# Closure of $E$ is the intersection of all $B$ containing $E$

Prove that the closure of $E$, $\overline{E}=\bigcap \{B:B\supseteq E\text{ and }B \text{ is closed in }\mathbb{R}^n\}$.

My attempt:

Consider $U:=\bigcap \{B:B\supseteq E\text{ and }B \text{ is closed in }\mathbb{R}^n\}$. The intersection of all $B\supseteq E$ will be the smallest closed set containing $E$, so $U$ will contain the interior of $E$, which is open in $\mathbb{R}^n$. Thus $U$ must also contain the boundary of $E$ so that the containment of $U$ is closed. Hence, $U=\overline{E}$.

I think that this is probably not quite rigorous, so I'd appreciate your evaluation of the proof and hints.

• True, it's a definition to me too, so I find this question in a preparatory problem set somewhat strange. – sequence Feb 20 '16 at 1:04
• I guess you should make it clear what is the definition for closure and smallest closed set containing $E$. One can see that $U$ is closed (by the definition of topology), $U$ contains $E$, and $U$ contains every closed set which contains $E$. I guess this implies that $U$ is the "smallest closed set" containing $E$. So if the definition of closure is the smallest closed set, it is a little bit weird to declare the intersection to be the smallest closed set in the first line, since it is just like saying the intersection is the closure. – k99731 Feb 20 '16 at 1:05
• Most people take this as a definition. I suppose you are starting with a different definition and want to show this condition is equivalent. But we need to know which definition of closure you are taking. – user310648 Feb 20 '16 at 1:25
• @Senpai: this is the definition I use, and this is the definition in the book An Introduction to Analysis by Wade. Nevertheless, it's asked to prove this (not in the book). – sequence Feb 20 '16 at 2:25
• @sequence you cannot prove a definition. So what exactly is asked? That it is well-defined? Or some other fact? – Henno Brandsma Feb 20 '16 at 8:30

$U$ is said to be the closure of $E$ if $U$ is the smallest closed set containing $E$. That is, if $B$ is closed and $E \subset B$, then $B\subset U$.
If so, what I would do, is show that the intersection is closed, and if $B$ is closed and $E \subset B$, then $B\subset U$.
Therefore if my definitions are not wrong, then I think it is quite weird to say the intersection is the smallest closed set containing $E$ in the first sentence.
• Since any intersection of closed sets is closed, and since the intersection of all sets containing $E$ must be the smallest set containing $E$, which is closed in this case, the intersection must be the smallest closed set containing $E$. – sequence Feb 20 '16 at 2:28