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

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    $\begingroup$ GCH implies AC. $\endgroup$ – André Nicolas Feb 26 '13 at 9:28
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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 $\mathfrak m\ge5$.


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.

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  • $\begingroup$ Is there an infinite counterexample to the Halbeisen-Shelah theorem? $\endgroup$ – Asaf Karagila Feb 27 '13 at 0:07
  • $\begingroup$ 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 Jech-Sochor to transfer this to ZF.) $\endgroup$ – Andrés E. Caicedo Feb 27 '13 at 0:57
  • $\begingroup$ Ah. Because the atoms are weakly o-minimal, we can encode every subset by a sequence of endpoints. $\endgroup$ – Asaf Karagila Feb 27 '13 at 1:00
  • $\begingroup$ The Halbeisen-Shelah 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 user-friendly way.) $\endgroup$ – Andrés E. Caicedo Feb 27 '13 at 1:05
  • $\begingroup$ 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! $\endgroup$ – Asaf Karagila Feb 27 '13 at 1:10
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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.


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
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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.

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  • $\begingroup$ To the OP: A proof of Sierpinski’s result may be found in Paul Cohen’s Set Theory and the Continuum Hypothesis. $\endgroup$ – Haskell Curry Feb 26 '13 at 9:34
  • $\begingroup$ Could you please be more specifically? Thanks. $\endgroup$ – user64030 Feb 26 '13 at 9:41
  • $\begingroup$ 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 $A$. $\endgroup$ – André Nicolas Feb 26 '13 at 9:47
  • $\begingroup$ What is the function that maps A*A into A? $\endgroup$ – user64030 Feb 26 '13 at 9:55
  • $\begingroup$ @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 well-order $A$ 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 well-order 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. $\endgroup$ – Asaf Karagila Feb 26 '13 at 10:53

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