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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