We say an infinite set $X$ is splittable if there are $X_1, X_2\subseteq X$ with $X_1\cap X_2 = \emptyset$, $X_1\cup X_2 = X$ and there are bijections $\varphi:X_1\to X_2$ and $\psi:X_1\to X$.

Does the statement "Every infinite set is splittable" imply $\mathsf{AC}$?


marked as duplicate by Asaf Karagila, Franz Lemmermeyer, Stefan Kohl, Jan-Christoph Schlage-Puchta, Wolfgang Apr 22 '16 at 7:51

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The answer is no and it follows from the following:

It is consistent that $AC$ fails but for all infinite cardinals $\kappa, 2 \cdot \kappa=\kappa.$

The above result is proved by Sageev:

Sageev, Gershon An independence result concerning the axiom of choice. Ann. Math. Logic 8 (1975), 1–184.

In a model as above, every infinite set is splittable but $AC$ fails in it.

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    $\begingroup$ Can you give a reference for this result, or at least some pointers for finding out more? Without that, it’s pretty much just a rephrasing of the question. $\endgroup$ – Peter LeFanu Lumsdaine Apr 19 '16 at 9:12
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    $\begingroup$ I added a reference $\endgroup$ – Mohammad Golshani Apr 19 '16 at 9:23
  • $\begingroup$ Thanks Mohammad -- this $2\cdot(\ldots)$ argument was really the correct starting point! $\endgroup$ – Dominic van der Zypen Apr 19 '16 at 18:32

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