Countable Complement Space is not First-Countable

Question

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 ${B}_x$?.

Answer 1

$x \in S$ 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 $V=S\setminus \{y\}$].

Answer 2

If $X$ is a $T_1$ space (like the cocountable topology is) and $\mathcal B_x$ is a local base for $X$ at some $x$ then $\bigcap \mathcal B_x=\{x\}$:

It's by definition that $x \in B$ for all $B \in \mathcal{B_x}$ so $x$ is in this intersection. If $y \neq x$ is any other point of $X$ it's not in the intersection: by $T_1$-ness, $X\setminus \{y\}$ is open and contains $x$ so for some $B_y \in \mathcal B_x$ we have $x \in B_y \subseteq X\setminus \{y\}$, hence $y \notin B_y$, so $y \notin \bigcap \mathcal B_x$; this shows the equality. $T_1$ is essential: look at the Sierpiński space or an indiscrete space for cases where this intersection does not equal $\{x\}$. You could say that it's a reformulation of $T_1$-ness of the space.