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A standard proof can be found here. Basically, the idea is to prove the contrapositive:

Let $A\subseteq X$. If $X$ is compact and $A$ doesn't have any limit point, then A is finite.

Since A has no limit points, for each $a\in A$, there exists a neighboorhood $U_a$ such that $U_a\cap A = \{a\}$. $\{X-A\}\cup\{U_a:a\in A\}$ forms an open cover of $X$. By compactness of $X$, only finitely many covers $X$. Because $X-A$ and $A$ are disjoint, A must be finite.

However, the above proof seems to rely on the axiom of choice, since we need to choose a neighborhood $U_a$ for each $a \in A$, and there isn't any definite way to do so.

So can somebody clarify whether AC is necessary here? If not needed, please provide an alternative proof. Thanks in advance.

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  • $\begingroup$ Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo. $\endgroup$ – Will M. Jan 18 at 7:58
  • $\begingroup$ Indeed it is. Topology is definitely pro-choice. $\endgroup$ – William Elliot Jan 18 at 7:58
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It does not require the Axiom of Choice.

For every $a\in A$, set $$ \mathcal W_a=\big\{U\subset X: U \,\text{open}\,\,\&\,\,U\cap A=\{a\} \big\} $$ clearly $\mathcal W_a\ne\varnothing$ and its definition DOES NOT require the Axiom of Choice.

Then set $U_a=\bigcup \mathcal W_a$.

Now, $\mathcal W=\{U_a: a\in A\}$ is a cover without a finite subcover, and it is defined without appealing to the AC.

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  • $\begingroup$ "Its definition does require the Axiom of choice." You mean without AC, $\mathcal W_a$ cannot be formed, right? $\endgroup$ – YuiTo Cheng Jan 18 at 8:09
  • $\begingroup$ Or you are intended to mean "doesn't"? $\endgroup$ – YuiTo Cheng Jan 18 at 8:19
  • $\begingroup$ @YuiToCheng Thanx! - I corrected my answer $\endgroup$ – Yiorgos S. Smyrlis Jan 18 at 8:21
  • $\begingroup$ Thanks, that's a perfect answer. $\endgroup$ – YuiTo Cheng Jan 18 at 8:21
  • $\begingroup$ You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover. $\endgroup$ – Henno Brandsma Jan 18 at 9:29

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