Prove that the following three metric space/subsequence boundedness conditions are equivalent.

Say I have a metric space $(M,d)$, and $A$ is a subset of $M$. I'm trying to prove that the following three conditions are equivalent.

$(\exists x \in M)$ $(\{d(x,y): y \in A\}$ is bounded)

$(\forall x \in M)$ $(\{d(x,y): y \in A\}$ is bounded)

$(\{d(x,y): x$ and $y \in A\}$ is bounded)

For the second condition leading to the first, if for all $x \in M$, $\{d(x,y): y \in A\}$ is bounded, then there certainly exists at least one $x \in M$ such that $(\{d(x,y): y \in A\}$ is bounded.

For the last condition leading to the first condition, I think if a distance function exists such that points in A are bounded, then there exists a point $x$ in M (that is also in A, presumably) such that distance function $d(x,y)$ is bounded for $y \in A$

Thoughts on how to prove equivalence formally?

Edit: The second condition being equivalent to the third is apparently more complicated than what I suggested in the comment below. (I had the third condition described to me as: For any arbitrary point (x,y) in A, the set of distances from (x,y) to other points in A is bounded.)

Solution 1:

It sounds like you know how to prove $(iii)\implies (i)$. For $(i)\implies (ii)$, let $x_0$ be the point under $(i)$. Then for all $x\in M$ and $y\in A$, we have $$ d(x,y)\leq d(x,x_0)+d(x_0,y)\leq d(x,x_0)+C, $$ where $C$ is some constant, independent of $y$, by hypothesis.

See if you can handle $(ii)\implies (iii)$.