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I can’t prove that a space having a countable basis and being sequentially compact is indeed compact. Can anyone help me with the prove? Thanks!

Brian Mac Guire
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  • I wrote a solution below. I suggest that Next time when you post another question, add your own thought so that we can help you specifically on the part where you get stucked. – Kevin.S Jan 31 '20 at 03:54
  • Why double post? You accepted the other answer, and the proof is there in the last part. – Henno Brandsma Jan 31 '20 at 07:21
  • Sorry that I broke the rules with a duplicate. At first I thought I had understood the proof, so I accepted it, and then re reading it, I realized I didn't. @KevinSong Thanks for the answer. – Brian Mac Guire Jan 31 '20 at 13:12

1 Answers1

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Let $X$ be a second countable and sequentially compact topological space.

Suppose $X$ is not compact, then we can find a sequence $(x_n)$ with no convergent subsequences in $X$. Now, let's take an open cover $\mathcal{B}$ without finite subcovers.

Since $X$ is second coutable, $\mathcal{B}$ always has countable subcovers $\{B_n|n\in\mathbb{N}\}$, and this subcover is infinite. For each $n$, $\exists \text{ some }x_n\in X-\cup_{i=1}^n B_i$. Now define a map $f:\mathbb{N}\to\mathbb{N}$ that is strictly increasing and a point $x\in B_N\subset X$, then all $x_{f(n)}\notin B_N$ where $f(n)\ge N$. therefore the sequence doesn't converge to $x\in X$. However, we assumed that $X$ is sequentially compact, which contradicts the result. So $X$ is compact.

Another way is to prove it by Lebesgue number Lemma, but it seems that this method only works when $X$ is a metric space instead of an arbitrary topological space.

Kevin.S
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