# A hereditarily Lindelöf, minimal KC-space is sequential

A space is said to have the finite derived set property if each infinite subset $A ⊂ X$ contains an infinite subset with only finitely many accumulation points in X.

A hereditarily Lindelöf, minimal KC-space is sequential.

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

I would like to know:

1: Why is $D$ discrete subset? is $D$ closed subset, due to there is no accumulation points in $D$?

2: Why is there a sequence in E converging out of E and hence out of A?( I mean $cl(E)$ is sequential, but there is a sequence in E converging out of E)

• A duplicate. Really? They don't even concern the same proof. And don't ask identical questions. – Lord_Farin Sep 5 '13 at 9:14

Under the assumption of Countable Choice, it can be shown that a $T_1$ space is countably-compact if and only if it has no countably-infinite closed subset which is discrete in the subspace topology (equivalently, no countably-infinite subset without accumulation points). A $KC$ space is hereditarily $T_1,$ so $A$ is $T_1$. Since $A$ is Lindelöf and non-compact, then it is not countably-compact, and so has a countably-infinite closed subset $D$ that is discrete as a subspace of $A$. Then $D$ has no accumulation points in $A$, for if such a point existed, it would be an element of $D$ (since $D$ is closed) that is not isolated in $D$, which is not possible since $D$ is discrete as a subspace of $A$.
As remarked in the proof, $D$ may well have accumulation points in $X$ (that is, $D$ need not be closed in $X$), but such points lie outside of $A$--and in particular, lie in $\operatorname{cl}(A)\setminus A$.