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** Lemma :Let $(X,\tau )$ be a KC-space which is not countably compact. Then X can be condensed onto a weaker KC-topology.**

Proof: Let new topology

$‎ ‎\tau‎^{‎\prime‎} = ‎\{U‎‎\in‎‎ ‎\tau:‎‎ ‎x‎_{0}\not\in U\}‎\cup ‎\{U‎\in \tau:‎‎ ‎‎‎x‎_{0} ‎‎‎\in U,U‎‎\in \mathcal{F}‎‎‎‎‎‎\}‎‎‎‎‎‎$

on $X$, where set $ \{ x_n : n \in \omega \} \subset X$ which has no accumulation points , and $\mathcal{F}‎‎‎‎‎‎\ $ is a uniform ultrafilter define over teh set $ \{ x_n : 0< n< \omega \}$. We will prove that $(X, \tau‎^{‎\prime‎})$ is a $KC$-space, and the proof will be a consequence of lemma (1).

For this we will show that there is an $F ∈ \mathcal{F}‎‎‎‎‎‎$ with $F ∩ K = ∅$ Indeed, suppose for a contradiction that there is $F_0 ∈ F$ such that $F_0 ⊂ \overline{K}_\tau$ . Let F1, F2 be subsets of F0 with $|F1| = |F2| = \omega $, $F_1 ∪F_2 = F_0$, and $F_1 ∩ F_2 = ∅$. Suppose that $F_1 ∈ \mathcal{F}$. We claim that $F_1 ∪ K$ is $\tau$-compact. Actually let $ \{ U_i : i ∈ I\}$ be a $\tau$-open cover of $F_1 ∪ K$. Then countably many of the $U^{\prime}_i$s, say, $\{‎U‎_{i‎_{n}‎} : n ∈ \omega \}$, cover the countable set $F_1$, and if we write

$U^{\prime}(x_0) = U(x_0) ∪ \bigcup \{‎U‎_{i‎_{n}‎} : n ∈ \omega \}$,

where $U(x_0)$ is a member of $ \{U_i : i ∈ I\}$ which contains $x_0$ then $U^{\prime}_(x_0)$ is a $\tau‎^{‎\prime‎}$-open neighborhood of $x_0$, and we will have

$ \bigcup \{U_i : i ∈ I\} = U^{\prime}(x_0) ∪ \bigcup \{V_j : j ∈ J \}$,

where $\{V_j : j ∈ J \}$ is a subcollection of $ \{U_i : i ∈ I \}$ which covers $ U^{\prime}(x0)^c ∩ K$. But $\{U_i : i ∈ I\}$ is also a $\tau‎^{‎\prime‎}$-open cover of $K$. So it contains a finite subcover. It turns out that finitely many $V^{\prime}_j$ , say, $‎V‎_{j‎_{1}‎}‎‎ ‎‎ , ‎V‎_{j‎_{2}‎}‎‎ ‎‎, . . . , ‎V‎_{j‎_{k}‎}‎‎ ‎‎$ , cover the set

$K ∩ (U(x_0) ∪ \bigcup \{U‎_{i_{n}‎}‎‎ ‎‎ : n ∈ \omega\})^c = K ∩ U^{\prime}(x_0)^c$.

Now

$ \bigcup \{‎V‎_{j‎_{m}‎} : m = 1, 2, . . . , k \} ∪ \bigcup \{U‎_{i_{n}‎}‎‎ : n ∈ \omega \} ∪ U(x_0)$

is a countable $\tau$-open cover of $K$. By lemma (2), it has a finite subcover. So $K ∪ F_1$ is $\tau$-compact and therefore $\tau$-closed. But this is impossible since every $x ∈ F_2$ is a $\tau$-accumulation point of$ K$. So there must be an $F \in \mathcal{F}$ with $F ∩ \overline{K}_{\tau} = ∅$ , so implies that $K$ is $\tau$-closed. Then $K$ is $\tau‎^{‎\prime‎}$-closed.

