Quick analysis question along with my attempt. Looking for feedback:
Let $X$ be a metric space and $E \subset X$.
Prove that the closure of $E$, $\overline{E}$, is closed.
Proof: By definition, $\overline{E} = E \cup E'$ where $E'$ is the set of all limit points of $E$. Take all subsequent neighborhoods as open.
Let there be an arbitrary $p \in X$ but $p \notin \overline{E}$. Then $p$ is neither a point of $E$ nor a limit point of $E$. Since $p$ is not a limit point of $E$, there exists a neighborhood $N$ centered at $p$ that contains no points of $E$. This implies that $N$ also contains no points of $E'$ since if there were such a point in $E'$, $N$ would be a neighborhood of such a point and thus contain a point in $E$ by definition of limit points, leading to a contradiction.
Thus, we have a neighborhood $N$ for point $p \in \overline{E}^c$ that contains points neither in $E$ nor in $E'$. In other words, $N \subset (E \cup E')^c$, or $N \subset \overline{E}^c$. Since $p$ was arbitrary, we have that every point $p \in \overline{E}^c$ is an interior point. Therefore $\overline{E}^c$ is open by definition. Hence $\overline{E}$ is closed. $\blacksquare$
Best Answer
Rephrasing:
$X$ is a metric space, $E \subset X$.
1) $\overline{E}= E \cup E'$.
Show that $\overline {E}$ is closed.
2) Let $p \not \in \overline {E}$, i.e $p \in (\overline{E})^c$.
$p \not \in \overline{E}$ implies:
$p\not \in E$, and $p \not \in E'$,
a) Since $p \not \in E'$ there is a $B_r (p)$ , a
$r$ - neighbourhood of $p$, with
$B_r(p) \cap E =\emptyset.$
b) Also: $B_r(p) \cap E' =\emptyset $.
Assume the contrary:
There is a $e' \in E'$ with $e' \in B_r(p).$
Consider the ball $B_\epsilon(e')$, where
$\epsilon:= r - ||p,e'|| >0.$
Then $B_\epsilon(e') \subset B_r(p)$. (Proof uses the triangle inequality).
Since $e' \in E'$,
$B_\epsilon(e') \cap E \not = \emptyset$,
implies $B_r(p) \cap E \not = \emptyset$,
a contradiction.
Hence $p \in (E \cup E')^c$, and
$B_r(p) \cap (E \cup E') =\emptyset$, i.e.
$B_r(p) \subset (E \cup E')^c$.
$\rightarrow $
$(E \cup E')^c$ open, or $\overline{E}$ closed.