# ultrafilter and KC minimalspace

By definition in a $KC$ space every compact set is closed.

A space $(X,‎\tau)‎‎‎$ is $KC$-minimal if $(X,‎\tau )‎‎‎$ is a $KC$ space and there is no $KC$ space $(X,\sigma)$ such that $\sigma\subsetneqq\tau$.

In the proof of the lemma below I have questions:

Lemma: Let $(X,‎\tau)‎‎‎$‎ be a $KC$-minimal space, $D$ a discrete subset with non-compact $\tau$-closure, $F$ an ultrafilter such that $D\in F$ and $F$ contains only sets with non-compact $\tau$-closures. Let $\sigma=\{U\in\tau:x_{0}\notin U\}\cup\{U\in\tau:x_{0}\in U\text{ and }U\in\mathcal{F}\}$ .Then every $\sigma$-compact subset is also $\tau$-compact.

Proof: Suppose for contradiction that $M$ is a $\sigma$-compact set which is not $\tau$-compact. Then there is $\tau$-open neighbourhood $U_0$ of $\{x‎_0\}$ such that $M‎ \setminus ‎U‎_0$‎ is not $\tau$-compact either. Let $N=(M\setminus U_0)\cup\{x_0\}$. Now we’ll prove that $N$ is $\tau$-closed. Let $x‎‎\in‎‎‎‎\overline{N}‎$.

Let $x \in V$ be such that ‎$K‎ =‎ ‎‎\overline{V\setminus U‎_{0}‎‎}‎‎$‎ is $\tau$-compact. Then the topologies $\tau$ and $\sigma$ agree on $K$. Since $V$ is a neighbourhood of $x$, we have ‎$x‎\in‎‎\overline{V\cap N}‎\subseteq\overline{K\cap N‎}\cup\{x‎_0\}‎‎‎‎$‎. Note that $N\cap K$ is $\sigma$-compact because it is a closed subset of a compact space $N$. But $\sigma$ and $\tau$ still agree on $K$, hence $N\cap K$ is $\tau$-compact, and so $\tau$-closed. This gives $x\in N$. Finally we have two possibilities:

(a) If $X\setminus N‎\in\mathcal{F}‎‎$ then the topologies $\tau$ and $\sigma$ agree on $N$, and hence it is $\tau$-compact.

(b) On the other hand, if $N‎ ‎\in\mathcal{F}‎‎‎$, then let $D‎' = D‎\cap ‎N‎$. Since $D‎'$ is discrete, we know that $\overline{D‎'}‎‎\setminus ‎D‎'$ is closed. Let $W$ be such an open set, that $\overline{D‎'\cap W} = D‎'$. Then $W‎\cup ‎U‎_0$ is a $\sigma$-open neighbourhood of $x‎_0$.

Now suppose that $\overline{D‎'}$ is not $\tau$-compact (otherwise $N$ would be $\tau$-compact). We know that there is a set $C$ without any complete $\tau$-accumulation points. But $C$ has a complete accumulation point in the topology $\sigma$, hence this point is $x‎_0$. Then ‎$|(‎ W‎ ‎‎\cup ‎U‎_{0} )‎ ‎‎\cap C‎| =‎ ‎‎|C‎|$, beacuse $W‎\cup ‎U‎_0$ is a $\sigma$-open neighbourhood of $x‎_0$. Since ‎$(‎W‎\cup ‎U‎_0)‎\cap C‎\subseteq ‎D‎'$‎, we can suppose without loss of generality that $C‎\subseteq D‎'$.

Let $D‎_0\cup D_1=‎ ‎C‎$, where $D‎_0$ and ‎$D‎_1$‎ ‎are disjoint and have the same cardinality as $C$. At most one of these sets can be in $F$. Without loss of generality assume that $D‎_1\notin\mathcal{F}‎‎$. Since $D$ is discrete, we get $\overline{‎(D‎_1‎)_\tau‎}\notin\mathcal{F}‎‎$ and so $D‎_1$‎ has no $\sigma$-accumulation points that are not $\tau$-accumulation points. Hence $‎D‎_1$ has no complete $\tau$-accumulation point. This contradicts $N$ is $\tau$-compact.

so, my problems are:

1. why in part (a) we can say: If $X\setminus N‎\in\mathcal{F}‎‎$ then the topologies $\tau$ and $\sigma$ agree on $N$, and hence it is $\tau$-compact.

2. why in part (b) we can say: $‎‎\overline{D‎'}‎‎\setminus ‎D‎'$ is closed and $W‎\cup ‎U‎_0$ is a $\sigma$-open neighborhood of $x‎_0$ and (otherwise $N$ would be $\tau$-compact)

3. why we can say: $C$ has a complete accumulation point in the topology $\sigma$, hence this point is $x‎_0$

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