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The definition of a bounded set $S$ in a metric space $X$, is simply that $$S\subset B_r(x) $$ for some $r>0$ and $x\in X$. But I've seen a few proofs where they use the fact that a set is bounded to say that it must be contained in some closed ball. I can't think of an example where this isnt true, but I wasn't sure if the two definitions are equivalent. Can someone give a proof? Thanks!

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    $\begingroup$ Every open ball is contained in a closed ball. Every closed ball is contained in an open ball. $\endgroup$ Jun 12 '17 at 5:58
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If $S$ is bounded then there exist $x\in X,r>0$ such that $S\subseteq B_r(x)$. But $B_r(x)\subseteq \overline{B_t}(x)$, where $t$ is any number you like with $t>r$.

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    $\begingroup$ It is not true in a general metric spaces that the closure of an open ball is the corresponding closed ball. For example, take the discrete metric on a set containing at least two points. Then the open ball of radius one about any given point contains only that point. The closure of that open ball is again the single point. However the closed ball of radius one is the whole space. $\endgroup$
    – SamM
    Jun 12 '17 at 6:31
  • $\begingroup$ I agree. And I edited my answer. $\endgroup$
    – Janitha357
    Jun 12 '17 at 6:57

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