If X is a countable KC-space, then every infinite $D\subset X$ contains an infinite subset with only a finite number of accumulation points (in X)

a topological space is called KC- space if every compact set is closed.

a topological space is called US, if every convergent sequence has unique limit. generally, KC- space imply US - space.

I have read bellow theorem from " Spaces in which compact subsets are closed and the lattice of T1-topologies on a set " by " Ofelia T. Alas, Richard G. Wilson ", dml.cz link.

But I have several difficulty with this. He said that:

" Lemma 8. If X is a countable KC-space, then every infinite $D ‎‎‎\subseteq ‎X‎$‎‎ contains an infinite subset with only a finite number of accumulation points (in $X$ )."

" Proof: Enumerate $X$ as $\{ ‎x‎_{n} : n ‎‎\in ‎\omega ‎‎‎‎‎\}$ and suppose that $D ‎‎‎\subseteq ‎X‎$‎‎ is infinite and every infinite subset of $D$ has infinitely many accumulation points. Let $‎n‎_{0} ‎‎‎\in ‎\omega ‎‎‎$‎ be the smallest integer such that $x‎_{n‎_{0}‎}‎$ is an accumulation point of $D$ . If each neighborhood V of $x‎_{n‎_{0}‎}‎$ has the property that $D ‎\setminus ‎V‎$‎ is finite, then any enumeration of $D$ converges to $x‎_{n‎_{0}‎}‎$‎ and hence $D$ has only one accumulation point, a contradiction. Thus we may choose an open neighborhood $V‎_{0}‎$ of $x‎_{n‎_{0}‎}‎$ such that $‎D‎_{1} =‎ D‎ ‎‎\setminus ‎V‎_{0}‎‎$‎ is infinite. Having chosen points ‎$‎x‎_{n‎_{0}‎}‎,‎‎‎ ‎x‎_{n‎_{‎1‎}‎}, ‎\ldots , ‎‎‎‎‎x‎_{n‎_{‎j-1‎}‎} ‎‎$‎ and open sets $‎V‎_{0}‎,‎ ‎V‎_{‎1‎‎}, ‎\ldots , ‎V‎_{‎j-1‎‎}‎‎‎$ such that ، ‎‎‎$x‎_{n‎_{k}‎}‎ ‎\in ‎V‎_{k}‎‎$‎‎ for each ‏‎$1‎ ‎‎\leq k‎ ‎‎\leq ‎j-1‎$‎‎ and $‎D‎_{j} = D ‎‎\setminus \big( ‎\bigcup \{ ‎V‎‎_{k} : ‎1‎\leq k ‎\leq j-1‎‎‎‎‎‎\} \big)‎‎‎‎‎$‎ is infinite, we let ‎$‎n‎_{j}‎‎$ be the least integer such that $‎x‎_{n‎_{j}‎}‎‎$‎ is an accumulation point of $D‎_{j}‎$‎ and we choose a neighborhood ‎$‎V‎_{j}‎‎$ of $‎x‎_{n‎_{j}‎}‎‎$‎ such that $‎D‎_{j+1} = ‎D‎_{j } ‎\setminus ‎V‎_{j} ‎‎‎‎‎‎$ is infinite. Such a choice is again possible for if every neighborhood V of $‎x‎_{n‎_{j}‎}‎‎$‎ is such that $‎D‎_{j}‎‎ ‎\setminus ‎V‎$ is finite, then any enumeration of $D‎_{j}‎$ is a sequence which converges to $‎x‎_{n‎_{j}‎}‎‎$‎ and hence $D‎_{j}‎$ has only one accumulation point.

Now for each $j ‎\in‎ ‎\omega‎$ we choose $‎y‎_{j} ‎\in ‎‎‎ ‎D‎_{j}‎‎ ‎\setminus ‎\{ ‎y‎_{0}‎,‎‎‎ ‎y‎_{‎1‎‎},‎ ‎\ldots,‎ ‎y‎_{‎j-1‎‎}‎ ‎‎ ‎\}‎$ and we denote the set $‎\{ ‎y‎_{n} : n ‎‎\in‎\omega‎‎‎‎‎‎\}‎$ ‎‎by $S$. It is clear that $S$. is infinite and all but finitely many points of S are contained in $D‎_{j}‎$ for each $j‎ ‎‎\in‎‎ ‎\omega‎$‎ and so an accumulation point of S is an accumulation point of $S‎ ‎\cap ‎D‎_{j}‎‎$‎ for each $j‎ ‎‎\in‎‎ ‎\omega‎$‎. Thus $S$. can have no accumulation point, since if p were such a point, then for some $k‎ ‎‎\in‎‎ ‎\omega‎$, $p‎ =‎ ‎x‎_{k}‎‎$ and from the construction, we would have that ‎‎$k‎ ‎‎\geq ‎n‎_{j}‎‎$ for each $j‎ ‎‎\in‎‎ ‎\omega‎$, which is absurd."

my questions are:

(a) : why in first paragraph he said: "any enumeration of $D$ converges to $x‎_{n‎_{0}‎}‎$ and hence $D$ has only one accumulation point, a contradiction."

