Let $M$ be a metric space. I'm asked to prove that the diameter of a set $A\subset M$ is the same diameter as its closure $\bar{A}\subset M,$ $$\text{diam}(A) = \text{diam}(\bar{A}).$$

My attempt:

Let $p\in\bar{A}-A, \ q\in\bar{A}-A.$

There exists $p',q' \in A \ s.t. \ \ D(p,p')<\epsilon, \ \ D(q,q')<\epsilon, \quad \epsilon > 0.$

Since $\bar{A}$ is a metric space in itself the triangle inequality gives: $$D(p,q) \leq D(p,p')+D(p',q') + D(q',q)<2\epsilon + D(p',q') < 2\epsilon + \text{diam}(A)$$ Since $p,q$ were chosen arbitrarily, it's true that $$\text{diam}(\bar{A}) < 2\epsilon + \text{diam}(A).$$ Since $\epsilon$ can be made arbritrarily small, $\text{diam}(\bar{A}) = \text{diam}(A).$

Is this correct?

  • 1
    $\begingroup$ The details are good, but you should probably point out that $\text{diam(}\overline{A}) \ge \text{diam})(A)$ for obvious reasons and explain why showing $D(p, q) \le \text{diam}(A)$ gives an upper bound on $\text{diam}(\overline{A})$. $\endgroup$
    – Rob Arthan
    Feb 17, 2019 at 20:21
  • $\begingroup$ I believe the most intuitive argument is to show that $ p \in \bar A \iff D(p, A) = 0 $, and then using the triangle inequality $\endgroup$
    – user359302
    Feb 17, 2019 at 21:15

2 Answers 2


Globally, yes. Only two remarks:

  1. When you wrote “There exists $p',q' \in A\ $ s.t. $\ D(p,p')<\varepsilon, \ D(q,q')<\varepsilon$, $ \ \varepsilon > 0$.”, it would have been better if you had written “Take $\varepsilon>0$. There exists $p',q' \in A \ $ s.t. $\ D(p,p')<\varepsilon$, $D(q,q')<\varepsilon$”.

  2. There is no need to write that $\bar A$ is a metric space in itself. You are working in the metric space $M$.


First note that it's clear that $\operatorname{diam}(A) \le \operatorname{diam}(\overline{A})$ because $A \subseteq \overline{A}$ and so we're taking the supremum of a larger subset.

To show the reverse your trick works, but start by picking $\varepsilon >0$. Then in your notation, for $p,q \in \overline{A}$ we find $p,q \in A$ such that $d(p,p') < \varepsilon$ and $d(q,q') < \varepsilon$.

Your small computation then indeed shows that $d(p,q) < \operatorname{diam}(A) + 2\varepsilon$ and so (as $p,q$ were arbitary in $\overline{A}$) that $\operatorname{diam}(\overline{A}) \le \operatorname{diam}(A)+ 2\varepsilon$ which together with the first trivial inequality and the fact that $\varepsilon>0$ was arbitrary shows the equality.


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