# If E is a subset of a metric space X then E closure is closed.

I am trying to find out if this proof is correct. The proof is in my text (Rudin) but without fine details as they are thought of as obvious. I am trying to make sure that I am filling in these "obvious" details correctly.

Let $E$ be a subset of a metric space $X$ and $p \in (\bar{E})^c$.
Then we have that $p \in X$ with $p \not \in \bar{E}$. This implies that $p \not \in E$ and $p \not \in E'$ (where $E'$ is the set of limit points of $E$.)
Since $p$ is not a limit point of $E$ there exists some neighborhood, $N_r(p)$ that contains no points of $E$ (this is where Rudin says $\therefore \; (\bar{E})^c$ is open and $\bar{E}$ is closed) and therefore $N_r(p) \subset E^c$. But since $E^c \subset (\bar{E})^c$ we have that $N_r(p) \subset (\bar{E})^c$ and therefore $p$ is an interior point of $(\bar{E})^c$. Since $p$ was arbitrary we have all points of $(\bar{E})^c$ are interior points and therefore $(\bar{E})^c$ is open which implies that $\bar{E}$ is closed.

Please let me know if all of this is completely correct. Thank you!!!!

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It’s not generally true that $E^c \subseteq (\bar{E})^c$: $E\subseteq\bar E$, so $E^c\supseteq(\bar E)^c$. Taking complements reverses the inclusion, much as multiplying a real number inequality reverses the inequality. It’s true that $N_r(p)\subseteq(\bar E)^c$, but not for the reason that you give. The point is that since $N_r(p)\cap E=\varnothing$, no point of $N_r(p)$ is a limit point of $E$: $N_r(p)\cap E'=\varnothing$.
If you want to beat this completely to death, let $q\in N_r(p)$, and let $s=r-d(p,q)>0$, then $N_s(q)\subseteq N_r(p)\subseteq E^c$, so $q\notin E'$.
Clearly also $N_r(p)\cap E=\varnothing$, so $N_r(p)\cap\bar E=\varnothing$, and therefore $N_r(p)\subseteq(\bar E)^c$.