# A question about Choice Functions.

Assume the axioms of ZFC. Suppose that X is an infinite set of infinite (and pairwise disjoint) sets, none of which has a cardinal number greater than that of X. Is the cardinal number of the set of all Choice Functions for X always equal to the cardinal number of the set of all mappings of X into itself?

If $X$ is a set of non-empty sets such that for all $y\in X$, $|y|\leq|X|$, then $|\bigcup X|=|X|$.
If all sets have at least two elements, then there are $2^{|X|}$ choice functions. Since you have added the assumption that all the sets in $X$ are infinite, we have that indeed this is $2^{|X|}$. On the other hand, there are exactly $|X|^{|X|}=2^{|X|}$ functions from $X$ to itself (because we assume the axiom of choice).
• Yes, but it's not quite surprising that when assuming the axiom of choice we have the maximal number of choice functions possible. What is nicer is that it is consistent with $\sf ZF$, for example, that there is a family of sets which admits a countable number of choice functions. – Asaf Karagila May 3 '14 at 18:59