A topological space is called $T_B$ if every compact subset is closed.
According to therem ( I, II , III), how does the below theorem proof??
Let $(X,\tau)$ be a $T_B$-space which is not countably compact. Is there $\mu \subset \tau$ s.t $(X,\mu)$ is $T_B$-topology.
(I):$Let (X,\tau)$ be a $T_B$-space which is not countably compact, $\{x_n :n \in \omega\}$ a set without accumulation points, $\mathscr{F}$ a uniform ultrafilter defined over $\{x_n : 0 < n < \omega\}$,the topology $\mu = \{ U \in \tau : x_0 \not\in U \} \cup \{ U \in \tau : x_0 U , U \in \mathscr{F} \}$ is on $X$ and $K$ a $\mu$-compact set. Then there is an $F \in \mathscr{F}$ , such that $F \cap K = \emptyset$.
(II):With the assumptions of Theorem if there exists an $F_0 \in \mathscr{F}$ such that $F_0 \cap \overline{K}_{\tau} = \emptyset$, then $K$ is $\mu$-closed.
(III):Let $(X,\tau)$ be a $T_B$-space which is not countably compact, $K$ a $\mu$-compact set, $x_0 \in K$ and $F_0 \in \mathscr{F}$ with $F_0 ⊂ (\overline{K}_{\tau} − K)$. Then $K$ is $\tau$-countably compact.
