I would like to understand how a certain statement below follows from the previous statements.
Theorem If $X$ is a metric space and $E \subset X$, then $\overline{E}$ is closed.
Proof. If $p \in X$ and $p \notin \overline{E}$ then $p$ is neither a point of $E$ nor a limit point of $E$. Hence $p$ has a neighborhood which does not intersect $E$. The complement of $\overline{E}$ is therefore open. Hence $\overline{E}$ is closed.
Since $p$ has a neighborhood which does not intersect $E$. I was wondering how it follows from "$p$ has a neighborhood which does not intersect $E$ " that the complement of $\overline{E}$ is open if and only if $(\overline{E})^{c}$ is open? At first I thought that since $p$ has a neighborhood, $N_{r}(p)$, which intersects $E^{c}$, then $N_{r}(p) \subset E^{c} \subset \overline{E^{c}}$ since $\overline{E^{c}} = E^{c} \cup (E^{c}$)'. But, I don't think that $\overline{E^{c}}$ necessarily equals $(\overline{E})^{c}$.
Best Answer
The argument goes like this. If $p$ is not in $\operatorname{cl}E$, then by definition $p\notin E$ and $p$ is not a limit point of $E$. This implies that $p$ has a nbhd $N_p$ that does not intersect $E$. We can do this for each $p\in X\setminus\operatorname{cl}E$, so
$$X\setminus\operatorname{cl}E\subseteq\bigcup_{p\in X\setminus\operatorname{cl}E}N_p\subseteq X\setminus\operatorname{cl}E\;.$$
But then
$$X\setminus\operatorname{cl}E=\bigcup_{p\in X\setminus\operatorname{cl}E}N_p\;,$$
which, being a union of open sets, is open. By definition the complement of $X\setminus\operatorname{cl}E$ is then closed, and
$$X\setminus(X\setminus\operatorname{cl}E)=\operatorname{cl}E\;,$$
so $\operatorname{cl}E$ is closed.