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?
2026-03-28 10:03:31.1774692211
A question about Choice Functions.
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That is one convoluted question.
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).
So the answer is yes.