Let $X$ be a topological space.

A subset $U$ of $X$ is sequentially open if each sequence $(x_n)$ in $X$ converging to a point of $U$ is eventually in $U$.

A subset $F$ of $X$ is sequentially closed if, whenever $(x_n)$ is a sequence in $F$ converging to $x$, then $x$ must also be in $F$.

(1):A sequential space is a space $X$ satisfying one of the following equivalent conditions:

Every sequentially open subset of $X$ is open.

Every sequentially closed subset of $X$ is closed.

(2):A KC space is a space that every compact subset is closed.

I have a question:

Is a compact, countable KC-space sequential? Why?


Yes, every compact, countable $KC$ space is sequential; this is Corollary $3$ in Ofelia T. Alas and Richard G. Wilson, ‘Spaces in which compact sets are closed and the lattice of $T_1$ topologies on a set’, Comment. Math. Univ. Carolinae, $43$ ($2002$), $641$-$652$.

  • $\begingroup$ What does it mean "If A is not closed, then it is not compact." ?which relation does have with "Theorem 2" in that paper? $\endgroup$ – fatemeh Sep 25 '13 at 11:02
  • 1
    $\begingroup$ @fatemeh: Let $A$ be a subset of $X$ that isn’t closed. $X$ is $KC$, so $A$ is not compact. By Theorem $2$ there is a sequence $\langle x_n:n\in\omega\rangle$ in $A$ converging to some $p\in X\setminus A$. This shows that $A$ is not sequentially closed. It follows that every sequentially closed set in $X$ is closed and hence that $X$ is sequential. $\endgroup$ – Brian M. Scott Sep 25 '13 at 11:09

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