>>13603726Here's what they mean by select. Given a collection of sets, can you define a function, f, that takes those sets as input and such that f(S) is always in S?
For example, let's say my collection of sets is A = {1, 2, 3}, B = {4, 5, 6}, and C = {7, 8, 9}. I can define f to be the function that returns the average of all the elements of a set. So f(A) = 2, f(B) = 5, and f(C) = 8. In this case, my function happens to always stay in the set. f(A) is in A, f(B) is in B, and f(C) is in C. If that happens, we say that f is a choice function.
Another example, let's say my collection of sets is all the non empty subsets of the natural numbers. So {6, 23, 43, 67} is in my collection, the set of primes is in my collection, and any other non empty collection of natural numbers that you can think of. Can you think of a choice function? If you define f(S) to be min(S), then you have a choice function! This works because every set (even infinite sets) of natural numbers has a smallest number. We would have f({6, 23, 43, 67}) = 6, f(the primes) = 2, and so on. No axiom of choice needed here, just use the min function.
Another example, let's say my collection of sets if all the non empty subsets of the real numbers. Can you think of a choice function? You can't define f(S) to be min(S), because not all sets of real numbers have a minimum, for example {x : x > 0} doesn't have a minimum. Turns out no one ever managed to define a choice function for that collection (if I recall correctly, it's actually impossible to define an explicit example). So we have no evidence that such a function even exists. The axiom of choice is needed here if ever you need such a choice function.
Note: You don't need the axiom of choice to find choice functions for collections of finite or even countably infinite sets. But things get weird once we look at collections of uncountably infinite sets. But we don't want things to get weird, so we pretend it isn't by assuming the axiom of choice.
