# Compact space implies Sequentially compact space?

In General Topology Class we proved that if X is a metric space, X is compact iff is sequentially compact.

In particular in the first implication we assumed there is some sequence that doesn't have any convergent subsequence converging to a point in X; let Z be the set of the elements of the sequence then for every point x belonging to X exists an open set containing x that either has no intersection with Z o has only x(if x belongs to Z);

then X\Z is open (so Z is closed) and every point in Z is an isolated point;

X is compact an Z is closed then Z is compact(for some prop.);

Z is a compact discret topology subspace so |Z| is finite and this is impossible because we assumed there is some sequence that doesn't have any convergent subsequence converging to a point in X

My question is:

when do i use the fact that X is a metric space? When i say that there is an open set etc. i can say that for every Hausdorff space but i know there are known counterexample to this

can you help me?

If something is not clear enough please ask me some clarification

• Jan 23 at 6:13

There are non-metrisable spaces that are compact but not sequentially compact, the most well-known of them are $$\beta \Bbb N$$, the Cech-Stone compactification of $$\Bbb N$$, and $$[0,1]^I$$ in the product topology where $$I$$ is an index set of size continuum or larger. A space like $$\omega_1$$ in the order topology is sequentially compact (and first countable) but not compact. So in general neither implication holds between sequential compactness and compactness; they can be quite different.