Let $(X,\tau)$ be a KC non-compact space. Then there is a discrete subset $D \subseteq X$ such that $\overline D$ is not compact. Furthermore there is an ultrafilter $F$ in $X$ such that $ D \in F $ and for every $C \in F$, $\overline C$ is not compact in $(X,\tau)$.
Proof. Let $ U = \{U_{i} : i < k \} $ be a strictly increasing open cover of $ X $, where $k$ is an infinite regular cardinal. We’ll construct sets $ D_{\beta } = \{ x_{i }: i < \beta \} $ by transfinite induction. First, let $ D_{0} = \{x_{0} \} $ for some $ x_{0 } ∈ U_{0} $. Let $ \beta $ is ordinal successor. If $ \overline{D}_{\beta - 1 } $ is compact, then there is $ \alpha_{\beta}$ such that $ \overline{D}_{\beta - 1 } \subseteq U_{\alpha_\beta}$. Let $x_\beta \in U_{\alpha+1} - U_{\alpha} $ and $ D_{\beta } = D_{\beta - 1 } \cup {x_{ \beta}}$. For limit ordinals $ \beta$, let $ D_{ \beta} = \bigcup_{i < \beta }D_{i} $. This process stops when $ \overline{D}_{\beta} $ is not compact, which holds at least for $ \beta = k $, because then the open cover $\mathcal{ U } $ witnesses that $ \overline{D}_{k} $ is not compact. It is easy to see that $ D_{ \beta} $ is discrete. The open set, which contains exactly one point $ x_{i+1 } $ is $ U_{ \alpha_{i+1}} - \overline{D_{i}}$. Let’s construct the ultrafilter. Let $ F $ a filter-base, such that $ D \in F $ and for any $ C \in F $, $ \overline {C } $ is not compact. Furthermore let $ F $ be maximal with these properties. If $ C \in F $ then for any $ C' \supseteq C $, holds $C'\in F $, because if $C'$ were compact, then it would be closed from $ KC $ property and $ \overline{C} \subseteq C' $, but $ \overline{C}$ is non-compact closed subset of a compact set $C'$, which is a contradiction. If $ D = D_{0 } \cup D_{1} $ and $ D_{0} \cap D_{1} = \varnothing $ then either $D_{0}$ or $ D_{1} $ is in $ F$. Otherwise both $ \overline{D_{0 }} $ and $ \overline{D_{1}} $ would be compact, hence $ \overline{D} = \overline{D_{0 }} ∪ \overline{D_{1 }} $ would be compact. Finally this gives that $ F $ is an ultrafilter.
(1): why we can say in the first paragraph" If $ \overline{D}_{\beta - 1 } $ is compact, then there is $ \alpha_{\beta}$ such that $ \overline{D}_{\beta - 1 } $ ⊆ $ U_{\alpha_{\beta}}$. Let $ x_{ \beta } ∈ U_{\alpha+1} - U_{\alpha} $ and $ D_{\beta } = D_{\beta - 1 } ∪ {x_{ \beta}}$"?
(2): why we can say " This process stops when $ \overline{D}_{\beta} $ is not compact, which holds at least for $ \beta = k $, because then the open cover $\mathcal{ U } $ witnesses that $ \overline{D}_{k} $ is not compact. It is easy to see that $ D_{ \beta} $ is discrete. The open set, which contains exactly one point $ x_{i+1 } $ is $ U_{ \alpha_{i+1}} -\overline{D_{i}}$."?
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. See the turorial on meta. $\endgroup$