# Does “Every infinite set is splittable” imply $\mathsf{AC}$? [duplicate]

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 $\mathsf{AC}$?

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.

• 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. – Peter LeFanu Lumsdaine Apr 19 '16 at 9:12
• I added a reference – Mohammad Golshani Apr 19 '16 at 9:23
• Thanks Mohammad -- this $2\cdot(\ldots)$ argument was really the correct starting point! – Dominic van der Zypen Apr 19 '16 at 18:32