Let $X$ be a$KC$ space. Then $X^{*}$ is $KC$ iff $X$ is a $K$ -space

Let $X$ be a$KC$ space. Then $X^{*}$ is $KC$ iff $X$ is a $K$ -space

The bellow theorem exist in " Between $T_{1} $ and $T_{2}$ " by " Albert
Wilansky ."
Let $ (ý ýXý^{*},ý\tauý^{*} ý)ý $ be topological space one- point
compatification of $ ( X. \tau)$. A topological space is called $k$ -
space if it has the property that any subset $S$ such that $ S\cap K$ is
closed for all closed compact $K$ is itself closed.
* Theorem : Let $X$ be a$KC$ space. Then $X^{*}$ is $KC$ iff $X$ is a $K$
-space*.
Proof: Let $X$ be a $K$ space .Let $S$ be a $\tau^{*}$- compact subset of
$X^{*}$ . If $ S\subset X$, is $\tau$- compact , hence $\tau$-closed, thus
$\tau^{*}$-closed. If on the other hand $\inftyý \in S$, let $ F = S -
\{\inftyý \}$ . Now let $K$ be an ar bitray $\tau $-closed compact subset
of $X$ . Then $K$ is $\tau^{*}$- closed. and so $ S \cap K$
is$\tau^{*}$-compact: hence $\tau $- compact, since it is a subset of $S$
. since $X$ is a $KC$ space,$ S \cap K$ is $\tau $-closed, and so $ S \cap
K = F \cap K$ is $\tau $-closed. since $X$ is $K$ space it follow that $F$
is $\tau $-closed.Then $S$ is $\tau^{*}$-closed because $X^{*} - S = X -
F$ is $ \tau$- open, hence $\tau^{*}$-open.
Conversly, let $X^{*}$ be $KC$. Let $F$ be a subset of $X$ which obeys the
test condition, namely $F \cap K$ is closed for every closed compact
compact set $ K \subset X $. Let $S = F \cup \{ \inftyý \}$. Let $
ý\Omegaý$ be a covering of $S$ by $\tau^{*}$-open sets. Choose $ G \in
\Omegaý$ whith $\inftyý \in \Omegaý$. Now $X - S$ is $\tau^{*}$- closed,
i.e. closed and compact: thus it meets $F$ in a closed(hence compact) set,
i.e . $F - G $ is $\tau$-compact; Let $\Omega^{\prime}ý = \{ O- \inftyý :
O \in \Omega \} $; then $\Omega^{\prime}ý$ is an open cover of $ F - G $.
Reducing $\Omega^{\prime}ý $ to a finite cover of $ F- G $ , restoring $
\inftyý $ to each of its members and adjoining $G$ gives a covering of $S$
which is a finite subcover of $\Omega$. We have now proved that $S$ is
$\tau^{*}$-compact. By hypothesise if follow that $s$ is
$\tau^{*}$-closed. But $ F = S \cap X $; thus $F$ is $\tau$-closed. it
folow that $X$ is a $K$- space.
(1): In the first paragraph, why we can say $ S \cap K = F \cap K$? and
$X^{*} - S = X - F$
(2) In the second paragraph, why " $X - S$ is $\tau^{*}$- closed, i.e.
closed and compact?and $F - G $ is $\tau$-compact" ?
(3)Why is it true that " $\Omega^{\prime}ý$ is an open cover of $ F - G $.
Reducing $\Omega^{\prime}ý $ to a finite cover of $ F- G $ , restoring $
\inftyý $ to each of its members and adjoining $G$ gives a covering of $S$
which is a finite subcover of $\Omega$" ?