# Is it possible to prove $|V_\kappa|=\kappa$ for strongly inaccessible $\kappa$ without AC?

First, let me note that my definition of "strongly inaccessible" is that a nonzero ordinal $\kappa$ is strongly inaccessible if ${\rm cf}(\kappa)=\kappa$ and $\forall\alpha<\kappa, {\cal P}(\alpha)\prec\kappa$ in the sense of an injection from ${\cal P}(\alpha)$ to $\kappa$. (Thus $\omega$ is strongly inaccessible, although it is probably not important.)

The only way I know to prove that $|V_\kappa|=\kappa$ is to prove that $V_\alpha\prec\kappa$ for all $\alpha<\kappa$ by induction. If $V_\beta\prec\kappa$, then $|V_\beta|=\gamma$ for some $\gamma<\kappa$, and so $|V_{\beta+1}|=|{\cal P}(V_\beta)|=|{\cal P}(\gamma)|<\kappa$ since $\kappa$ is a strong limit. For limit ordinals $\alpha<\kappa$, $V_\alpha=\bigcup_{\beta<\alpha}V_\beta$, so since $V_\beta\prec\kappa$, $V_\alpha\preceq\alpha\times\kappa\prec\kappa\times\kappa\approx\kappa$ (where the inequalities correspond to injections and $\alpha\times\kappa$ is a cross product) by AC. Then we can finish by noting again that $V_\alpha\prec\kappa$ implies $V_\kappa=\bigcup_{\alpha<\kappa}V_\alpha\preceq\bigsqcup_{\alpha<\kappa}\kappa=\kappa\times\kappa\approx\kappa$, and $\kappa\subseteq V_\kappa$ so that $V_\kappa\approx\kappa$ by Schroder-Bernstein.

In the proof above, I used AC twice, in the claim that $\forall \alpha<\kappa,A_\alpha\preceq B$ implies $\bigcup_{\alpha<\kappa}A_\alpha\preceq\kappa\times B$. Is it possible to generalize the approach of this question to avoid AC altogether? Also, it appears I only used the strong limit property above, without ever using the fact that $\kappa$ is regular. That can't be right; is it true that $|V(\beth_\omega)|=\beth_\omega$?

Edit: The above transfinite induction proof has a flaw in it in the limit step. For limit ordinals $\alpha<\kappa$, $V_\alpha=\bigcup_{\beta<\alpha}V_\beta$, so since $V_\beta\prec\kappa$, $V_\alpha\preceq\alpha\times\kappa\preceq\kappa\times\kappa\approx\kappa$ by AC, but the inequality is not strict. To make it strict, we note that if $|V_\alpha|=\kappa$, then the function $f:\alpha\to\kappa$ defined by $f(\beta)=|V_\beta|$ is a cofinal map, so ${\rm cf}(\kappa)=\kappa\le\alpha$, a contradiction. Thus $|V_\alpha|<\kappa$. So it is not true that strong limits are sufficient to ensure that $|V_\kappa|=\kappa$.

• How is this "elementary"? :-) Jun 6, 2013 at 4:17
• @Asaf I feel like I just graduated. ;) Jun 6, 2013 at 4:18
• You’ve gone beyond elementary aleph and reached not-so-elementary beth. Just wait till you reach thoroughly-mysterious gimel! :-) Jun 6, 2013 at 4:27
• @Brian: $\LaTeX$ has $\gimel$ and also $\daleth$, but I'm not aware of notational uses of the latter... (and it kinda looks like Resh anyway) Jun 7, 2013 at 2:55
• @Asaf: Did you check the link? It has a notational use of $\gimel$ that can be seen, for instance, in Jech’s Set Theory (p. $56$). Jun 7, 2013 at 3:47

First of all, assuming choice one can show that $|V_{\omega+\alpha}|=\beth_\alpha$ (this addition is ordinal addition!), so we have $|V_\alpha|=\alpha$ if and only if $|\omega+\alpha|=|\alpha|=\beth_\alpha$ (note that this implies $\omega+\alpha=\alpha$ as ordinals). Inaccessible cardinals are indeed $\beth$ fixed points, but $\beth_\omega$ is certainly not.

So while there are fixed points which are singular, you still can't do that for any strong limit cardinal.

Now, to your main question. You can't adapt the proof from $\omega$ to an uncountable inaccessible $\kappa$, because that would mean that $V_\kappa\models\sf ZF+\text{Global Choice}$, because we can find a well-ordering of $V_\kappa$ internally definable within $V_\kappa$.

If we assume that $V_{\kappa+1}$ can be well-ordered, then this is not very difficult. The proof follows from Theorem 11 and Proposition 4 from the paper:

Blass, Andreas, Ioanna Dimitriou, and Benedikt Lowe. Inaccessible cardinals without the axiom of choice. Fundamenta mathematicae 194.2 (2007): 179-189.

