ – Wojowu Oct 18 at 12:09
  • $\begingroup$ @Wojowu $\omega_1$ is also the least upper bound, so $\mathcal{P}(\omega)\subseteq L_\alpha$ if and only if $\alpha\ge \omega_1$. It follows from a simple cardinal comparison. $\endgroup$ – Hanul Jeon 2 days ago
  • 5
    $\begingroup$ It seems worthwhile to mention that the answer given by @Wojowu is the main point in Gödel's proof that V=L implies the continuum hypothesis. $\endgroup$ – Andreas Blass 2 days ago
4
$\begingroup$

For simplicity, assume $\mathsf{V=L}$ below.


In fact the situation is as simple as it could possibly be:

For each ordinal $\alpha$, we have $\mathcal{P}(\alpha)\subseteq L_{\vert\alpha\vert^+}$, and moreover for each $\beta<\vert\alpha\vert^+$ there is some $X\subset\alpha$ with $X\in L_{\vert\alpha\vert^+}\setminus L_\beta$.

The second clause follows from the first clause by a simple counting argument (think about the size of $L_\beta$); the first clause is where all the action is. This follows from the condensation lemma, which is really the key fact about $L$:

Suppose $M$ is an elementary submodel of $L_\kappa$ for some uncountable cardinal $\kappa$. Then the Mostowski collapse of $M$ is $L_\gamma$ for some $\gamma\le\kappa$.

(Actually this is a weak version of the condensation lemma, but it's enough for us.) To see how this can be applied, let's use it to show $\mathbb{R}^L\subseteq L_{\omega_1}$ (which will answer the question in the OP). Fix a real $r\in L$. Taking $\kappa$ large enough, let $M$ be a countable elementary submodel of $L_\kappa$ with $r\in M$. By condensation, let $L_\gamma$ be the Mostowski collapse of $M$. Since $M$ is countable we have $\gamma<\omega_1$. Moreover, the Mostowski collapse map $\mu:M\cong L_\gamma$ doesn't move $r$ (this is a good exercise) so we have $r=\mu(r)\in \mu[M]=L_\gamma$.

More generally, suppose $X\subseteq\alpha$. Fix some uncountable cardinal $\kappa$ with $X\in L_\kappa$, and let $M$ be an elementary submodel of $L_\kappa$ with $\alpha\subseteq M$, $\vert M\vert=\vert L_\alpha\vert$, and $X\in M$. By condensation let $\mu: M\cong L_\gamma$ be the Mostowski collapse map for $M$. By choice of $M$ we have $\mu(X)=X$, so again we get $X=\mu(X)\in \mu[M]=L_\gamma$.

Note that as a corollary this gives $\mathsf{ZFC+V=L\vdash GCH}$.

| cite | improve this answer | |
$\endgroup$

Your Answer

user27887 is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.