ultrafilter and $KC$-minimal space

ultrafilter and $KC$-minimal space

By definition in a $KC$ space every compact set is closed.
A space $(X,ý\tau)ýýý$ is $KC$-minimal if $(X,ý\tau )ýýý$ 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 $(X,ý\tau)ýýý$ý 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
$\{xý_0\}$ such that $Mý \setminus ýUý_0$ý is not $\tau$-compact either.
Let $N=(M\setminus U_0)\cup\{x_0\}$. Now we'll prove that $N$ is
$\tau$-closed. Let $xýý\inýýýý\overline{N}ý$.
Let $x \in V$ be such that ý$Ký =ý ýý\overline{V\setminus Uý_{0}ýý}ýý$ý is
$\tau$-compact. Then the topologies $\tau$ and $\sigma$ agree on $K$.
Since $V$ is a neighbourhood of $x$, we have ý$xý\inýý\overline{V\cap
N}ý\subseteq\overline{K\cap Ný}\cup\{xý_0\}ýýýý$ý. 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 $X\setminus Ný\in\mathcal{F}ýý$ then the topologies $\tau$ and
$\sigma$ agree on $N$, and hence it is $\tau$-compact.
(b) On the other hand, if $Ný ý\in\mathcal{F}ýýý$, then let $Dý' = Dý\cap
ýNý$. Since $Dý'$ is discrete, we know that $\overline{Dý'}ýý\setminus
ýDý'$ is closed. Let $W$ be such an open set, that $\overline{Dý'\cap W} =
Dý'$. Then $Wý\cup ýUý_0$ is a $\sigma$-open neighbourhood of $xý_0$.
Now suppose that $\overline{Dý'}$ 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 $ xý_0$. Then ý$|(ý Wý
ýý\cup ýUý_{0} )ý ýý\cap Cý| =ý ýý|Cý|$, beacuse $Wý\cup ýUý_0$ is a
$\sigma$-open neighbourhood of $xý_0$. Since ý$(ýWý\cup ýUý_0)ý\cap
Cý\subseteq ýDý'$ý, we can suppose without loss of generality that
$Cý\subseteq Dý'$.
Let $Dý_0\cup D_1=ý ýCý$, where $Dý_0$ and ý$Dý_1$ý ýare disjoint and have
the same cardinality as $C$. At most one of these sets can be in $F$.
Without loss of generality assume that $Dý_1\notin\mathcal{F}ýý$. Since
$D$ is discrete, we get $\overline{ý(Dý_1ý)_\tauý}\notin\mathcal{F}ýý$ and
so $Dý_1$ý has no $\sigma$-accumulation points that are not
$\tau$-accumulation points. Hence $ýDý_1$ has no complete
$\tau$-accumulation point. This contradicts $N$ is $\tau$-compact.
so, my problems are:
why in part (a) we can say: If $X\setminus Ný\in\mathcal{F}ýý$ then the
topologies $\tau$ and $\sigma$ agree on $N$, and hence it is
$\tau$-compact.
why in part (b) we can say: $ýý\overline{Dý'}ýý\setminus ýDý'$ is closed
and $Wý\cup ýUý_0$ is a $\sigma$-open neighborhood of $xý_0$ and
(otherwise $N$ would be $\tau$-compact)?
why we can say: $C$ has a complete accumulation point in the topology
$\sigma$, hence this point is $xý_0$?