You have to slightly adjust your construction as presented there, but it works. Alternatively, you can use the same proof as Jech gives (and attributes to Rubin & Rubin) in Theorem 9.1 (b) in his Axiom of Choice book.

Now, I haven't sat down to verify all the details, but it should probably work as a counterexample. And a contrived one too. The idea is based on Joel David Hamkins proof that $\sf ZFC$ does not prove the universe can be linearly ordered.

Let $V$ be a model of $\sf ZFC$ and $\kappa\in V$ is inaccessible (for simplicity, assume $\sf GCH$ holds below $\kappa$). We consider the forcing $\Bbb P$ which add two subsets to each power set of a regular $\lambda<\kappa$ with functions whose domains are of cardinality $<\lambda$.

So a condition in $\Bbb P$ is a function $p\colon\kappa\times2\times\kappa\to 2$ with the following properties:

1. If $(\alpha,i,\beta)\in\operatorname{dom}(p)$ then $\beta<\aleph_{\alpha+1}$, and $i<2$.
2. The set $\{\beta<\aleph_{\alpha+1}\mid (\alpha,i,\beta)\in\operatorname{dom}(p)\}$ has size of at most $\aleph_\alpha$.
3. $|p|<\kappa$.

Now consider the automorphisms which "switch" the middle coordinate, $i$, separately between each $\alpha$. Informally, if $A_\alpha$ and $B_\alpha$ are the new subsets of $2^{\aleph_{\alpha+1}}$ then we just switch these two.

And let $\cal F$ be the filter of subgroups of bounded permutations, that is things which are fixed by a permutations which leave the $i$ coordinate intact for $i>i_0$ for some $i_0$. And let $N$ be the symmetric extension induced by $\cal F$.

It is not hard to show that $\kappa$ is still inaccessible in $N$. And it is not hard to show that the set $R=\{A_\alpha,B_\alpha\mid\alpha<\kappa\}$ is also in $N$. But it cannot be linearly ordered in $N$.

Since $R\subseteq V_\kappa$ we have that it is impossible that $|V_\kappa|=\kappa$ in $N$.

• I might be screwing up with the definition of the forcing, and it should be functions whose domain is $(\alpha,i,\beta,\gamma)$. Or maybe something else? I'm not sure. Let me know if you catch a mistake there (it should be inconsequential to the final result, though). Jun 6, 2013 at 5:17
• I'm still getting the hang of forcing arguments, which I find to be of a rather different character than standard set thoeretical arguments (it's always harder to prove something is unprovable than provable...). I did want to ask about your first paragraph, though. I'm not claiming that $\beth_\omega$ is strongly inaccessible; obviously it isn't. I claimed instead that my ZFC proof that $|V_\alpha|=\alpha$ apparently works for strong limits, not just strong inaccessibles, and $\beth_\omega$ is a strong limit, so $|V_{\beth_\omega}|=\beth_\omega$. Do you disagree? Jun 6, 2013 at 5:32
• @Mario: Yes, I very much disagree. You can prove by transfinite recursion the formula that I gave is true, and therefore $|V_\alpha|=\alpha$ holds only for fixed points of the $\beth$ function, not for strong limits in general. Jun 6, 2013 at 5:33
• I don't follow. In ZFC, I get that $|V_{\omega+\alpha}|=\beth_\alpha$, and I also get that if $\kappa$ is strongly inaccessible, then $\beth_\kappa=\kappa$ (so $\beth_\omega$ is not inaccessible). But is there a fault in my proof, assuming AC? I didn't prove that $\alpha$ is inaccessible iff $|V_\alpha|=\alpha$ -- I only proved one direction of implication, and I realized later that I didn't even use the full strength of the hypothesis, so I can say now that "if $\alpha$ is a strong limit, then $|V_\alpha|=\alpha$". The only way this could be false is if there is a flaw in my proof. Jun 6, 2013 at 5:38
• Oh, I was completely missing your point. You are right, and there is a flaw in my proof: $\alpha<\kappa$ does not imply $\alpha\times\kappa\prec\kappa\times\kappa$. I believe the correct argument at limit stages is to note that since $V_\beta\prec\kappa$ for all $\beta<\alpha$, if $\bigcup_{\beta<\alpha}|V_\beta|=\kappa$, then $f(\beta)=|V_\beta|$ is a cofinal map onto $\kappa$, a contradiction (here we use that $\kappa$ is regular). But how do we know that $|V_\alpha|=\bigcup_{\beta<\alpha}|V_\beta|$? Jun 6, 2013 at 5:56