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.

  • 3
    $\begingroup$ You also need that basis element to contain $x$. $\endgroup$ Apr 24 '19 at 4:44
  • $\begingroup$ Yes Sir Thanks a lot $\endgroup$ 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 $x \in B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.