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)

  • 1
    $\begingroup$ A duplicate. Really? They don't even concern the same proof. And don't ask identical questions. $\endgroup$ – Lord_Farin Sep 5 '13 at 9:14

I can give you a partial answer, but I must admit that I'm having trouble parsing the grammar in the last sentence of your proof. Perhaps there was a transcription mistake, or perhaps you've simply got results I've never seen before, but either way, it doesn't make sense to me. What is the source of this proof? If I knew that, I could take a look, and perhaps be better equipped to clear up your confusion.

I will continue to think on it, though, and add to my answer if I come up with anything.

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 $\operatorname{cl}(A)\setminus A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.