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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 $\mathsf{AC}$?

Dominic van der Zypen
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  • You might also be interested in this question, concerning two inequivalent concepts of even cardinalities: http://mathoverflow.net/q/69461/1946 – Joel David Hamkins Apr 19 '16 at 11:24
  • I am almost sure that you asked this question not too long ago. – Asaf Karagila Apr 19 '16 at 14:43
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    I was not wrong. http://mathoverflow.net/questions/230260/does-x-times-0-cup-x-times-1-leq-x-for-x-infinite-imply-s and also a follow up question, http://mathoverflow.net/questions/230271/some-very-weak-statements-on-choice – Asaf Karagila Apr 19 '16 at 15:02

1 Answers1

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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.

Mohammad Golshani
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