# uniform ultrafilter

‎‎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 $V_0\supseteq F_0$.
In the original article suppose $X_{0} \not\in K$. " Minimal $KC$ - spaces are countably compact" by " T. Vidalis" –  fatemeh Aug 13 '13 at 20:44
@fatemeh: Exactly: $x_0$ is not in $K$. –  Brian M. Scott Aug 13 '13 at 20:51
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. –  fatemeh Aug 13 '13 at 21:07