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Usually, the equality $\mathfrak{m}^\mathfrak{m} = 2^\mathfrak{m}$ for infinite cardinal $\mathfrak{m}$ is proved like that: $$ 2^\mathfrak{m} \leqslant \mathfrak{m}^\mathfrak{m} \leqslant (2^\mathfrak{m})^\mathfrak{m} = 2^{\mathfrak{m} \cdot \mathfrak{m}} = 2^\mathfrak{m} $$ The last step relies much on the axiom of choice, because the fact $\mathfrak{m}\cdot\mathfrak{m} = \mathfrak{m}$ does. Moreover, Tarski's theorem states that in $\sf{ZF}$ this fact is equivalent to $\sf{AC}$.

Note, by a cardinal I mean an equipollence class of an arbitrary set, not necessarily admitting a well-ordering.

That has motivated my question: what can we say about relationship of the initial equality and the axiom of choice in pure $\sf{ZF}$?


To avoid question about definition of cardinals without $\sf{AC}$, we may consider the following statement instead:

For any infinite set $X$ exists a bijection $\; F \, \colon X^X \rightarrow 2^X$

where $X^X$ is the set of all functions from $X$ to $X$, $\; 2^X = \mathcal P(X)$ is the power set, and an infinite set is such that cannot be bijectively mapped onto a finite ordinal.

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  • $\begingroup$ Tarski's theorem states that $X \times X \approx X$ for every infinite set $X$ implies AC, but $\kappa.\kappa = \kappa$ for any infinite cardinal $\kappa$ can be proven in ZF. $\endgroup$ – nombre Feb 17 '16 at 21:03
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    $\begingroup$ @nombre, you probably call cardinal thing I call aleph, that is cardinality of an ordinal. In that sense you are right, but I don't want to debate on the definition of a cardinal. $\endgroup$ – quartermind Feb 17 '16 at 21:55
  • $\begingroup$ Ok. I really don't know whether the second statement implies AC but it is an interesting question. $\endgroup$ – nombre Feb 17 '16 at 22:20
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After more intense search, I've found the answer from this mathoverflow and math.se questions.

The relation of the subject statement to the axiom of choice seems poor. So, the subject statement is consistent with $\sf ZF + \neg\sf AC$, as in final remark in this paper. More, its negation also may hold in models of $\sf ZF$ without choice, as in this paper!

For reference, permutation models of $\sf ZF$ with atoms (or urelements) used in the above papers are described in book "The Axiom of Choice" by Thomas J. Jech.

I hope there may be another interesting contributions to the question.

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