Lemma(1):Let $(X,\tau )$ be a KC-space which is not countably compact, $\{ x_n : n \in \omega \}$ a set without accumulation points, $\mathcal{F}$ a uniform ultrafilter defined over $\{ x_n : 0 < n <\omega \}$, and $\tau‎^{‎\prime‎}$ is a new topology as above, and $K$ a $\tau‎^{‎\prime‎}$-compact set . If there exist an $F_0 \in \mathcal{F}$ such that $F_0 \cap \overline{K}_{\tau} = \emptyset $ , then $K$ is $\tau‎^{‎\prime‎}$- closed.

Lemma(2): Let $(X,\tau )$ be a KC-space which is not countably compact,and $\tau‎^{‎\prime‎}$ is a new topology as above, and $K$ a $\tau‎^{‎\prime‎}$-compact set,$x_0 \in K$ and $F_0 \in \mathcal{F}$ with $F_0 \subset ( \overline{K}_{\tau} - K ) $. Then $K$ is $\tau$- countably compact.

My questions are:

(1): Why can we say:

" $ \bigcup \{U_i : i ∈ I\} = U^{\prime}(x_0) ∪ \bigcup \{V_j : j ∈ J \}$, where $\{V_j : j ∈ J \}$ is a subcollection of $ \{U_i : i ∈ I \}$? which covers $ U^{\prime}(x0)^c ∩ K$?

and $\{U_i : i ∈ I\}$ is also a $\tau‎^{‎\prime‎}$-open cover of $K$."?

(2): Is it right to say:

"finitely many $V^{\prime}_j$ , say, $‎V‎_{j‎_{1}‎}‎‎ ‎‎ , ‎V‎_{j‎_{2}‎}‎‎ ‎‎, . . . , ‎V‎_{j‎_{k}‎}‎‎ ‎‎$ , cover the set $K ∩ (U(x_0) ∪ \bigcup \{U‎_{i_{n}‎}‎‎ ‎‎ : n ∈ \omega\})^c = K ∩ U^{\prime}(x_0)^c$."?

(3): Why

"this is impossible since every $x ∈ F_2$ is a $\tau$-accumulation point of$ K$.? and there must be an $F ∈ \mathcal{F}$ with $F ∩ \overline{K}_{\tau}$?

But in the article of " Minimal KC spaces" by " Jakub Opršal" was proved that this proof contains an error.He show that this finite subcover is not always cover of $ K ∪ F_1$, the only thing we can say is that this subcover covers $K$.

Indeed, with assumptions of the proof let’s construct a $\tau$-open cover of $K ∪ F_1$ with no finite subcover. Let $U = K - F_1$, then $U$ is $\tau$-open set in $K$, because $F_1$ has no accumulation points. For each $x_i ∈ F_1$ let $V (x_i)$ be such open set, that contains exactly the point $x_i$ of $F_1$, i.e. such an open set, that $V (x_i) ∩ F_1 = \{x_i\}$ (it exists because $F_1$ is discrete). Finally $\mathcal{U} = \{U\} ∪ \{V (x_i) : x_i ∈ F_1 \}$ is an open cover of $K ∪ F_1$, but has no finite subcover, because every finite subcover covers only finite number of points of the infinite set $F_1$

can you help me and give the best result to me? Thanks alot.

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$\newcommand{\cl}{\operatorname{cl}}$This is imperfectly quoted from Theodoros Vidalis, ‘Minimal $KC$-spaces are countably compact’, Commentationes Mathematicae Universitatis Carolinae, Vol. $45$ ($2004$), No. $3$, $543$-$547$, freely available here; specifically, it’s Lemma $3.5$. The proof contains an error, so it’s no wonder that you were having trouble with it. The argument needs to show that $F_1\cup K$ is $\tau$-countably compact, but Lemma $3.4$ does not apply to $F_1\cup K$. It does apply to $K$, but there is no obvious way to reduce a $\tau$-open cover of $K$ to a countable one.

In the meantime the stronger result that every minimal $KC$-space is compact has been proved by Angelo Bella and Camillo Costantini in ‘Minimal $KC$ spaces are compact’, Topology and its Applications $155$ $(2008)$ $1426$-$1429$.

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