Suppose that is a separable metric space. Then there exists a sequence such that it is dense in X. Prove that any open set is a union of elements from the set .
My idea so far.
So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence .
So say that is an nonempty open subset in X, then there should exist at least one element from the sequence, say , that is contained in U. And since U is open, we can find some such that .
Also, I believe that I can pick a point, say , such that for some small . Then I should be able to find another point since it is open. Then I can get another ...
But how should I proceed to get ?