Question

Prove that a sequentially compact metric space x is separable

Answer 1

know that if X is separable then X contains a countable dense subset.

And that a subset E⊂X is dense in X if the closure of E equals X.

I also know that since X is compact then there are two definitions we can use: 1. X is closed and bounded or 2. every open cover of X admits a finite subcover.

I think because we are trying to show X is separable so there is some dense subset - we want to use the definition of compact implies closed and bounded. (My thinking behind that is we want to show a subset is dense and dense means that the closure of the subset equals X - the closure is the union of the set and its limit points. If X is compact then a subset of X would be compact, so that subset is closed and bounded. The subset is closed so it contains all it limit points....)

Proof. So X is compact. Now let E⊂X. Since X is compact, then E is a compact subset of X. Now we need to show that E is a countable dense subset...

I don't know where to go from here.

I have all these definitions of terms, but I don't know how to quite use them to prove that X is separable.

OOH I also have the hint to cover X with neighborhoods of radius 1n - so for every positive integer n, there are finitely many neighborhoods of radius 1n whose union covers X. So maybe it would be better to work with the open cover definition of compact.