# Is this proof that $\text{diam}(A) = \text{diam}(\bar{A})$ correct?

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?

• 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 $\text{diam}(\overline{A})$. Feb 17, 2019 at 20:21
• 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
– user359302
Feb 17, 2019 at 21:15

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.