Prove that the set of accumulation points of a bounded subset of $\mathbb{R}^n$ is compact.

Question

We are allowed use of the Heine-Borel theorem that states that a set is compact in $\mathbb{R}^n$ iff the set is closed and bounded. I know that the set of accumulation points is closed, but I am having trouble expressing that a set being bounded means that the set of its accumulation points are also bounded.

Our definition of bounded states that if a set $A$ is bounded, there exists an $M$ such that for all $x$ and $y$ in $A$, $d(x,y) < M$.

My thought is that this $M$ can be extended to bound the set of accumulation points, but I am unsure on how to express this.

Answer 1

Let $w,z$ be accumulation points of $A$. Then, there exists some $x,y \in A$ such that $d(w,x) < 1$ and $d(z,y) < 1$. Then $$d(w,z) \leq d(w,x) + d(x,y) + d(y,z) < 1 + d(x,y) + 1 < M + 2.$$

Thus the set of accumulation points of $A$ is bounded.