# Question about Generalized Continuum Hypothesis [closed]

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

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.