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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.

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    $\begingroup$ You also need that basis element to contain $x$. $\endgroup$ Apr 24 '19 at 4:44
  • $\begingroup$ Yes Sir Thanks a lot $\endgroup$
    – Math Lover
    Apr 24 '19 at 9:05
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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 $x \in B$.

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