(b): why in the last line he said:" Thus $S$. can have no accumulation point, since if $p$. were such a point, then for some $k‎ ‎‎\in‎‎ ‎\omega‎$, $p‎ =‎ ‎x‎_{k}‎‎$ and from the construction, we would have that ‎‎$k‎ ‎‎\geq ‎n‎_{j}‎‎$ for each $j‎ ‎‎\in‎‎ ‎\omega‎$, which is absurd."

(c) : can we say that $D$ is discreet and closed subset of $X$.?

plese help me.

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I see that you have suggested an edit using a different account. (It does not seem probable that someone with the same username edited the post.) I think it would be useful for you to register. You will be able to follow your posts easier and it won't happen that you are not able to edit your own post. – Martin Sleziak Aug 1 '13 at 7:34

(a) Suppose that $D\setminus V$ is finite for each nbhd $V$ of $x_{n_0}$, and let $D=\{y_k:k\in\omega\}$ be any enumeration of $D$; I claim that the sequence $\langle y_k:k\in\omega\rangle$ converges to $x_{n_0}$. To prove this, let $V$ be any open nbhd of $x_{n_0}$; we must show that there is an $m\in\omega$ such that $y_k\in V$ for all $k\ge m$. By hypothesis $D\setminus V$ is finite. If $D\setminus V\ne\varnothing$, let $m=1+\max\{k\in\omega:y_k\in F\}$; otherwise let $m=0$. In either case $y_k\in V$ for every $k\ge m$, and $\langle y_k:k\in\omega\rangle\to x_{n_0}$.

(b) Suppose that $p$ is an accumulation point of $S$. Recall that $X=\{x_k:k\in\omega\}$, and of course $p\in X$, so $p=x_k$ for some $k\in\omega$. Now let $j\in\omega$ be arbitrary; $p$ is an accumulation point of $S\cap D_j$. By construction

$$n_j=\min\{\ell\in\omega:x_\ell\text{ is an accumulation point of }D_j\}\;,$$

and $k\in\{\ell\in\omega:x_\ell\text{ is an accumulation point of }D_j\}$, so $k\ge n_j$. Since $j$ was arbitrary, we have $k\ge n_j$ for each $j\in\omega$. For each $j$ we have $x_{n_j}\in V_j$ and $D_{j+1}\cap V_j=\varnothing$, and $x_{n_\ell}\in D_{j+1}$ if $\ell>j$, so the points $x_{n_j}$ are distinct: if $j<\ell<\omega$, $x_{n_j}\ne x_{n_\ell}$. This means that $\{x_{n_j}:j\in\omega\}$ is infinite, so $\{n_j:j\in\omega\}$ is infinite, and there is no natural number $k$ such that $k\ge n_j$ for all $j\in\omega$.

(c) No, definitely not. $\Bbb Q$ is a countable $KC$ space with lots of infinite subsets that are neither closed nor discrete.

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The OP did not ask this, but I wonder why in (a) the fact that any enumeration of $D$ converges to $x_0$ is sufficient to claim that $x_0$ is the only accumulation point of $D$. How can I be sure that there is not a net consisting of some elements of $D$ converging to a different point? (I think that the quotient space obtained from Arens-Fort space and contains a countable subset $D$ such that every enumeration of this set converges to the same point $x_0$, but there is a different point, which is the accumulation point of $D$. However, this example is not $T_1$ and thus not KC.) – Martin Sleziak Aug 1 '13 at 12:39
@Martin: $\{x_{n_0}\}\cup D$ is compact, therefore closed, so any other accumulation pt. would have to be in $D$. But if $x\in D$, then $\{x_{n_0}\}\cup(D\setminus\{x\})$ is compact, therefore closed, so $x$ is isolated in $\{x_{n_0}\}\cup D$. Therefore $x_{n_0}$ is the only accum. pt. of $D$. – Brian M. Scott Aug 2 '13 at 3:55