# Prove that if $A \subset C$ is bounded and closed, then $A$ is compact. (Proof guidance)

So I'm working on a problem with the following statement:

Problem Statement: Assume metric space $C$ is complete and has this property:

If $(x_n)_{n \ge 0}$ is a sequence in $C$ and there exists $\varepsilon > 0$ such that $d(x_i,x_j) > \varepsilon$ for all $i \neq j$, then the sequence $(x_n)$ is not bounded.

Prove that if $A \subset C$ bounded and closed, then $A$ is compact.

Here is what I have done so far/my thoughts...

Proof: From the statement, we know that $C$ is complete thus this means that every Cauchy sequence converges to a point in metric space $C$. Now, compactness means that a subset $A$ in metric space $C$ is compact when the set $A$ is contained in the union of a collection of open subsets of $C$, then $A$ is contained in a finite number of these subsets.

From the property given, we know that the metric space $C$ contains a divergent (non-convergent) sequence that is not bounded. Hence, a subset $A$ of $C$ must be select points which are bounded thus we let $\varepsilon > 0$ such that $d(x,y) < \varepsilon$ $\forall$ $x,y \in A\subset C$. Since $A$ is bounded and closed, we select $x < y$ such that A is monotonically increasing, thus $A$ is a convergent sequence because it is a bounded monotonically increasing sequence. Thus, since $A$ contains a convergent subsequence it is compact. Thus, metric space $C$ is compact.

To me, I feel like I'm in the right direction but am struggling to have that closing logic for this proof. I appreciate the help!

• $x<y$ doesn't mean anything in a general metric space. – aschepler Nov 11 '17 at 22:57
• @aschepler can you elaborate on that? Is that from the definition of a metric space? – Inti Nov 11 '17 at 23:02
• @Inti There doesn't have to be an order defined on the metric space. What if for example $C = \mathbb{R}^2$ with the Euclidean metric? – Demophilus Nov 11 '17 at 23:38
• You assumed x_n in C and then out of the blue, you are talking about unrelated x_i and x_j. – William Elliot Nov 12 '17 at 4:06

You lost me when you say $x<y$: as pointed out in the comment, it is not clear what the ordering here is referring to. In a general metric space, there is no such ordering.
Since $A$ is closed and $C$ is complete, $A$ is also complete. From the answer to this question, it suffices to show that $A$ is totally bounded. That is, for all $\epsilon >0$, there are finitely many open balls of radius $\epsilon$ which covers $A$.
This can be shown easily: Fix an $\epsilon$. If $A$ is not covered by any finite number of $\epsilon$-open balls, then there is a sequence $x_1, x_2, x_3\cdots$ in $A$ so that $d(x_i, x_j) >\epsilon$ (Try to construct this sequence inductively, start with any point $x_1\in A$). But then from the condition on $C$, this sequence is unbounded and this contradicts the fact that $A$ is bounded. Hence $A$ is totally bounded and thus is compact.