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

general-topologymetric-spacesproof-verification

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?

Best Answer

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 $M$.

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