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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 $|V_\kappa|=\kappa$.

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Mario Carneiro
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1 Answers1

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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 $N$.

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Asaf Karagila
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  • 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). – Asaf Karagila Jun 06 '13 at 05:17
  • 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? – Mario Carneiro Jun 06 '13 at 05:32
  • @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. – Asaf Karagila Jun 06 '13 at 05:33
  • 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. – Mario Carneiro Jun 06 '13 at 05:38
  • 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 $|V_\alpha|=\bigcup_{\beta<\alpha}|V_\beta|$? – Mario Carneiro Jun 06 '13 at 05:56
  • @Mario: I don't understand your last comment. (And I missed your previous one, too.) – Asaf Karagila Jun 06 '13 at 05:58
  • @Mario: I think that the flaw in your argument is assuming that without the axiom of choice it holds that $|\bigcup_{\beta<\alpha} A_\beta|=\alpha\times\sup{|A_\beta|\mid\beta<\alpha}$. – Asaf Karagila Jun 06 '13 at 06:04
  • I'm trying to patch up my proof in the OP, which has a problem in it which was making me think that $|V_{\beth_\omega}|=\beth_\omega$ (in fact $|V_{\beth_\omega}|=\beth_{\beth_\omega}$). I know that the union bound needs to use AC, which I recognize in the OP, but I have a mistake in there that says $\alpha\prec\kappa$ implies $\alpha\times\kappa\prec\kappa\times\kappa$, which is incorrect: $2<\omega$, but $2\times\omega=\omega\times\omega$. (con't) – Mario Carneiro Jun 06 '13 at 06:07
  • The correct proof needs to use that $\kappa$ is regular, and for limit $\alpha$ we know that $|V_\alpha|=|\bigcup_{\beta<\alpha}V_\beta|\le|\kappa\times\alpha|=\kappa$ (I used the union bound there). But if $|V_\alpha|=\kappa$, then the sizes of $|V_\beta|$ less than $\kappa$ make a cofinal map, so ${\rm cf}(\kappa)\le\alpha<\kappa$, a contradiction. Thus $|V_\alpha|<\kappa$, and the transfinite induction continues all the way to $\kappa$. – Mario Carneiro Jun 06 '13 at 06:10
  • @Mario: I'm not sure if it's the extremely odd sleep pattern of the last three days, or something else. I still don't understand fully what you're talking about... – Asaf Karagila Jun 06 '13 at 06:15
  • I think I'm going to need to sleep on your forcing argument to understand it, so I guess we're even. ;-) – Mario Carneiro Jun 06 '13 at 06:22
  • Good luck with that. Sleeping on that seems to be very uncomfortable. I'd only do that if someone forced me to do it... :-P – Asaf Karagila Jun 06 '13 at 06:24