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 $T_1$?)

  • $\begingroup$ The glaring suggestion is to examine a proof of your opening statement to see where the hypothesis of $T_1$ is used, right? $\endgroup$ – MPW Sep 3 '16 at 3:45
  • $\begingroup$ @MPW I know where the hypothesis is used. It is precisely when we construct the appropriate subsequence, as I mentioned. However, it does not suggest me in any way an idea of a construction/example of the kind of space I want. $\endgroup$ – Aloizio Macedo Sep 3 '16 at 3:50
  • $\begingroup$ See math.stackexchange.com/questions/44907/… $\endgroup$ – MPW Sep 3 '16 at 3:51
  • $\begingroup$ And +1 for an interesting question. $\endgroup$ – MPW Sep 3 '16 at 3:52
  • $\begingroup$ @MPW Thanks for the link! It is an interesting example, but it is not what I need, since it is not first-countable. $\endgroup$ – Aloizio Macedo Sep 3 '16 at 3:55

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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