# Proving that sequentially compact spaces are compact.

I remember seeing this proof somewhere (perhaps here, but I don't remember where) that goes something like this.

Suppose $X$ is sequentially compact, and by contradiction suppose $\{U_n\}$ is a countable open cover with no finite subcover. Then for any positive integer $n$, the set $\{U_i : i \le n\}$ is not an open cover, so there exists $x_n \notin \bigcup_{i \le n} U_i$. Hence, we obtain sequence, and by sequential compactness, there exists a subsequence $x_{n_j}$ that converges to $a \in X$. However, $a \in U_k$ for some positive integer $k$ and by construction, $x_{n_j} \notin U_k$ if $n_j \ge k$. This is a contradiction.

Doesn't this only prove every countable open cover must have a finite subcover?

• In general, sequential compactness neither implies, nor is implied by, compactness; but for metric spaces, they are equivalent (Wikipedia reference). – Zev Chonoles Jun 29 '12 at 6:34
• Now that I think of it, the result was probably "Sequential compactness implies countable compactness" - not this. – Dom Jun 29 '12 at 6:41
• Observe that for metric spaces the notions are indeed equivalent, since we can prove that sequentially compact (and compact) metric spaces are second-countable, so every open cover can be replaced with basic cover, but we have a countable basis so we have the open cover is countable - and the proof you gave follows. – Asaf Karagila Jun 29 '12 at 7:02
• See form [8 AE] at consequences.emich.edu/conseq.htm. $\:$ – user57159 Jun 29 '12 at 7:32

For example, the topological space $\omega_1$ with the order topology is a sequentially compact space but not a compact space. In the first countable space (in fact, it's only need sequential space), sequentially compace = countably compact. As we know, the space is a first countable space and countably compact, therefore, it is a sequentially compact. But, it is not a compact space:)