# accumulation points , ultrafilter

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?

(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.
• 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 – fatemeh Aug 8 '13 at 19:10
• 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 $....\subset W(F)$ ? – fatemeh Aug 9 '13 at 12:19
• @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$; I've fixed it now. – Brian M. Scott Aug 9 '13 at 12:26 • In (a), can we say$z$is the only$\tau‎^{\prime}‎‎$‎‎ -accumulation point due to$ KC ‎\Longrightarrow‎ US$? are free and uniform ultrafilter, the same on the infinit set? – fatemeh Aug 9 '13 at 12:42 • @fatemeh: No, because you don't know that$\tau'$is$KC\$. A filter on a countably infinite set is free iff it is uniform. – Brian M. Scott Aug 9 '13 at 12:48