Question about Generalized Continuum Hypothesis

Question

I wonder how the Generalized Continuum Hypothesis reveal that $A\times A$ is equivalent to $A$? $A$ is any infinite set.

Answer 1

It is relatively easy to see the outline of the proof (over $\mathsf{ZF}$) that $\mathsf{GCH}$ implies $A\times A\sim A$ for all infinite sets $A$. This is usually presented as an intermediate step towards the meatier result that $\mathsf{GCH}$ gives us choice. Let me give a sketch.

Assume $\mathsf{GCH}$. Note that if $\mathfrak m$ is an infinite cardinality, and there are no intermediate sizes between $\mathfrak m$ and $2^{\mathfrak m}$, then $\mathfrak m+1=\mathfrak m$. This is because one can see directly that $\mathfrak m+1<2^{\mathfrak m}$ for all $\mathfrak m>1$ (generalizing slightly Cantor's argument for $2^{\mathfrak m}>\mathfrak m$).

But then we have that $\mathfrak m+\mathfrak m=\mathfrak m$, because $\mathfrak m\le\mathfrak m+\mathfrak m\le 2^{\mathfrak m}+2^{\mathfrak m}=2^{{\mathfrak m}+1}=2^{\mathfrak m}$. This is because $2\mathfrak m<2^{\mathfrak m}$. (In fact, over $\mathsf{ZF}$, we have $n\mathfrak m<2^{\mathfrak m}$ for all finite $n$. This is a result of Specker. A stronger result is that if $\aleph_0$ injects into $X$, then $\mathcal P(X)$ cannot inject into the set of finite sequences of elements of $X$. This was proved by Halbeisen and Shelah, see for example this blog post of mine.)

Finally, $\mathfrak m^2=\mathfrak m$, since $\mathfrak m\le \mathfrak m^2\le (2^{\mathfrak m})^2=2^{2\mathfrak m}=2^{\mathfrak m}$, and $2^{\mathfrak m}\not\le\mathfrak m^2$ (by the Halbeisen-Shelah result, for example.)

Using the Halbeisen-Shelah result here is an overkill, of course. Specker's original argument established $2^{\mathfrak m}\not\le \mathfrak m^2$ directly from the assumption $\mathfrak m\ge5$.

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Note that the argument above is "local" in the sense that it concludes $\mathfrak m^2=\mathfrak m$ from the sole assumption of $\mathsf{GCH}$ at $\mathfrak m$. Since the assumption that $\mathfrak m^2=\mathfrak m$ for all $\mathfrak m$ implies choice, the ideal result here would be to show that from the $\mathsf{GCH}$ at $\mathfrak m$ it follows that $\mathfrak m$ is well-orderable (that is, an aleph). Sierpiński proved that the well-orderability of $\mathfrak m$ follows from $\mathsf{GCH}$ at $\mathfrak m$, $2^{\mathfrak m}$, and $2^{2^{\mathfrak m}}$. Specker improved this, showing that assuming $\mathsf{GCH}$ at $\mathfrak m$ and $2^{\mathfrak m}$ suffices. Whether $\mathsf{GCH}$ at $\mathfrak m$ is already enough is an open problem.

Answer 2

Generally speaking in ZFC, even without GCH, every infinite set can be well-ordered, and it is true for infinite well-ordered sets that $A\times A\sim A$ (where $\sim$ denotes equinumerosity).

In ZF itself, it is consistent that some infinite set $A$ it holds that $A\times A\nsim A$. But such $A$ cannot be well-ordered of course.

GCH can be formulated in two seemingly non-equivalent ways:

1. For every $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+1}$.
2. For every infinite set $A$, if $A\lesssim B\lesssim\mathcal P(A)$ then either $B\sim A$ or $B\sim\mathcal P(A)$. That is to say, there is no intermediate cardinality between an infinite set and its power set.

While both statements are clearly different, one speaks on power sets of every set and the other only refers to well-ordered cardinals, it turns out that either one imply the axiom of choice, and therefore the other as well. It follows if we assume GCH it must hold that $A\times A\sim A$ for every infinite $A$.

It should be remarked that $A\times A\sim A$ implies the axiom of choice on its own accord, although not GCH in any of the forms.

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Related:

1. For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
2. Is the axiom of choice needed to show that $a^2=a$?
3. About a paper of Zermelo

Answer 3

It is an old result of Sierpinski that GCH implies AC, and therefore all the nice results of cardinal arithmetic that are consequences of AC.