# If for every cover of open balls there exists a finite sub-cover, then $K$ is sequentially compact

Let $(X,d)$ be a metric space, and let $K\subseteq X$. Suppose that every cover by open balls of $K$ has a finite sub-cover. I need to prove that $K$ is sequentially compact.

I thought on trying to take an arbitrary open cover, and maybe somehow use the fact that every open subset can be represented by a union of open balls, but that failed. Any hints?

• what is your definition of compact? The usual definition of "compact" is exactly what you stated in topology – Siddharth Bhat Dec 7 '16 at 11:20
• Sequential compactness. Updated. – Joshhh Dec 7 '16 at 11:23
• @SiddharthBhat The usual definition is that for every open cover there is a finite subcover. Here, it is only given that for open covers consisting of balls, there is a finite subcover. – астон вілла олоф мэллбэрг Dec 7 '16 at 11:23
• @астонвіллаолофмэллбэрг it doesn't matter for compactness, right? whatever your cover has been made of, compactness guarantees a finite subcover? – Siddharth Bhat Dec 7 '16 at 11:25
• @SiddharthBhat I think there is a confusion, we will sort it out. Compactness means: every open cover has a finite subcover. It is "not" given that $K$ is compact. What is only given is that if the open cover consists of balls, then there is a finite subcover. To show that $K$ is compact, we would need to show that open covers consisting of all kinds of shapes and sizes, have finite subcovers. This is a stronger statement, because the open sets have no structure that you can exploit here.So while you can go from open sets to balls, the reverse is non-trivial. – астон вілла олоф мэллбэрг Dec 7 '16 at 11:29

You want to show that every sequence in $K$ has a limit point. Suppose the sequence $(a_n)$ has none. For each $x\in K$, there exists $\delta_x$ such that the sequence $(a_n)$ visits $B(x,\delta_x)$ only finitely many times.

The set of balls $\{B(x,\delta_x):x\in K\}$ is an open cover of $K$ by balls.

Can you finish and fill in the details?

• If it had a finite sub-cover then at least one ball would contain infinitely many elements of $a_n$. Is that the idea? – Joshhh Dec 7 '16 at 11:48
• @Joshhh Essentially so; however here it's not the elements, but the terms of the sequence. – egreg Dec 7 '16 at 13:09
• Yes, I meant that infinitely many terms of the sequence. Thank you – Joshhh Dec 7 '16 at 13:13

Show first that any infinite countable set $A$ in $X$ has a limit point. If not, every $X \in X$ is not a limit point which means that every $x \in X$ has an open neighbourhood $O_x$ such that $O_x \cap A \subseteq \{x\}$. THis gives an open cover of $X$, which has a finite subcover $O_{x_1},\ldots O_{X_N}$. But then

$$A = A \cap X \subset A \cap (\cup_{i=1}^N O_{x_i}) = \cup_{i=1}^N (O_{x_i} \cap A) \subseteq \{x_1, \ldots x_N\}$$

which implies $A$ is finite, a contradiction.

(this is the general proof that a compact set is limit point compact).

Next, if $(x_n)$ is a sequence in $X$, suppose that $A = \{x_n: n \in \mathbb{N}\}$ is infinite, otherwise some value $x_k$ occurs infnintely many often, yielding a constant (hence convergent) subsequence and we'd be done. But then there is some limit point $p$ of $A$, i.e. a point $p$ such that for every $\varepsilon >0$, $B(p,\varepsilon) \cap A$ has some point unequal to $p$. In fact every ball intersects $A$ in an infinite set.

This allows one to construct a subsequence converging to $p$: Pick $n_1$ with $x_{n_1} \in B(p,1)$. Pick $n_1 > n_2$ with $x_{n_2} \in B(p, \frac{1}{2})$ and $n_2 > n_1$ with $x_{n_3}\in B(p, \frac{1}{3}$, etc.

THis argument works for any first countable $T_1$ space.