Show that, in $\mathbb{R}$ with the usual metric, $\overline{(a, b)}$ = [$a, b$] for all $a < b$.

In order to show the equality, you need to show that LHS $\subseteq$ RHS and RHS $\subseteq$ LHS, right? This is what I was thinking:

To show $\overline{(a, b)}$ $\subseteq$ [$a, b$]: we know [$a, b$] is closed. And, since the closure of a set is the smallest closed set containing the set, we must have $\overline{(a, b)}$ $\subseteq$ [$a, b$].

To show [$a, b$] $\subseteq \overline{(a, b)}$: this is where I got stuck. Since $\overline{(a, b)}$ = $(a, b) \bigcup$ {accumulation points of $(a, b)$}, we need to show that [$a, b$] consists of only those elements, and everything outside of it is not an accumulation point, right? How would I go about doing that? Should I assume $x > b$ and $x < a$ is an accumulation point and show $x$ cannot be an accumulation point through contradiction? How should I go about doing that? I considered $x$ infinitesimally bigger than $b$, but don't know how to there is a neighborhood that doesn't contain an element from [$a, b$]. Or is there a much simpler way to prove this statement? Thank you.


While it is true that every $x>b$ and every $x<a$ is not an accumulation point of $(a,b)$, it is not necessary to prove this claim in order to prove that $[a,b]\subset\overline{(a,b)}$. The fact that every $x>b$ and every $x<a$ is not an accumulation point of $(a,b)$ follows from the direction you've already proved.

Instead, to show that $[a,b]\subset\overline{(a,b)}$, it is enough to prove that $a$ and $b$ are accumulation points of $(a,b)$.

  • $\begingroup$ Thank you. How should I proceed to show $a$ and $b$ are accumulation points of $(a, b)$? Is it through contradiction? I'm a bit stuck on this part. $\endgroup$ – Max Feb 22 '18 at 5:57
  • 1
    $\begingroup$ Well, if $U$ is an open set containing $b$, then there is some $\varepsilon>0$ such that $(b-\varepsilon,b+\varepsilon)\subset U$, and so the intersection of $U$ and $(a,b)$ is non-empty. The same argument works for $a$. $\endgroup$ – carmichael561 Feb 22 '18 at 6:01
  • $\begingroup$ Ah, got it. Thank you. $\endgroup$ – Max Feb 22 '18 at 6:33

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.