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Suppose M is a metric space. Assume $A\subseteq M$. $A$ is bounded if $\exists R>0$ such that $\forall x,y \in A: d(x,y) \le R$.

I would like to show that the following are equivalent:

1) $A$ is bounded.

2) $A$ is contained in some closed ball.

Proof: Let $x,y\in A$. Since $\forall x,y \in A$, $d(x,y) \leq R$, it follows that if $x \in A$ then $x \in N_R(y)$. Hence A is contained in the closed neighbourhood (ball). (Note: $N_R(y)$ means closed ball (with equality))

Assume $A$ is contained in some closed ball. $A \subseteq N_r(x)$ for some $x \in M$. Hence $\forall x,y \in A: d(x,y) \le r$. Is my proof correct? What can I do to improve it?

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Slightly better:

Suppose $A$ is bounded. If $A= \emptyset$, any closed ball of $M$ will do. If not, pick $p \in A$. Then let $R$ be given as in the definition of boundedness. As for all $x \in A$, in particular $d(x,p) \le R$ we have that $A \subseteq N_R(p)$ and $A$ is contained in a closed ball.

On the other hand if $A \subseteq N_R(p)$ for some $p \in M$ and $R>0$, then for all $x,y \in A$ we have $d(x,y) \le d(x,p) + d(p,y) \le R + R=2R$. So then $2R$ works in the definition of boundedness and $A$ is thus bounded.

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The first part is correct, but in the second part you didn't finish the proof. There is a specific $x$ such that $A\subset B_r(x)$ when $B$ is a closed ball. So yes, it follows that for this specific $x$ we have $d(x,y)\leq r$ for any $y\in A$. But this is not enough, you need to bound the distance between any two points $y,z\in A$. But it easily follows from the triangle inequality. Let $y,z\in A$. Then:

$d(y,z)\leq d(y,x)+d(x,z)\leq r+r=2r$

So for any $y,z\in A$ we have $d(x,y)\leq 2r$. Hence the set is bounded.

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