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‎‎Lemma:‎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 a‎‎ ‎new ‎topology‎ $ \tau‎^{‎\prime} ‎$ ‎define on ‎$‎X‎$ ‎as follow:‎

$‎ ‎\tau‎^{‎\prime‎} = ‎\{U‎ ‎‎\in‎‎ ‎\tau :‎‎ ‎x‎_{0} ‎\not‎\in U \} ‎\cup ‎\{U‎ ‎‎\in ‎\tau:‎‎ ‎x‎_{0} ‎‎‎\in U ,‎U‎‎\in ‎\mathcal{F}‎‎‎‎‎‎\}‎‎‎‎‎‎$ and K a $ \tau‎^{‎\prime} ‎$ ‎‎‎-compact set. Then there is an ‎$ ‎F‎ ‎\in ‎‎\mathcal{F} ‎$‎ ‎, such that ‎$ F ‎\cap‎ K =‎\emptyset‎‎.‎‎‎ $‎

‎ ‎ Lemma : ‎With the assumptions of ‎abone ‎Lemma if there exists an ‎$ ‎F_{‎0} ‎\in ‎‎\mathcal{F}‎‎ $‎ such that ‎$ ‎F_{‎0} ‎\cap‎ ‎\overline{‎K‎}‎ =‎‎ ‎‎\emptyset‎‎ $‎ , then K is ‎$ \tau‎^{‎\prime}‎ $‎‎‎‎‎‎-closed.

Proof: ‎Since ‎ ‎$ x_{0} ‎\in‎K $ ‎i‎‎‎t suffices to show that ‎‎$‎K‎$‎ is ‎$ ‎\tau‎ $‎-closed. Let‎ $ ‎ \{ U_i : i ‎\in‎ I \} ‎$‎ , be a ‎$ ‎\tau‎ $‎‎‎-open cover of ‎‎$‎K‎$‎ and let ‎$ ‎V‎_{0‎}‎ ‎$‎‎‎ be an open set containing ‎$ ‎F‎_{0}‎‎ $‎ such that ‎$ ‎F‎_‎0 ∩ K = ‎\emptyset‎‎ $‎ . Then the collection ‎$ \{‎U_‎i ‎\cup ‎V_‎0 : i ∈ I \} $‎, is a $ \tau‎^{‎\prime} ‎$‎-open cover of ‎‎$‎K‎$‎ and thus it has a finite subcover, say,‎$ ‎U_‎i_1 ‎\cup ‎U_‎i_2 ‎\cup‎ . . . ‎\cup ‎U_‎i_n ‎\cup ‎V_‎0‎ $‎ . The set ‎$ ‎\cup \{‎U_‎i_k : k = 1, 2, . . . , n ‎‎\}$‎ covers ‎$‎K‎$‎, so ‎$‎K‎$‎ is $ ‎\tau ‎$‎-compact and therefore $ ‎\tau ‎$‎-‎closed.‎‎

(1) ‎We ‎can ‎say ‎"‎ ‎Since ‎ ‎$ x_{0} ‎\in‎K $ ‎i‎‎‎t suffices to show that ‎‎$‎K‎$‎ is ‎$ ‎\tau‎ $‎-‎closed." ‎is ‎it ‎due ‎to‎ ‎$ K‎_{‎\tau‎}‎ = K‎_{‎\sigma‎}‎$‎?

‎(2)is ‎the ‎exsistence ‎of ‎$ ‎F‎_‎0 ∩ K = ‎\emptyset‎‎ ‎‎‎ $‎ ‎proved ‎by ‎abov ‎lemma?‎

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  1. Yes, except that you copied it incorrectly: $K_\tau=K_\sigma$ because $x_0$ is not in $K$. As usual with a new topology defined in this way, the new topology differs from the original one only at the special point $x_0$.

  2. No, it's much more trivial: just take $V_0=X\setminus\operatorname{cl}K$. Since $F_0\cap\operatorname{cl}K=\varnothing$, clearly $V_0\supseteq F_0$.

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  • $\begingroup$ In the original article suppose $ X_{0} \not\in K$. " Minimal $KC$ - spaces are countably compact" by " T. Vidalis" $\endgroup$ – fatemeh Aug 13 '13 at 20:44
  • $\begingroup$ @fatemeh: Exactly: $x_0$ is not in $K$. $\endgroup$ – Brian M. Scott Aug 13 '13 at 20:51
  • $\begingroup$ I review it again and the proof is divided into 2 part. $ x_{0} \in K$ and $ x_{o} \not\in K$, and this lemma is in the part $ x_{0} \in K$, but I get confused. $\endgroup$ – fatemeh Aug 13 '13 at 21:07

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