Do choice functions have the cardinality of the largest set in their domain?

[nomadreid]

Is the following a theorem from ZFC?
Given a collection C of non-empty sets that includes at least one infinite set, the cardinality of the collection of distinct choice functions on C (as defined in AC) equals the cardinality of the largest element of C.
My feeling that this is true is from generalizing the case when the largest cardinality is $\aleph$₀, where it seems that a simple proof is possible, but I am not sure whether it is true and, if so, provable (from ZFC) for higher cardinalities.

 

[yossell]

Won't the claim fail when the cardinality of C is greater than the cardinality of the largest element of C?

That is, if C is a collection of aleph1 sets of aleph0 elements, there will be at least aleph1 choice functions.

 

[nomadreid]

yossell, thanks for the answer. Good point. So, if I were to amend it, would the theorem be that the number of choice functions is max (|C|, |S|) with S being a set with the largest cardinality? Whereas this seems intuitively clear, is it provable in ZFC?

 

[micromass]

Given sets $(X_i)_{i\in I}$, what you want is the cardinality of

$$\prod_{i\in I} X_i$$

For convenience, we set $I$ to be equal to a cardinal number, so we put $I = |I| = \lambda$. It is a theorem in ZFC that if $\lambda$ is infinite and if $|X_i|$ are nondecreasing and nonzero, then

$$\prod_{i<\lambda} |X_i| = (\sup_{i<\lambda} |X_i| )^\lambda$$

See "Set Theory" by Jech for a proof.

 

[nomadreid]

Thanks, micromass. Lemma 5.9, to be exact.