# First countable separable spaces are not second countable.

I know that $$\Bbb R_l$$ is first countable and separable but not second countable.

But I had earlier tried to prove that first countable separable spaces are second countable. So I wrote a proof. Now I know a counterexample, but still I can not find a mistake in my proof.

My "proof" for the wrong fact:

First countable separable spaces are second countable.

Proof : As we have separable space there is countable dense set, say $$A$$.

Now for each point in $$A$$ we can find a countable basis.

Countable unions of countable set is countable.

Claim: Set formed by taking the union of all countable basis at the dense set is required basis.

Take any $$x\in X$$. Choose any neighborhood of $$x$$. Now as $$A$$ is dense, there exists some point in $$A$$ that is in the given neighborhood and so choose one of the countable basis element that contained in that neighborhood.

So we are done.

Actually, I was thinking maybe the last statement has a problem. Is it so? Why not countable basis element at that point should not contain in given neighborhood?

Any help will be appreciated.

• You also need that basis element to contain $x$. Apr 24 '19 at 4:44
• Yes Sir Thanks a lot Apr 24 '19 at 9:05

Let $$\mathcal B$$ denote the union of the countable neigborhood bases at all $$a \in A$$. To prove that it is a basis for $$X$$ you have to show that for any $$x \in X$$ and any open neighborhhod $$U$$ of $$x$$ in $$X$$ there exists $$B \in \mathcal B$$ such that $$x \in B \subset U$$.
You argue as follows: Take any $$a \in U \cap A$$ and choose $$B$$ such that $$a \in B \subset U$$. But unfortunately there is no reason why $$x \in B$$.