Let $(\mathcal M,d)$ be a metric space and $A\subset\mathcal M$. The following types of compactness are equivalent:

(i) Each open cover of $A$ contains a finite subcover.

(ii) $A$ is sequentially compact, i.e. each sequence in $A$ contains a subsequence which converges in $A$.

In the lecture, I've seen proofs for (i) $\Rightarrow$ (ii) and (ii) $\Rightarrow$ (i). I have a problem with the first one, (i) $\Rightarrow$ (ii), which reads as follows:

If an (infinite) sequence $(x_k)_{k\in\mathbb N}\subset A$ does not have any cluster points, then $\forall\ y\in A$ we can find a ball of radius $r_y>0$ around $y$ which contains only finitely many elements of $(x_k)_k$.

We can use these balls to construct an (infinite) open cover of $A$, $A\subset\bigcup_{y\in A}B_{r_y}(y)$. By (i), there exists a finite subcover, i.e.

$\exists\ y_1,\dots,y_n\in A$, such that $A\subset \bigcup_{i=1}^nB_{r_{y_i}}(y_i)$.

Since $(x_k)_k\subset A$ has infinitely many elements but each of the finitely many balls contains only finitely many elements of $(x_k)_k$, we have a contradiction and, thus, $(x_k)_k$ has to have at least one cluster point. So far, I agree.

Now the argument is that, $(x_k)_k$ has to contain a subsequence which converges and, thus, we've shown (ii). My point is: we are required to show that the subsequence should converge in $A$ but I don't seen why the cluster point of the original sequence should be in $A$. For this to hold, $A$ would have to be closed. In the lecture, we had a lemma stating that $A$ is closed if it is compact, but we used the equivalence of (i) and (ii) in the process.


2 Answers 2


The contradiction shows that there is at least one $y \in A$ such that every ball with centre $y$ contains infinitely many $x_k$. Then there is a subsequence converging to $y$.

Alternatively, we can show that a compact set is closed without the equivalence:

Let $A\neq \overline{A}$. Let $x \in \overline{A}\setminus A$. Then the open cover

$$\{ \mathcal{M}\setminus \{ x : d(x,y) \leqslant 2^{-n}\} : n \in \mathbb{N}\}$$

of $A$ has no finite subcover.


If no $y\in A$ is a limit point, then for each $y \in A$, there is an $r_y > 0$ such that $B(y;r_y)$ contains only finitely many points of the sequence ...

The argument implies that a limit point must be in $A$. In fact it begins with "If an (infinite) sequence $(x_k)_{k\in \mathbb{N}} \subset A$ does not have any cluster points in $A$, then ..


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