Properties if the one-point compactification of an uncountable discrete space Let $D ( \tau )$ be an uncountable discrete space, and $\alpha D ( \tau )=D ( \tau )\cup\{\alpha\}$ the one-point compactification of $D ( \tau )$.
I want to show that if $U$ is any countably infinite subset of $D ( \tau )$, then $\overline{U} = U \cup \{ \alpha \}$.
Also, is it true that $\{ \alpha \}$ is a $G_\delta$-subset of $\alpha D ( \tau )$? 
 A: To answer these questions you must really determine what the open neighbourhoods of $\alpha$ are. Since $X = D ( \tau )$ is discrete, the compact subsets are exactly the finite sets. From this it follows that the open neighbourhoods of $\alpha$ are of the form $\{ \alpha \} \cup B$ where $B \subseteq X$ is cofinite (i.e., $X \setminus B$ is finite).



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*To show that $\overline{U} = U \cup \{ \alpha \}$ for any countably infinite $U \subseteq \alpha X$, first note that as the points of $X$ are isolated in $\alpha X$, then $\operatorname{cl}_{\alpha X} ( U ) \subseteq U \cup \{ \alpha \}$. Now, given any open neighbourhood $V$ of $\alpha$ in $\alpha X$ since $X \setminus V$ is finite and $U$ is infinite it must be that $V \cap U \neq \varnothing$. Thus $\alpha \in \operatorname{cl}_{\alpha X} ( U )$.

*To see that $\{ \alpha \}$ is not a Gδ-subset of $\alpha X$, let $\{ V_n  = \{ \alpha \} \cup B_n : n \}$ be any countable family of neighbourhoods of $\alpha$ in $\alpha X$. (So each $B_n$ is a co-finite subset of $X$.) Note that $X \setminus \bigcap_n B_n = \bigcup_n ( X \setminus B_n )$ is countable, and so $\bigcap_n B_n = X \setminus ( X \setminus \bigcap_n B_n )$ is uncountable. Therefore $\bigcap_n V_n$ contains uncountably many points from $X$.
