# Explicit choice functions for finite sets in topological spaces

When dealing with finite nonempty sets of real or natural numbers it is always possible to define a explicit choice function, that choose one (arbitrary, but well defined) element out of that set: Take for example the smallest element; or the greatest.

My question is: If we are in a very much more abstract setting, like a (arbitrary) topological space $X$, is then there a way to explicitely define a choice function as above (i.e. a function that for every finite nonempty set of elements in this space returns one element of this sets) ?

I'm aware that this may be a somewhat open question, since the probable answer which awaits me is, I think "there isn't any known explicit choice function" - but that doesn't mean it's proven that no such function exists. The use of the axiom of choice isn't allowed! Also answers that depend on the set-theoretical construction of $X$ don't count, like always taking the element that has least amount of $\emptyset$'s in it - this is just an (incorrect - as we may consider different elements with the same amount of $\emptyset$'s "in" them) example of what I want to avoid. Side question: Would such a function, that depends on the set-theoretic construction of $X$, even be define in the ZFC that I'm working in, can I only define it in metamathematics ?

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There is not: it’s consistent that there be a space $X$ such that no choice function for the non-empty finite subsets of $X$ exists. –  Brian M. Scott Dec 21 '12 at 15:35
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## 1 Answer

It is consistent that there are sets which can be expressed as a countable union of pairs, but there is no choice function from the pairs.

Given such set, how could you expect to find a choice function on its finite subsets? Regardless to what the topology is, that would be impossible.

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What dou you mean with "it is consistent that [...]" ? Do you mean "it follows from ZFC that "there are sets which can be expressed [...]" or do you mean, the statement "there are sets which can be expressed [...]" is independent of ZFC ? –  temo Dec 22 '12 at 8:53
First note this is not about ZFC, but rather about ZF. if you decline the use of choice, might as well not assume it. Secondly what I meant is that if ZF is consistent then there is a model of ZF in which such set exists. However not in every model there is such a set, e.g. in models of ZFC this is impossible. –  Asaf Karagila Dec 22 '12 at 8:56
Ok, I understood, it depends on the model whether such a set exists or not. But what happens in those models in which such a set (which has the property that no choice function for the non-empty finite subsets of X exists) doesn't exist ? Is it maybe there possible to define for every set an explicit choice function ? (My level of knowledge in logic is very limited, so I know that my lay-men objections may seem out of place.) –  temo Dec 22 '12 at 10:07
@temo: But what does "explicit" choice function mean? Does it mean definable with variables? Well, in this case obviously we can because there is some choice function and we can use it as a variable. Does it mean define without variables? In this case it's even simpler to show that the answer is negative. Simply forcing enough new real numbers would turn out a model in which we cannot do this (even though the axiom of choice does hold). –  Asaf Karagila Dec 22 '12 at 10:11
Last question of mine: With "explicit" I only meant, that there is an explicit rule, that tells us, which element to pick (like in the case of the real number, we can always pick the smallest element out of your subset); in which of the two categories (with variables/without variables) you specified would this fall ? And why ? (If it falls in the first category our explanation seems to me to be a little bit of a cheat, since we don't really know how the choice functions whose existence we know, looks like, but we use it, to get a choice function with variables, which we need) –  temo Dec 22 '12 at 11:01
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