Relative Compactness and equivalent conditions

Question

Let $(X,d)$ be a metric space. In the text I am using $A\subset X$ is said to be relatively compact iff $\bar{A}$ is compact.

Show that $A$ is relatively compact iff every sequence $\{a_n\}\subset A$ has a convergent subsequence.

My first question is that shouldn't the subsequence converge in $\bar{A}$? Or is that a given as in metric spaces sequential compacteness $\Leftrightarrow$ compactness?

The "only if" direction is obvious.

For the "if" direction I was only able to come up with a contrapositive argument. So please let me know if it is correct or if you have a neater or easier proof, please can you pass it on.

So suppose that $\bar{A}$ is not compact for and let $\{a_n\}\subset \bar{A}$. So then for each $n\in\mathbb{N}$ define $A_n:=A\cap B(a_n,\frac{1}{n})\neq\emptyset$ where $B(a_n,\frac{1}{n})$ is the open ball centred at $a_n$ with radius $1/n$. Then pick a $b_n$ from each $A_n$ so that $\{b_n\}$ is a sequence in $A$, where for any $\epsilon>0$, there is a $N\in\mathbb{N}$ such that for any $n>N$ $d(a_n,b_n)<1/n<\epsilon$.

Then I claim $\{b_n\}$ has no convergent subsequence as if it did, say $\{b_{n_k}\}$ with limit $b$, there will be a $N'\in\mathbb{N}$ such that for all $n_k>N'$,

$d(a_{n_k},b)\leq d(a_{n_k},b_{n_k})+d(b_{n_k,b})<\epsilon/2+\epsilon/2$,

where $\{a_{n_k}\}$ is the corresponding subsequnce of $\{a_n\}$, thus $a_{n_k}\rightarrow b $, implying sequential compactness(hence compactness) and contradicting the supposition.

So I would appreciate any feedback to clear up my confusion on where the sequence's limit should be and to give me a better proof than the one I could only think of or correct any errors I have made. If there are other equivalents of relative compactness please let me know them as well. Thanks in advance.

Answer 1

First, to expand and explain Crostul's comment. I assume your course has already developed that a metric space $C$ is compact if and only if it is sequentially compact: every sequence in $C$ has a convergent subsequence. At first glance the definition of $A$ being relatively compact appears to be the same. But there is a difference: the subsequences are required to converge in $X$, not necessarily in $A$. That is, there is a limit point, but that point need not be in $A$, while actual compactness does require it to be in $A$.

Now if $B \subseteq A$, then $\overline B \subseteq \overline A$, so if $B$ is any sequence in $A$, any of its limits points has to be in $\overline A$. This holds for all sequences, not just convergent subsequences of other sequences.

Overall, your argument is correct, but there is an error in how you've stated it. Nowhere do you say that $\{a_n\}$ should have no convergent subsequence. You just introduce it as an arbitrary sequence in $\overline A$. So the fact that $\{a_n\}$ has a subsequence converging to $b$ contradicts nothing. But this is a minor oversight. You should have started off with:

"Assume $\overline A$ is not compact, then there is at least one sequence $\{a_n\} \subseteq \overline A$ which has no convergent subsequences."

From there on, the argument is correct.

By the way, this argument can be be recast to be direct, rather than by contradiction: "Suppose that every sequence in $A$ has a convergent subsequence. Let $\{a_n\} \subseteq \overline A$." Then proceed to develop $\{b_n\}$ and show that $\{a_n\}$ has a convergent subsequence as in your version. Then you conclude "Thus the arbitrary sequence $\{a_n\}$ must have a convergent subsequence. Hence $\overline A$ is compact."