Let X be a a non empty metric space and suppose A $\subseteq$ X. Then A is bounded if $\exists$ $z$ $\exists$ $r>0$ so that $A \subseteq B(z,r)$.
WTS The following are equivalent:
1) A is bounded
2) for each x in X there exists a radius r>0 so that A is contained within $B(x,r)$.
3)$diam(A)$ is finite
My proof:
(1) implies (2):
Suppose (1) holds. This means that there exists a $z \in X$ and $r>0$ so that whenever $b \in A$ we have $d(b,z)<r$. So suppose x in X is arbitrary. We observe that whenever $a \in A$, $d(a,x) \leq d(a,z) +d(z,x)$ $< r + d(z,x)$. And so for $s=r +d(z,x)$ we have that whenever $a \in A$, $a \in B(x, r+ d(z,x))$.
(2) implies (3): Suppose 2 holds. Let $a,b \in A$ be arbitrary. That means A is contained within $B(a,r_1)$ and also contained within $B(b,r_2)$ for some $r_1,r_2 >0$, and so by the triangle inequality, $d(a,b) < r_1+r_1=2r_1$ and the set {d(a,b) | a,b $\in A$ } is bounded above and so by the axiom of completeness, must have a supremum. Therefore the diameter of A is finite.
(3) implies (1): Suppose that $diam(A)$ is finite. Assume for the sake of contradiction that for each $ x\in X$ and for every $r>0$. A is not contained in $B(x,r)$. That means there exists an element $ a\in A$ so that $d(a,x)>r$. Since that holds for each x in X and A is a subset of X, it holds for each element of A, and so the diameter is not finite, hence a contradiction.
Please tell me if this proof is correct. Also, I would like as much feedback on this proof as possible. I've worked on this and put effort, so I'd very much love feedback, please.
1 to 2 seems OK enough, though I'd formulate it more succinctly. Suppose $A \subseteq B(z,R)$ for some $z \in X, R>0$. Then if $x \in X$ is arbitrary,
$$A \subseteq B(x, d(x,z)+R)$$ by the triangle inequality and so the radius for $x$ can be taken as $d(x,z)+R$.
2 to 3. is similar: Fix $a_0 \in A$ and use $2$ to find $R>0$ such that $ A \subseteq B(a_0,R)$ and if $a,b\in A$ then $d(a,b) \le d(a_0,a) + d(a_0,b) < 2R$ and so $2R$ is an upper bound for $\{d(x,y): x \in A, y \in A\}$ and so $\operatorname{diam}(A) \le 2R$ is finite.
You cannot take $r$'s depending on the variable $a,b$ as you do, you need a fixed point and its fixed radius to get an unvarying upperbound. So the direct definition of bounded already suffices.
3 to 1 is easier than you think. Let $R = \operatorname{diam}(A)$, which is a well-defined real number, and pick any $a_0 \in A$. Then for $a \in A$, $d(a,a_0) \le R < R+1$ so $a_0$ and the fact that $$A \subseteq B(a_0,R+1)$$ shows that $A$ is bounded.
Don't go for contradictions where a straight application of the defintion is enough.