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.
-
5$\begingroup$ Your desired upper bound is $\omega_1$, as follows from the condensation lemma $\endgroup$– WojowuOct 18, 2020 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 JeonOct 18, 2020 at 14:59
-
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 BlassOct 18, 2020 at 16:29
-
$\begingroup$ Now reacting very late for a suspicion of misunderstanding. I already knew before about the comparison in terms of cardinality, that is the fact that constructibility implies the continuum hypothesis. My question was in terms of set inclusion, which looked a priori quite different. Namely, my wonder was about whether there could be "a break" in the timeline of construction of subsets of $\omega$, with some subsets only constructible "much later", at a higher cardinality step. Sorry I have not the level to decipher the given argument to check which question it answers. $\endgroup$– user27887Jun 4, 2022 at 20:43
1 Answer
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 $\mathsf{ZFC+V=L\vdash GCH}$.