On page 19 of Counterexamples in Topology, the authors say "sequential compactness clearly implies countable compactness" without explaining.
I feel dumb, why is this so obvious? Does anyone have nice explanation or reference for this fact?
On page 19 of Counterexamples in Topology, the authors say "sequential compactness clearly implies countable compactness" without explaining.
I feel dumb, why is this so obvious? Does anyone have nice explanation or reference for this fact?
Suppose $X$ is sequentially compact but $\{U_n: n \in {\mathbb N}\}$ is a countable open cover with no finite subcover. For every $N$, since $\{U_n: n \le N\}$ is not a subcover, there must be $x_N \notin \bigcup_{n \le N} U_n$. By sequential compactness, some subsequence $x_{N_j}$ converges to some point $x \in X$. But $x \in U_m$ for some $m$, and since $x_{N_j} \notin U_m$ when $N_j > m$ we get a contradiction.