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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 \} $ . WE define a new topology $‎\tau‎^{‎\prime‎} = ‎\{ U‎‎\in‎‎\tau :‎‎ ‎x‎_{0} \not‎\in U \} \cup ‎\{U‎ ‎‎\in \tau:‎‎‎‎‎x‎_{0}\in U‎‎‎,U‎\in ‎\mathcal{F}‎‎‎‎‎‎\}$ ank $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 countably compact.

proof: Let $ F_{0} \in ‎\mathcal{F}‎‎‎‎‎‎$ be such that $ F_{0} \in ‎\mathcal{F}‎‎‎‎‎‎$ with $ F_{0}‎‎‎\subset‎‎ (\overline{K_{ ‎\tau‎}} - K ) $, with $ F_{0} = \{ x_{n_{k}} :k \in \omega \}$ and sppose for a contradiction that $K$ is not countably compact. Then there is a set $ \{ y_{n} : n \in \omega \} \subset‎‎ K $ without $‎\tau‎ $-accumulation points in$K$ and since $ x_{0} \in K $. there is a $\tau‎ $- open neighbourhood $U ( x_{0}) $ of $ x_{0}$ with $ U ( x_{0}) \cap \{ y_{n} : n \in \omega \} = ‎\emptyset‎ $. We claim that for every infinite subset $ \{ y_{n_{k}} : k \in \omega \}$ of $ \{ y_{n} : n \in \omega \}$ and for every $ z \in F_{0}$ there is a $\tau $- open neighbourhood of $ z$, $ U(z)$, such that $\mid U(z)^c \cap \{ y_{n_{k}} : k \in \omega \} \mid = \omega $.

Actually, for otherwise $\{ y_{n_{k}} : k \in \omega \} ‎\longrightarrow‎ z$ and sinse $\tau‎ $ is a$KC$ space , $z$ will be the unique $\tau‎ $-accumulation point of $\{ y_{n_{k}} : k \in \omega \}$.

But , there is an $ F \in‎\mathcal{F} $ with $z \not\in F$, thus there is an open set $ W(F)$ containing $F$ with $ z \not\in W( F) $. so $ z \not\in U(x_{0}) \cup W(F)$, and consequently $ X_{0} $ is not a $\tau‎^{‎\prime‎} $ -accumulation point of $\{ y_{n_{k}} : k \in \omega \}$.

It follow that $\{ y_{n_{k}} : k \in \omega \}$ is an infinite subset of $K$ with no $\tau‎^{‎\prime‎} $ -accumulation point in$K$ which is impossible , since $K$ is$\tau‎^{‎\prime‎} $- compact.

My questions are:

(a): Why will $z$ be the unique $\tau‎ $-accumulation point of $\{ y_{n_{k}} : k \in \omega \}$ ?

(b): why is there an $ F \in‎\mathcal{F} $ with $z \not\in F$? and thus there is an open set $ W(F)$ containing $F$ with $ z \not\in W( F) $? and $ z \not\in U(x_{0}) \cup W(F)$?

(c):We know that $\tau‎ $ and $\tau‎^{‎\prime‎} $ are $T_{1}$ space. why can we say that $ x_{0}$ is the unique point which can be $\tau‎^{‎\prime‎} $ -accumulation point for a set $ K \subset X$ while it is not $\tau‎ $-accumulation point of it?

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(a) This is exactly the same thing that you've seen several times already. If $\{y_{n_k}:k\in\omega\}\setminus U$ is finite for every open nbhd $U$ of $z$, then every enumeration of $\{y_{n_k}:k\in\omega\}$ converges to $z$. In particular, $\langle y_{n_k}:k\in\omega\rangle\to z$. Since $\langle X,\tau\rangle$ is $KC$ and $\{y_{n_k}:k\in\omega\}\cup\{z\}$ is compact, $\{y_{n_k}:k\in\omega\}\cup\{z\}$ is $\tau$-closed. Thus, no point of $X\setminus\big(\{y_{n_k}:k\in\omega\}\cup\{z\}\big)$ is a $\tau$-accumulation point of $\{y_{n_k}:k\in\omega\}$. And no point of $K$ is a $\tau$-accumulation point of $\{y_n:n\in\omega\}$, so no point of $\{y_{n_k}:k\in\omega\}$ is a $\tau$-accumulation point of $\{y_{n_k}:k\in\omega\}$, and $z$ is therefore the only $\tau$-accumulation point of $\{y_{n_k}:k\in\omega\}$.

