Just as the title explains, does the Sorgenfrey Line have point-countable base? Thanks ahead.
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It does not.
Let $\Bbb S$ be the Sorgenfrey line. Suppose that $\mathscr{B}$ is a point-countable base for $\Bbb S$. $\Bbb Q$ is dense in $\Bbb S$, so each $B\in\mathscr{B}$ contains some $q_B\in\Bbb Q$. For each $q\in\Bbb Q$ let $\mathscr{B}(q)=\{B\in\mathscr{B}:q_B=q\}$; $\mathscr{B}(q)$ is countable, since $q\in\bigcap\mathscr{B}(q)$, and $\mathscr{B}=\bigcup_{q\in\Bbb Q}\mathscr{B}(q)$, so $\mathscr{B}$ is countable. But $\Bbb S$ is not second countable.
Brian M. Scott
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So could we deduce that every separable space with point countable base must be second countable? – Paul Feb 19 '13 at 13:28
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@Paul: Yes: the same argument goes through in general. – Brian M. Scott Feb 19 '13 at 13:33