I have been starting to study set theory and cardinal arithmetic and I have come across the following question which I am not very sure on how to tackle:

Assuming ZF and the axiom of constructibility, $\alpha > \omega, V_\alpha = L_\alpha \iff \alpha = \aleph_\alpha$.

Here, $\alpha$ is an ordinal, $\omega =\{0,1,...\}$ is the first infinite ordinal and the $V_\alpha$ and $L_\alpha$ correspond to the von Neumann universe and the constructible universe as in the following recursive definitions:

$V_\emptyset = \emptyset$.

$V_{\alpha+1} = P(V_\alpha)$ for all ordinals $\alpha$.

$V_\gamma = \cup_{\beta < \gamma} V_\beta$.


$L_\emptyset = \emptyset$.

$L_{\alpha+1} = Def(L_\alpha)$ for all ordinals $\alpha$.

$L_\gamma = \cup_{\beta < \gamma} L_\beta$.

I know that since ZF+constructibility imply choice, I can use the fact that the cardinal of $L_\alpha$ is the same as that of $\alpha$, but I am not sure on how to use that since I don't know what the cardinal of $V_\alpha$ is.

Could someone lend me a hand?

  • $\begingroup$ What is the size of $V_\alpha$? $\endgroup$ – Andrés E. Caicedo Feb 27 '18 at 1:01
  • $\begingroup$ If $\alpha$ is a successor ordinal, say $\alpha = \beta + 1$, it should be $2^{|V_\beta|}$ but I don't know if that's the case when $\alpha$ is a limit ordinal. $\endgroup$ – Kevin Smith Feb 27 '18 at 1:10
  • $\begingroup$ No, don't compute it as a recursive function, but rather as a specific formula. $\endgroup$ – Andrés E. Caicedo Feb 27 '18 at 1:21
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    $\begingroup$ $\mathsf{GCH}$ does not follow from $\mathsf{ZFC}$. $\endgroup$ – Andrés E. Caicedo Feb 27 '18 at 2:26
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    $\begingroup$ @KevinSmith (Hopefully you meant follows from V=L, not from ZFC :)) $\endgroup$ – spaceisdarkgreen Feb 27 '18 at 2:26

Define the sequence of beth numbers by $\beth_0=\aleph_0$, $\beth_{\alpha+1}=2^{\beth_\alpha}$ and $\beth_\lambda=\sup_{\beta<\lambda}\beth_\beta$ for all ordinals $\alpha$ and all limit ordinals $\lambda$. It should be immediate that $|V_{\omega+\alpha}|=\beth_\alpha$ for all $\alpha$. Also, an easy recursion establishes that $|L_\alpha|=|\alpha|$ for $\alpha$ infinite. If $V=L$, then $\mathsf{GCH}$ holds, and we also have that $\beth_\alpha=\aleph_\alpha$ for all $\alpha$.

It follows that if $\alpha$ is infinite and $V_\alpha=L_\alpha$, then either $\alpha=\omega$ or $\aleph_\alpha=|\alpha|$: Note that if $\alpha\ge\omega^2$ then $\omega+\alpha=\alpha$ so, for $\alpha\ge\omega^2$, $|V_\alpha|=\aleph_\alpha$. On the other hand, if $\omega\le\alpha\le\omega^2$ then $|\alpha|=\aleph_0$ while $|V_\alpha|$ is uncountable, unless $\alpha=\omega$. Now, for any $\beta$, $\beta\le\aleph_\beta$ so $|\alpha|\le\aleph_{|\alpha|}\le\aleph_\alpha$, and equality holds if and only if $|\alpha|=\alpha=\aleph_\alpha$ (which simplifies to $\alpha=\aleph_\alpha$, since $\aleph_\alpha$ is a cardinal).

We now argue that, conversely, if $\alpha=\aleph_\alpha$, then $L_\alpha=V_\alpha$. Clearly $L_\alpha\subseteq V_\alpha$. For the converse, note that $V_\alpha$ is closed under transitive closures, and that if the transitive closure of a set $x$ belongs to $L_\alpha$, then so does $x$ itself. From this, it suffices to show that if $x\in V_\alpha$ is transitive, then $x\in L_\alpha$.

Now, let $x\in V_\alpha$ be transitive. Find a limit ordinal $\theta$ large enough that $x\in L_\theta$ and consider an elementary substructure $N\prec L_\theta$ as small as possible subject to the condition that $\{x\}\cup x\subset N$. Note that $|x|<|V_\alpha|$, so $|N|<\aleph_\alpha$. The Mostowski collapse of $N$ is an $L_\beta$ with $x\in L_\beta$ and $\beta<\aleph_\alpha$, so $\beta<\alpha$ and $x\in L_\alpha$, as required.

(For completeness, it should probably be mentioned that $V_\alpha=L_\alpha$ also holds for all $\alpha\le\omega$.)

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    $\begingroup$ (For even more completeness, it should be remarked that $V_\alpha=L_\alpha$ for all fixed point $\alpha$ implies $V=L$ holds. :)) $\endgroup$ – Asaf Karagila Feb 28 '18 at 15:33
  • $\begingroup$ Great answer, thank you very much! $\endgroup$ – Kevin Smith Mar 4 '18 at 17:44

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