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I read there that the fact about equivalence of $A$ and $A^2$ for any infinite set $A$ and the axiom of choice are equivalent. But what if we prove it only for sets that have cardinality of $\aleph_n, n \in \mathbb{N}\cup\lbrace 0 \rbrace$.

Let's prove it with mathematical induction.

  1. For $n = 0$ it's evident, we can use the same proof that was used for the countability of $\mathbb{Q}$.
  2. Suppose that it's true for some $n \in \mathbb{N}$. Let's prove it for $n + 1$. Consider a set $A$ with cardinality of $\aleph_{n + 1}$. Then there exists a set $B$ with cardinality of $\aleph_n$ which powerset is equivalent to $A$: $A \sim 2^B$. Then every element of $A$ corresponds to a subset of $B$. We can construct an injection between $(2^B)^2$ and $2^B$. First, we construct an injection between $(2^B)^2$ and $2^{B^2}$: $f(S_1, S_2) = S_1 \times S_2$. Further we can construct an injection between $2^{B^2}$ and $2^B$ by using, as we assume, existing injection between $B^2$ and $B$. Also we can easily can construct an injection between $2^B$ and $(2^B)^2$: $f(S) = (S, S)$. Then by the Cantor-Bernstein theorem we have that $A \sim 2^B \sim (2^B)^2 \sim A^2$

As far as I'm concerned, that proof doesn't use the axiom of choice. But it doesn't work for such cardinals as $\aleph_\omega, \aleph_{\omega + 1}$ and so on, because there we should use transfinite induction which uses the axiom of choice. Did I get it right?

nyekitka
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2 Answers2

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No, transfinite induction has nothing to do with the axiom of choice.

It is a theorem of ZF that if $|A| = \aleph_\alpha$ for some ordinal $\alpha$, then $|A|=|A^2|$. The issue is that without assuming choice, there are sets $A$ such that $|A|$ is not an aleph (i.e., $A$ is not well-orderable, so $A$ is not in bijection with any ordinal).

Also, your proof doesn't work. You assume that if $|A| = \aleph_{n+1}$, then $|A| = |2^{B}|$ where $|B| = \aleph_n$. This is only true if the generalized continuum hypothesis (GCH) holds at $\aleph_n$, i.e., if $2^{\aleph_n} = \aleph_{n+1}$. This is not provable in ZFC.

Alex Kruckman
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There are three points here.

  1. For $\aleph$ numbers, the Axiom of Choice is not at all required to prove that $\aleph_\alpha^2=\aleph_\alpha$. This is done by constructing a well-ordering of $\omega_\alpha\times\omega_\alpha$ which is isomorphic to $\omega_\alpha$ with its standard order. You can see more here or here, for example. This holds for much more than $\aleph$ numbers indexed by $n<\omega$, indeed, it holds for all of them.

  2. The assumption that $\aleph_{\alpha+1}$ is the cardinality of $2^{\aleph_\alpha}$, is known as the Generalised Continuum Hypothesis. It implies the Axiom of Choice, so by making that assumption, you invariably (and unnecessarily, in this instance) introduce the Axiom of Choice back into your system.

  3. We can prove, in $\sf ZF$ alone, that if $X^2=X$, or even just $X\cdot 2=X$, then $2^X\cdot 2^X=2^X$, simply because of power distribution laws: $2^X\cdot 2^X=2^{X+X}$. Since this fact holds for $\aleph_0$, it follows that any and all of its iterated power sets will also satisfy these equalities as well.

Asaf Karagila
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