# Confusion about GCH and the von Neumann universe construction

I am getting slightly confused over von Neumann universe. According to what I see, as each stage can be constructed using power set, generalized continuum hypothesis must be true!

Can anyone show me how I am wrong?

Thanks very much.

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The reals, as constructed, have the cardinality of a power set. But that doesn't rule out the existence of subsets of the reals of intermediate cardinality. – André Nicolas Mar 25 '12 at 18:37

## 4 Answers

Why must it be true?

Suppose $2^{\aleph_0}=\aleph_3$. Since $V_\omega$ is countable, this means that $V_{\omega+1}$ has cardinality $\aleph_3$. In particular this means that we have added collections of size $\aleph_1$ and $\aleph_2$ at this very stage, they will be materialized as actual sets in $V_{\omega+2}$ (namely, in $V_{\omega+1}$ we only have countable sets, but in $V_{\omega+2}$ we will have sets of sizes $\aleph_1,\aleph_2,\aleph_3$ as well).

Furthermore, if $2^{\aleph_0}$ cannot be well ordered, then $V_{\omega+1}$ cannot be well ordered (recall that the von Neumann universe is a universe of ZF, not just ZFC!) and in particular this means that we are adding new and strange sets to the universe.

The general claim in ZFC is that $V_{\omega+\alpha}$ has cardinality $\beth_\alpha$. The assumption of GCH is indeed that $\beth_\alpha=\aleph_\alpha$ for all $\alpha$, but the first part is true just by assuming the axiom of choice.

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Look at the first, finite sets in the cumulative hierarchy. The set $V_0$ contains $0$ elements. The set $V_1$ contains $1$ element. The set $V_2$ contains $2$ elements, the set $V_3$ contains $4$ elements. But there exists a set with $3$ elements, so it has cardinality between the cardinality of $V_2$ and $V_3$. And you can find a $3$ element set in $V_4$.

Now if the GCH fails, something similar happens at an infinite stage. If there is set with cardinality between the cardinality of $V_\alpha$ and $V_{\alpha+1}$, a subset of $V_{\alpha+1}$ must have exactly the same cardinality, and you can find it in $V_{\alpha+2}$.

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This is an intuitively good answer, but its not quite right on the technical details... in particular, you write : "If there is set with cardinality between the cardinality of $V_α$ and $V_{α+1}$, a subset of $V_{α+1}$ must be of exactly this cardinality, so you can find it in $V_{α+2}$." This isn't right, for instance $X:=\{\beth_0,\beth_{\omega},\beth_{\omega2}\}$ has cardinality between $V_1$ and $V_2$, but its rank is way high, so $X \notin V_2$. – goblin Apr 25 '13 at 22:55
@user18921 Tank you, I'll change it. – Michael Greinecker Apr 25 '13 at 23:51

As others have stated, the Continuum Hypothesis might be false for the Von Neumann construction, but I think it's important to note that there's no reason that has to be the case. Put another way, I could construct a model very much like the Von Neumann universe, which models ZFC but in which the Continuum Hypothesis is clearly true. That would not be a contradiction.

It sounds like you might not be understanding what it means to be a model of an axiomatic system. The Von Neumann universe is just one model of ZFC. There are many others. There are some where the Continuum Hypothesis is true, and some where it is false. That is what it means when we say that the Continuum Hypothesis is "independent of ZFC."

I hope I'm not saying something you already knew, but it's better to be safe than sorry.

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I think I understand your confusion. Let me try to lift it.

Let $H$ denote the hereditarily finite sets (note that $H$ is countable). Now consider the following ordinal-indexed sequence.

$$(\alpha \mapsto V_\alpha) = (\emptyset,\mathcal{P}(\emptyset),\mathcal{P}^2(\emptyset),\mathcal{P}^3(\emptyset),...H, \mathcal{P}(H), \mathcal{P}^2(H),\mathcal{P}^3(H)...)$$

Now if every set occurred somewhere in this sequence, then yes, GCH would be true. However, this is nonsense! For instance, there no set of cardinality $3$ occurs in this sequence, because the associated cardinality sequence goes $(0,1,2,4...).$

So what I'm saying is this. Its not the case that for every set $X$ there exists an ordinal $\alpha$ such that $X=V_\alpha$, nor is it the case that for every set $X$ there exists an ordinal $\alpha$ such that $|X|=|V_\alpha|$. It is the case that for every set $X$, there exists an ordinal $\alpha$ such that $X \in V_\alpha$.

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But $\sf GCH$ doesn't speak about finite cardinals, and your sequence does. Your answer implies that $\sf GCH$ is false because $2^2=4$ and $2^+=3$. – Asaf Karagila Apr 26 '13 at 10:07
@AsafKaragila, I edited my answer. I write: "So I guess what I'm saying is this. Its not the case that for every set $X$ there exists an ordinal $\alpha$ such that $X=V_\alpha$, nor is it the case that for every set $X$ there exists an ordinal $\alpha$ such that $|X|=|V_\alpha|$. It is the case that for every set $X$, there exists an ordinal $\alpha$ such that $X \in V_\alpha$." – goblin Apr 26 '13 at 11:31