If $\|x-y\|\ge c$ for all $x,y\in A$, then $A$ is unbounded Let $A\subset\mathbb{R}^n$ an infinite set and $c\in\mathbb{R}^+$. Prove that if $\|x-y\|\ge c$ for all $x,y\in A$, then $A$ is unbounded.
I tried using inequalities, but I'm stuck. Any hint? I must proceed with the definition of bounded set: $A\subset\mathbb{R}^n$ is bounded if $\exists M>0:\|x\|\le M$ $\forall x\in A$.
 A: Hint. Suppose otherwise. Choose an infinite sequence of distinct elements in the set $A$. Since the set is bounded, the sequence here is bounded, so you can apply Bolzano-Weierstrass theorem. Can you get a contradiction from here?
A: Pick $k\in \mathbb Z^+$ such that $\sqrt{n\frac{1}{k^2}} < c/4$. Notice that the balls of radius $c/4$ around each point of $A$ are all disjoint and contain a point such that all it's coordinates are of the form $a/k$. It follows these balls contain an infinite amount of points with all coordinates of the form $\frac{a}{k}$.
If $A$ is bounded then it is contained by a ball of radius $R$, it follows that the ball with center of radius $R+c/4$ contains an infinite number of points with all coordinates of the form $\frac{a}{k}$.
However a ball of radius $r$ can contain at most $(2\lfloor{r/k} \rfloor+1)^n$ points with coordinates of that form.
A: Suppose that $A$ is bounded and place it inside a hypercube $B$ (having finite edge length). Using finitely many hyperplanes, slice $B$ into small hypercubes with edge length less than ${c\over\sqrt n}$.
Since its diameter is less than $c,$ each small hypercube can contain at most one point of $A.$ Therefore, $A$ is finite, a contradiction.
