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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.

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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 $(a,b)$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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