# A compact KC-space of cardinality less than c has the FDS-property

we say that a space $X$ has the finite derived set property (which we abbreviate as the $FDS$ property) if whenever $A$ is infinite, there is an infinite subset $B\subseteq A$ such that $B$ has only a finite number of accumulation points in $X$.

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

Below Theorem comes from "THE FDS-PROPERTY AND SPACES IN WHICH COMPACT SETS ARE CLOSED " by O.T. Alas, M.G. Tkachenko, V.V. Tkachuk, and R.G. Wilson. I have questions from it.

Theorem $\bf1.7.$ A compact $KC$ space of cardinality less than $\mathfrak{c}$ has the $FDS$ property (and hence is sequentially compact).

Proof: Suppose that $(X,\tau)$ is a compact $KC$ space and $|X| < \mathfrak{c}$. We assume to the contrary that $X$ does not have the $FDS$ property and so there is some countably infinite subset $A \subseteq X$ such that every infinite subset of $A$ has infinitely many accumulation points in $X$. By Lemma $1.2$, without loss of generality, we can assume that $A$ is discrete and that $\operatorname{cl}_X(A) = X$. Let $x\in X \ A$; if every neighbourhood $U$ of $x$ is such that $A\setminus U$ is finite, then $A\cup\{x\}$ is compact, hence closed in $X$ and so $x$ is the unique accumulation point of $A$, a contradiction. Thus we can choose an open neighbourhood $V$ of $x$, such that $A \setminus V$ is infinite. The closed subspace $X \setminus V$ of $X$ is compact and $A$ is countable, and hence Lindelöf, and so $A\cup (X \setminus V )$ is Lindelöf. This in its turn implies that $A \cup (X \setminus V )$ is not countably compact, for otherwise it would be compact, but being a proper dense subspace of $X$, it is not closed in $X$, a contradiction. Thus there is a discrete set $D \subseteq A \cup (X \setminus V )$ which is closed in $A \cup (X \setminus V )$. However, since $X \setminus V$ is compact, only a finite number of points of $D$ lie in $X \setminus V$ and hence $D \cap V$ is infinite and all its accumulation points (by our hypothesis, an infinite number in $X$) must lie in $V$; that is to say, $\operatorname{cl}_X(D \cap V )\subseteq V$.

Thus we have constructed two infinite sets $D \cap V$ and $A \setminus V$ whose closures are disjoint in $X$. Since $D \cap V \subseteq A$, each of these sets has the property that every infinite subset has an infinite number of accumulation points and the above argument can be repeated using $D \cap V$ and $A \subseteq V$ in place of $A$. A standard binary tree argument can now be used to show that $|X| \ge\mathfrak{c}$.

my questions are:
(1) why he can say :

However, since $X \setminus V$ is compact, only a finite number of points of $D$ lie in $X \setminus V$ and hence $D \cap V$ is infinite and $\operatorname{cl}_X(D \cap V) \subseteq V$?

(2) how he reach to contradiction in the last line

"A standard binary tree argument can now be used to show that $|X|\ge\mathfrak{c}$.

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You really ought to proofread when you cut and paste: the fi-ligature generally doesn’t copy, which is why this has nite instead of finite and innite instead of infinite. – Brian M. Scott Jul 31 '13 at 7:45

$X\setminus V$ is compact because it is a closed subset of the $KC$ space $X$. $D$ is discrete and closed in $A\cup(X\setminus V)$, so $D\cap(X\setminus V)$ is closed and discrete in $X\setminus V$. A compact set has no infinite closed discrete subset, so $D\setminus V=D\cap(X\setminus V)$ must be finite. But $D$ is infinite, so $D\cap V$ must also be infinite. Finally, $D$ is closed in $A\cup(X\setminus V)$, so $D$ has no accumulation points in $A\cup(X\setminus V)$ and therefore none in $X\setminus V$. Thus, all accumulation points of $D$ are in $V$, and it follows immediately that $\operatorname{cl}_X(D\cap V)\subseteq V$.
In the first paragraph they construct the infinite sets $D\cap V$ and $A\setminus V$ whose closures in $X$ are disjoint; call these sets $A_0$ and $A_1$. They point out that $A_0$ and $A_1$ satisfy the same hypotheses as $A$, so the same construction can be applied to each of them. This produces infinite subsets $A_{00}$ and $A_{01}$ of $A_0$ and infinite subsets $A_{10}$ and $A_{11}$ of $A_1$ whose closures in $X$ are pairwise disjoint and which again satisfy the same hypotheses as $A$. Continuing recursively in this fashion we can construct sets $A_\sigma$ for each finite sequence $\sigma$ of zeroes and ones in such a way that for each such $\sigma$, $A_{\sigma^\frown0}$ and $A_{\sigma^\frown1}$ are infinite subsets of $A_\sigma$ whose closures in $X$ are disjoint. (Here $\sigma^\frown i$ is the sequence obtained by appending $i$ to the sequence $\sigma$. For instance, $01101^\frown0=011010$.)
Now let $\alpha$ be any infinite sequence of zeroes and ones. For each $n\in\omega$ let $\sigma_n$ be the finite sequence consisting of the first $n$ terms of $\alpha$. For each $n\in\omega$ the set $\operatorname{cl}_XA_{\sigma_n}$ is closed in $X$ and therefore compact. Let $$K_\alpha=\bigcap_{n\in\omega}\operatorname{cl}_XA_{\sigma_n}\;;$$ $\operatorname{cl}_XA_{\sigma_n}\supseteq\operatorname{cl}_XA_{\sigma_{n+1}}$ for each $n\in\omega$, so $K_\alpha\ne\varnothing$. You can easily check that if $\alpha$ and $\beta$ are distinct infinite sequences of zeroes and ones, then $K_\alpha\cap K_\beta=\varnothing$. Finally, there are $\mathfrak{c}$ infinite sequences of zeroes and ones, so $X$ has $\mathfrak{c}$ pairwise disjoint, non-empty subsets $K_\alpha$. Clearly this implies that $|X|\ge\mathfrak{c}$.