Compact Sets, Finite Subcover,Axiom of Choice

Question

Definition: A subset $A$ of a metric space $X$ is compact , if every open cover $(U_i)_{i \in I}$ of $A$ has a finite subcover, i.e. finitely many indexes $i_1,i_2,..., I_k$, $\in I $ exist, s.t.

$A \subset U_1\cup U_2 ....\cup U_k.$

$I$ is an arbitrary index set.

Question: This is a statement about the existence of such finite $i_l$ , $l=1,2,..k.$

Does this statement involve the axiom of choice? The index set $I$ can be uncountably infinite,

True, we do not actually have to pick $i_l$ out of $I$ (axiom of choice?), the statement is about the existencence of finitely many.

Please clarify. Thanks.

Answer 1

It will start to involve choice, if you have infinitely many open covers and want to pick a finite subcover for each of them simultaneously, e.g.

Just picking one finite subcover does not involve choice at all. If one exists (by the definition) we can pick one. If we know that $A \neq \emptyset$ it does not invoke choice when we pick some $x \in A$.

Answer 2

The statement asserts existence. Nothing about the axiom of choice is related here. You might ask if you need the axiom of choice to prove that an infinite set has a finite subset. The answer, of course, is negative.

A set is infinite if and only if it has arbitrarily large finite subsets. The definition of compactness tells us that there is at least one finite subset with a certain property. Choosing one is just applying existential instantiation, not appealing to choice.