Second countable and sequentially compact topological spaces are compact. Proof

Question

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!

Answer 1

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.