A countable, compact KC-subspace of a hereditarily Lindelöf minimal KC-space

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A space in which all compact subsets are closed is called KC-space.

A space in which every infinite set contains an infinite subset with only a finite number of accumulation points is said to have the finite derived set property.

Theorem: A hereditarily Lindelöf minimal KC-space is sequential.

Proof: Suppose that $(X, ‎\tau‎)$ is a hereditarily Lindelof minimal $KC$-space; suppose that $A ‎\subset‎ X $ is not closed and hence not compact. Since $X$ is hereditarily Lindelof, $A$ is not countably compact and hence we can find a countable discrete subset $D = \{x_n : n ∈ \omega \} ‎\subset‎ A$ which is closed in $A$; that is to say, all of the accumulation points of $D$ lie outside of $A$. $X$ has the FDS-property, and so there is some countably infinite set $E ‎\subset‎ D$ with only a finite number of accumulation points in $X$, all of which lie in $‎\overline{A}‎ - A$. Thus $‎\overline{E}‎$ is a countable, compact $KC$-space and $‎\overline{E}‎$ is sequential; thus there is a sequence in E converging out of $E$ and hence out of $A$.

Why is $‎\overline{E}‎$ is a countable, compact $KC$-space?

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  • Since $E$ is countable and has only finitely many accumulation points, it is clear that $\overline{E}$ is countable.
  • By Lemma 2.1 of

    O.T. Alas, M.G. Tkachenko, V.V. Tkachuk, and R.G. Wilson, The FDS property and spaces in which compact sets are closed, Sci. Math. Jap. 2004-46, volume link

    (which appears to be the paper you are reading) we have that every hereditarily Lindelöf, minimal KC-space is compact. Therefore $X$ is compact, and thus so, too, is any closed subspace of $X$, such as $\overline{E}$.

    (As an aside, note that in

    A. Bella, and C. Costantini, Minimal KC spaces are compact, Top. Appl. vol. 155 (2008), No. 13, pp. 1426–1429, link

    it was subsequently shown that all minimal KC spaces are compact.)

  • It is fairly easy to see that subspaces of KC-spaces are themselves KC.