Closure of a subset Show that the closure $\overline E$ of a subset $E \subset \mathbb R^n$ is the smallest closed set containing $E$.
I am new to topology so I was wondering how to establish a solid proof of this statement that seems basic.
Here is my attempt:
First we can show that $\overline E$ is a closed set containing $E$ :

*

*$\overline E = E \cup \partial E $ so it surely contains $E$.

*$\overline E^C = E^C \cap (\partial E)^C$. I need to show that $\forall \vec{x} \in \overline E^C, \exists \delta_x > 0 \text{ such that } B(\vec{x},\delta_x) \subset \overline E^C$ but I struggle to show it properly.

*Then I need to show that $\overline E$ is the smallest set with the previous conditions. I wanted to do it by contradiction by supposing that $\exists F \subset \overline E$ (so smaller than $\overline E$) and then find a contradiction but it seems too strange to prove.


For your attempt toward contradiction, you assume that $F$ contains $E$, is closed, and is strictly contained in $\overline{E}$. If such a set could exist, then $\overline{E}$ would not be the smallest closed set containing $E$. There are several equivalent definitions of the closure, this being one of them, but I will proceed with the fact that $\overline{E}$ is the set $E$ together with all its limit points.
Assuming $E \subset F \subset \overline{E}$ with $F$ closed, then every sequence in $E$ is also a sequence in $F$. Using the fact that closed sets contain their limit points and the aforementioned definition of the closure, can you arrive at a contradiction from here?
