Having a bit of trouble with this question. Would appreciate a basic outline of what needs to be done to prove it, as although I have an idea of the required definitions I've no clue how to apply them.
Let $(X, d)$ be a metric space. Prove that $A ⊂ X$ is bounded if and only if
$A ⊂ Br(a)$ for some $a ∈ X$ and $r > 0$.
Any help would be very much appreciated.
**EDIT
From my notes ;
$A$ is a subset of a metric space and diameter of a A is defined as $diam(A)=sup$ {$d$ $(x,y)| x,y∈A$}. $A$ is bounded if $diam(A)<∞$.
AND
A subset $A$ of a metric space is only bounded if there exists an open ball $Br(a)$ such that $A⊂Br(a)$.
Best Answer
Suppose $A \subseteq B_r(a)$, for some $r \in \mathbb{R}^{\geq 0}, a \in X$. Then $\text{diam}(A) = \sup \{ d(x, y) \mid x, y \in A \}$. But $$d(x, y) \leq d(x, a) + d(a, y) \leq r+r = 2r$$ so $A$ is bounded.
Conversely, suppose $A$ is bounded. Then $D := \sup \{ d(x, y) \mid x, y \in A \}$ is finite. If $A = \emptyset$, we're done instantly as it's contained in every ball; so wlog we're free to pick some $a \in A$, and let $r = D$.
Then every point in $A$ is at most $D$ away from $a$.