# Metric spaces and boundedness

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?

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.
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.