(b) $\mathscr{F}$ is a free ultrafilter on $\{x_n:0<n<\omega\}$, so for any $F\in\mathscr{F}$ and any finite subset $S$ of $F$, $F\setminus S\in\mathscr{F}$. In particular, $z\in F_0\in\mathscr{F}$, so $F_0\setminus\{z\}\in\mathscr{F}$. Let $F=F_0\setminus\{z\}$. $\langle X,\tau\rangle$ is $T_1$, so each $x_n\in F$ has an open nbhd that does not contain $z$; the union of these nbhds is an open set $W(F)$ such that $\subseteq W(F)$ and $z\notin W(F)$. Now recall that $\langle y_{n_k}:k\in\omega\rangle\to z$. This implies that every open nbhd of $z$ contains infinitely many of the $y_{n_k}$, but $U(x_0)$ contains none of them, so $z\notin U(x_0)$, and therefore $z\notin U(x_0)\cup W(F)$. And $U(x_0)\cup W(F)\in\tau'$, so it's a $\tau'$-nbhd of $x_0$ that does not contain $z$ and therefore cannot be a $\tau'$-accumulation point of $\{y_{n_k}:k\in\omega\}$.

(c) The topologies $\tau$ and $\tau'$ are identical on $X\setminus\{x_0\}$, so a point of $X\setminus\{x_0\}$ is a $\tau'$-accumulation point of $\{y_{n_k}:k\in\omega\}$ iff it is a $\tau$-accumulation point of $\{y_{n_k}:k\in\omega\}$. Thus, the point $x_0$ is the only point of $X$ that can possibly be a $\tau'$-accumulation point of $\{y_{n_k}:k\in\omega\}$ without also being a $\tau$-accumulation point of the set.

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  • $\begingroup$ can we say : for every $ K \subset X $, we have $ cl_{‎\tau‎‎‎ }( (K) \subset cl_{\tau^{\prime}} (K) $ and $ cl_{‎\tau‎^{\prime}} ( K) \subset cl_{\tau}(K) \cup \{ x_{0} \}$ so $ \{ x_{0} \} $ is the only $ \tau‎^{\prime}‎‎$ -accumulation point such that is not $\tau‎ $-accumulation point $\endgroup$ – fatemeh Aug 8 '13 at 19:10
  • $\begingroup$ In part (b), what is $S$? what does it mean $ F - F \in \mathcal{F} $ ?In the third line of part b : what does it mean $ ....\subset W(F)$ ? $\endgroup$ – fatemeh Aug 9 '13 at 12:19
  • $\begingroup$ @fatemeh: For the first question: yes. In (b) $S$ is any finite subset of $F$. That was supposed to be $F\setminus S$, not $F\setminus $F$; I've fixed it now. $\endgroup$ – Brian M. Scott Aug 9 '13 at 12:26
  • $\begingroup$ In (a), can we say $z$ is the only $\tau‎^{\prime}‎‎$ ‎‎ -accumulation point due to $ KC ‎\Longrightarrow‎ US$ ? are free and uniform ultrafilter, the same on the infinit set? $\endgroup$ – fatemeh Aug 9 '13 at 12:42
  • $\begingroup$ @fatemeh: No, because you don't know that $\tau'$ is $KC$. A filter on a countably infinite set is free iff it is uniform. $\endgroup$ – Brian M. Scott Aug 9 '13 at 12:48

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