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I am trying to prove the above claim. My question is on the bolded part and if I am correct in making that statement. A metric space is bounded if there exists an $M \in \mathbf{N}$ such that $\rho(x, y) \leq M$ for all $x, y \in X$. The negation would be, a metric space is not bounded if for every $n \in \mathbf{N}$, there exist $x, y \in X$ such that $\rho(x, y) > n$.

I'm not sure if what I have written in my proof is correct.

Edit: The $X$ should be an $E$.

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  • $\begingroup$ You haven't defined $X$ anywhere in your proof, so no, I would not call it correct. $\endgroup$ – 5xum Feb 11 '14 at 7:03
  • $\begingroup$ The $X$ is a typo. It should be $E$. I will edit the OP. $\endgroup$ – Joseph DiNatale Feb 11 '14 at 7:09
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Here is a more direct proof. Pick $x \in E$, and let $U_n = B(x, n)$. Then $U_n$ is an open cover, and so has a finite subcover. In particular, $E \subset B(x,n)$ for some $n$. Hence $\rho(x,y) < 2n$ for all $x,y \in E$.

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    $\begingroup$ Nice proof!! +1 $\endgroup$ – ZFR Nov 1 '19 at 18:06
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Let $e\in E$ be a point and let $f:x\in E\mapsto d(x,e)\in\mathbb R$, with $d$ the distance function. This is continuos, so Weierstrass' theorem tells that $f$ is bounded. This means, exactly, that $E$ is bounded.

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  • $\begingroup$ The argument in the image is really needlessly elaborate! $\endgroup$ – Mariano Suárez-Álvarez Feb 11 '14 at 7:08
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An equivalent definition of a bounded subset $E$ of a metric space $X$ is that there exist $x \in X$ and $\epsilon > 0$ such that $E \subset B(x,\epsilon)$, that is, $E$ is contained in some ball. The negation would be that for any $x \in X$ and any $\epsilon>0$ there is a $y \in E$ such that $\rho(x,y)>\epsilon$, that is $y \notin B(x,\epsilon)$ (that means $E$ is not contained in any ball). Fix $x \in E \subseteq X$, choose a $y$ (call it $x_n$) for every $\epsilon=n$ and you get what you want.

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