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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

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You’re using first countability (or sequential-ness) , mostly, because you’re using that a point is in the closure of a set iff there is a sequence from the set converging to the point. This allows you to conclude that the sequence without a convergent subsequence is closed and discrete as a set. That, together with Hausdorffness is what you use from metrisability. Plus the fact that sequential compactness gives separability and thus (!) Lindelöfness, for the reverse implication.

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.

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