Let $X$ be a compact metric space with topology $\tau$ generated by the metric. Consider a new point $x_1\notin X$ and a non-isolated point $x_0\in X$. Set $\overline{X}=X\cup \{x_1\}$, equipped with the topology

\begin{equation} \overline{\tau}= \tau \cup \left\{W\cup \{x_1\}: x_0\in W, W\in\tau\right\}\cup \left\{(W\setminus \{x_0\})\cup \{x_1\}: x_0\in W, W\in \tau\right\}. \end{equation} I think that $(\overline{X}, \overline{\tau})$ is a compact space. Also It is not metrizable, because $\overline{X}$ is not a Hausdorff space.

Can we say that $\overline{X}$ is sequentially compact?

Please help me to know it.

  • $\begingroup$ @evaristegd, Thanks for your reply. I would like to know that $\overline{X}$ is a sequentially compact? $\endgroup$
    – user479859
    Jul 17 '19 at 4:59

Yes, $\overline{X}$ is sequentially compact. Given a sequence $(a_n)$ in $\overline{X}$, if $a_n\in X$ for infinitely many $n$, then we can find a subsequence which converges in $X$ and hence in $\overline{X}$. The only other possibility is that $a_n=x_0$ for all but finitely many $n$, and then $(a_n)$ converges to $x_0$.

More generally, any space which is a finite union of sequentially compact subspaces is sequentially compact (here the subspaces are $X$ and $\{x_0\}$). The proof is similar: given a sequence, it must have infinitely many terms in one of the subspaces, and so it has a convergent subsequence in that subspace.


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