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Prove that if a subset $A$ of a metric space is bounded then the closure of $A$ is bounded and the diameter of $A$ is equal to the diameter of the closure of $A$.

This is the question I am working on at the moment. I think I have proven that cl(A) is bounded as follows:

If A is bounded then $\exists x_0 \in X$ and $K$ constant s.t. $d(x_0,x)\leq K$ $\forall x\in X$.

Let $a \in$ cl(A), then $B_\epsilon (a) \cap A \neq \emptyset$ $\forall \epsilon > 0$.

Now let $a'\in B_\epsilon (a) \cap A$, then $d(a,x_0)\leq d(a,a')+d(a',x_0)\leq \epsilon +K$.

As this is true $\forall \epsilon > 0$ this means $d(a, x_0) \leq K$ and cl(A) is hence bounded.

So now I need to show that diam(A) = diam(cl(A)).

The diameter is defined as diam(A) = sup($d(x,y)$) $\forall x,y \in A$.

I want to do this by letting $K$ be the least upper bound of A and then I would like to just say that as both A and cl(A) are bounded by the same thing, the supremum of the distance between any two points in either A or cl(A), which is $K$, is equal to diam(A) and diam(cl(A)). Is this the correct or right way to go about it?

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  • $\begingroup$ I would encourage you to write down your proof with proper mathematical sentences, ideally with the whole $\epsilon, \delta, \exists, \forall$ verbiage. This will tell you immediately if your idea works or not. Plain english "proofs" tend to hide away the complex details and to be misleading. $\endgroup$ – Alexandre Halm Mar 29 '15 at 10:47
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Here is an outline of a proof:

  • $D = \text{diam}(A) \le D' = \text{diam}(\bar A)$

    • by def. of $D$, there are sequences $x_n,y_n$ in $A$ with $d(x_n,y_n) \to D$
    • since $A$ is metric and bounded, it satisfies BW, so you can assume $x_n \to x$ and $y_n \to y$, with (by def. of $\bar A$), $x,y \in \bar A$)
    • conclude using continuity of $d$
  • $D' \le D$

    • take $\epsilon > 0$ and try to prove $D' \le D+\epsilon$ using arguments similar to the one above
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