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

5GCH implies AC. – André Nicolas Feb 26 '13 at 09:28
3 Answers
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 HalbeisenShelah result, for example.)
Using the HalbeisenShelah result here is an overkill, of course. Specker's original argument established $2^{\mathfrak m}\not\le \mathfrak m^2$ directly from the assumption $\mathfrak m\ge5$.
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 wellorderable (that is, an aleph). Sierpiński proved that the wellorderability 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.
 75,806
 9
 204
 327

Is there an infinite counterexample to the HalbeisenShelah theorem? – Asaf Karagila Feb 27 '13 at 00:07

Do you mean, if $X$ is not Dedekind infinite? Yes, in the Mostowski model, if $A$ is the set of atoms, then $\mathcal P(A)$ injects into the set of finite sequences of elements of $A$. (And then one can use JechSochor to transfer this to ZF.) – Andrés E. Caicedo Feb 27 '13 at 00:57

Ah. Because the atoms are weakly ominimal, we can encode every subset by a sequence of endpoints. – Asaf Karagila Feb 27 '13 at 01:00

The HalbeisenShelah paper has several nice results I'm sure you'll appreciate, you should look at it. (And at Halbeisen's book, where some of the proofs are written in a slightly more userfriendly way.) – Andrés E. Caicedo Feb 27 '13 at 01:05

I have looked at it, but my impressionistic memory didn't paint the whole picture on the consistency results part. I did not know these appear in the book, perhaps those will stick better! Thanks! – Asaf Karagila Feb 27 '13 at 01:10
Generally speaking in ZFC, even without GCH, every infinite set can be wellordered, and it is true for infinite wellordered 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 wellordered of course.
GCH can be formulated in two seemingly nonequivalent ways:
 For every $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+1}$.
 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 wellordered 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.
Related:
 370,314
 41
 552
 949
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.
 491,093
 43
 517
 948

To the OP: A proof of Sierpinski’s result may be found in Paul Cohen’s *Set Theory and the Continuum Hypothesis*. – Haskell Curry Feb 26 '13 at 09:34


It is a consequence of the Axiom of Choice that for any infinite set $A$, the set $A\times A$ can be put in one to one correspondence with $A$. – André Nicolas Feb 26 '13 at 09:47


1@user64030: The existence of such function relies on the axiom of choice, and therefore it cannot be written out explicitly. But assuming that you can wellorder $A$ you can define the function. It's not "pretty" in any way, and writing it explicitly would probably be a pain. The idea is to define a wellorder on $\omega_\alpha\times\omega_\alpha$ whose order type is $\omega_\alpha$. So the collapse would result in a bijection between $\omega_\alpha\times\omega_\alpha$ and $\omega_\alpha$. So we can compose this with a bijection between $A$ and $\omega_\alpha$ and then with the inverse map. – Asaf Karagila Feb 26 '13 at 10:53