# Does this choice require Axiom of Choice? [duplicate]

This theorem is from Munkres Topology 2nd ed.:

In the last para, it is said that 'choose an element' $n$ of $D$. Why we do not need axiom of choice here? (Here $D$ may not be finite.)

• The axiom of choice is only needed for simultaneous choices of elements from infinitely many sets, i.e. it is required to be able to choose $a_i\in A_i$ for $i$ in some infinite set $I$. The axiom of choice really just assures that the cartesian product of infinitely many sets is non-empty. If you already have a set which you know is non-empty ($D$, in your example), you may just pick any element. This picking is not what the "choice" in the axiom of choice refers to. I'm writing this as a comment, not an answer, because I have no formal training in set theory – Bananach Nov 13 '16 at 10:56
• There might be better duplicates. But I think this one should cover it nicely. Choosing one element from a provably non-empty set has nothing to do with the axiom of choice, and everything to do with the inference rules of your logic. – Asaf Karagila Nov 13 '16 at 11:24

The fact that we can choose an $n\in D$ is just the fact that $D$ is assumed nonempty. This doesn't require choice. As a heuristic, you don't need the axioms of choice to choose a sock from a pair of socks (or one sock from each pair of finitely many pairs of socks), but if I give you an infinite collection of pairs, you need some form of the axiom of choice to select one from each pair simultaneously. You can grab one element from each set in a finite collection of sets from the axioms of set theory. If you want to grab one from infinitely many sets, you need a stronger axiom in general.
• You said it clearly right at the beginning, but I just want to emphasize that "D is nonempty" means "there exists an x such that x is in D," and so "Choose an element n of D, then the set $A=D\cap\{1,\dots,n\}$ is nonempty" could logically mean "there exists an x such that x is in D, such that there exists a set A which consists of all y in $\{1,\dots,n\}$ such that y is in D, and such that there exists a z in A." – Kyle Miller Nov 13 '16 at 11:16