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A space $X$ is called scattered provided every non-empty subspace $Y$ has an isolated point( with respect to the subspace topology on $Y$).

How to prove that:

Every countably compact, scattered $T_3$-space is sequentially compact?

Thanks for your help.

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  • $\begingroup$ It would be nice if the space $X$ was $T_1$ also, because it seems that in this case I can easily obtain the proof. $\endgroup$ May 17 '13 at 15:00
  • $\begingroup$ @AlexRavsky: The $T_3$ space implies $T_1$ when I denote it. $\endgroup$
    – Paul
    May 18 '13 at 0:37
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It seems the following. Let $X$ be a countably compact, scattered regular (that is $T_3$ and $T_1$) space and $S=\{x_n:n\in\omega\}\subset X$ be a sequence. If the set $\overline S\backslash S$ is empty then $S$ is closed in $X$ and we put $Y=S$. In the opposite case the subspace $\overline S\backslash S$ has an isolated point $x_0$. Therefore there exists a neighborhood $U_0$ of the point $x_0$ in the space $\overline S$ such that $U_0\subset\{x_0\}\cup S$. Since the space $\overline S$ is regular, there exists an open neighborhood $V_0$ of the point $x_0$ in the space $\overline S$ such that $\operatorname{cl}_{\overline S} V_0\subset U_0$. Put $Y=\operatorname{cl}_{\overline S} V_0$. In both cases $Y$ is a countable countably compact space, containing infinitely many members of the sequence $S$ (some of them may coincide as points of $X$). Because $Y$ is a regular countably compact space with $G_\delta$-diagonal, by Chaber Theorem is a metrizable compact (see, for instance, “Handbook of Set-Theoretic Topology 1984”, p.422 of the file or p.433 of the book). Because $Y$ contains infinitely many members of the sequence $S$, we can construct a convergent subsequence of $S$ by them.

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    $\begingroup$ Since $Y$ is countable, $Y$ is Lindelof, and hence it is compact metrizable. $\endgroup$
    – Paul
    May 19 '13 at 9:17

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