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?


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.

  • $\begingroup$ Many thanks for your answer. It seems to me that when mathematicians talk about choice functions they are mostly interested in whether or not they exist. They do not seem much interested in the set of all choice fuctions. So I became curious about this set and wondered what its cardinal number would be. $\endgroup$ – Garabed Gulbenkian May 3 '14 at 18:32
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila May 3 '14 at 18:59
  • $\begingroup$ That is a very interesting fact about ZF which I didn't know. It is just one more example of the strange things that become possible when the axiom of choice is not around to keep them from occuring. $\endgroup$ – Garabed Gulbenkian May 5 '14 at 19:03
  • $\begingroup$ Quite. I don't think most people are aware of this. Not even people working on axiom of choice related questions. It's one of these things you notice once, and then you forget, and then you notice again, and forget again, and so on and so forth. Pretty cute detail, though. $\endgroup$ – Asaf Karagila May 5 '14 at 19:07

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