I am tring to understand the proof given in Countable Complement Space is not First-Countable
What I don't understand is that how is it that the intersection of all members of the countable local basis of $x$ equals $\{x\}$? Are we assuming that $\{x\}$ is a member of ${B}_x$?.
Best Answer
$x \in S$ and $S$ is open in $S$ so there is a member $B$ of the base such that $x \in B$. $\{x\}$ is contained the intersection of all members of the base that contain $x$. For the reverse inclusion you need the fact that if $y \neq x$ then there exists an open set $V$ containing $x$ but not $y$ and so there is a member of the base that contains $x$ but not $y$. [We can take $V=S\setminus \{y\}$].