A sequential minimal KC-space is compact A space $(X,\tau )$ is said to
be minimal KC , if $(X,\tau )$ is KC but no topology
on X which is strictly smaller  than $ \tau$  will be KC
Theorem : A sequential minimal KC-space is compact.
Proof: Let $(X, τ)$ be a non-compact space satisfying the hypothesis of the theorem. Fix
$a ∈ X$ and define a new topology $σ$ on $X$ as follows:
$$σ = \{U ∈ τ : a \not\in U \} ∪ \{U ∈ τ : a ∈ U\text{ and }X - U\text{ is compact}\}.$$
$(X, σ)$ is a compact $T_{1}$-space and $σ ⊂ τ$. Thus to complete the proof, it suffices
to show that $(X, σ)$ is a KC-space. To this end, suppose that $S ⊆ X $ is a compact subset
of $(X, σ)$. It is clear that $cl_{σ}(S) ⊆ cl_{τ} (S) ∪ \{a\}$ and that if $a \not\in S$ , then $S_{σ} = S_{τ}$. There
are then two possibilities:
(i) If $a \not\in S$, then by the preceding remarks, $S$ is compact, and hence closed, in $(X, τ)$
and so $X - S$ is an open $σ$-neighbourhood of $a$. Thus $a \not\in cl_{σ}(S)$ and so $cl_{σ}(S) = cl_{τ} (S) = S$.
(ii) If $a ∈ S$ then $cl_{σ}(S) = cl_{τ }(S)$ and so if $S$ is not closed in $(X, σ)$, then it is not closed
in $(X, τ)$ either. Thus there is some $x ∈ cl_{τ} (S) - S$ and a sequence $ \{x_{n}\}_{n∈ω} $ in $S$ convergent
to $x$. Since $a \not =  x$, we may assume that $x_{n} \not =  a$ for all $n ∈ ω$. Then $K = \{x_{n} : n ∈ ω \} ∪ \{x\}$ is compact in $(X, τ)$, hence closed in $(X, τ)$ and since $a \not\in K$, it is closed in $(X, σ)$. Thus
$K ∩S = \{x_{n} : n ∈ ω\}$ is a closed subset of the compact space $(S, S_{\sigma})$ and thus is compact.
Since $a \not\in K ∩ S$, we have $(K ∩ S)_{σ} = (K ∩ S)_{τ}$ so $K ∩ S$ is compact in $(X, τ)$ and hence
closed in $(X, τ)$. However, $x ∈ cl_{τ} (K ∩ S) - (K ∩ S)$ which is a contradiction.
So, can you give me a hand?

(1) why $(X, σ)$ and T_{1}?
(2) : Since $a \not =  x$, we may assume that $x_{n} \not =  a$ for all $n ∈ ω$?
(3)$K ∩S = \{x_{n} : n ∈ ω\}$ is a closed subset of the compact space $(S, S_{\sigma})$ ,( why in $(S, S_{\sigma})$? why $(K ∩ S)_{σ} = (K ∩ S)_{τ}$?

 A: *

*$\langle X,\tau\rangle$ is $T_1$, since it's $KC$. Let $x\in X\setminus\{a\}$; then $X\setminus\{x\}$ is a $\tau$-open nbhd of $a$, so $X\setminus\{x\}\in\sigma$ and is therefore a $\sigma$-open nbhd of $a$ that does not contain $x$. Now let $y$ be any other point of $X\setminus\{x\}$, and let $U=X\setminus\{a,y\}$. Then $a\notin U\in\tau$, so $U\in\sigma$, and clearly $x\in U$ and $y\notin U$. Thus, for any distinct $x,y\in X$ there is a $U\in\sigma$ such that $x\in U$ and $y\notin U$, and $\langle X,\sigma\rangle$ is therefore $T_1$.

*Since $x\ne a$, $X\setminus\{a\}$ is a $\tau$0-open nbhd of $x$. The sequence $\langle x_n:n\in\omega\rangle$ converges to $x$, so there is an $m\in\omega$ such that $x_n\in X\setminus\{a\}$ for all $n\ge m$. Without loss of generality replace $\langle x_n:n\in\omega\rangle$ by $\langle x_n:n\ge m\rangle$: this is still a sequence in $S$ converging to $x$, and no term of it is equal to $a$.

*$K$ is closed in $\langle X,\sigma\rangle$, and $S$ is compact in $\langle X,\sigma\rangle$, so $K\cap S$ is closed in the compact space $\langle S,S_\sigma\rangle$. A closed subset of a compact space is always compact, so $K\cap S$ is compact in $\langle S,S_\sigma\rangle$.

*The only point of $X$ at which the topologies $\sigma$ and $\tau$ differ is $a$. This means that if $Y$ is any subset of $X$, and $a\notin Y$, then $Y_\sigma=Y_\tau$. Taking $Y=K\cap S$, we see that $(K\cap S)_\sigma=(K\cap S)_\tau$.
