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It is known that a first-countable, compact $T_1$ topological space is sequentially compact.

Now, the way I know the proof, you pass through limit point compacity to, given a sequence, construct an appropriate subsequence using the fact that the space is $T_1$ and first-countable.

In this post, it is shown an example of a first-countable, limit point compact, non-$T_1$ topological space which is not sequentially compact. But the space used as example is not compact.

Therefore, I ask: what is an example of a first-countable, compact non-$T_1$ topological space which is not sequentially compact? (or do we really not necessarily need $T_1$?)

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Aloizio Macedo
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1 Answers1

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Suppose that $X$ is a first-countable compact topological space. Consider an infinite sequence $x_1,x_2,\dots,x_n,\dots$ of points in $X.$

Claim. There is a point $x\in X$ such that, for every neighborhood $U$ of $x,$ we have $x_n\in U$ for infinitely many $n.$

Proof of claim. Assume the contrary. Each point $x$ is covered by an open set $U$ such that $x_n\notin U$ for all sufficiently large $n.$ By compactness, the space $X$ is covered by finitely many such sets. It follows that $x_n\notin X$ for sufficiently large $n,$ which is absurd.

Choose a point $x$ as claimed above. Let $V_1\supseteq V_2\supseteq\cdots\supseteq V_n\supseteq\cdots$ be a countable decreasing neighborhood base for $x.$ Choose $n_1\lt n_2\lt n_3\lt\cdots$ with $x_{n_k}\in V_k.$ The subsequence $x_{n_1},x_{n_2},\dots$ converges to $x.$

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