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.

  • $\begingroup$ True, it's a definition to me too, so I find this question in a preparatory problem set somewhat strange. $\endgroup$ – sequence Feb 20 '16 at 1:04
  • $\begingroup$ 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. $\endgroup$ – k99731 Feb 20 '16 at 1:05
  • 2
    $\begingroup$ 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. $\endgroup$ – user310648 Feb 20 '16 at 1:25
  • $\begingroup$ @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). $\endgroup$ – sequence Feb 20 '16 at 2:25
  • $\begingroup$ @sequence you cannot prove a definition. So what exactly is asked? That it is well-defined? Or some other fact? $\endgroup$ – Henno Brandsma Feb 20 '16 at 8:30

Maybe I should make this an answer.

Here are the definitions I know.

$U$ 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.

  • 1
    $\begingroup$ 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 $E$. $\endgroup$ – sequence Feb 20 '16 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.