Will $\{B_n(x)\}_n$ be a local base of point $x$?

Question

If $X$ is a Hausdorff compact space and $\{B_n(x)\}_n$ is a family of open sets of $X$ for each $x\in X$ such that $\cap \{B_n(x)\} =\{x\}$.

Will $\{B_n(x)\}_n$ be a local base of point $x$?

Thanks ahead.

Answer 1

The answer in general (all spaces) is no. Any countable $T_1$ space $X$ has the property that there is a countable family of open sets $B_n(x)$ such that $\cap_n B_n(x) = \{x\}$ for every $x$. Just use $B_n(x) = X\setminus \{y_n\}$ where the $y_n$ enumerate $X\setminus \{x\}$.

But many countable spaces are not first countable: Arens-Fort space, an ultrafilter space, a countable dense subset of $[0,1]^c$ are some of my favourites. So the above family cannot be a local base for a point where first-countability fails, obviously. All of these spaces are Tychonov, an ultrafilter space is even (hereditarily) normal.

[added] I later saw you were asking for compact Hausdorff spaces. There a variant is true: suppose that $O_n(x)$, $n \in \mathbb{N}$ exist such that $\cap_n O_n(x) =\{x\}$, then $x$ has a countable local base (which need not be the same $O_n$, see Littefield's example. This is known as $\psi\chi(x,X) = \chi(x,X)$ in terms of cardinal invariants. You first make the $O_n$ decreasing in the standard way ,then find open $B_n$ with $x \in B_n \subseteq \overline{B_n} \subseteq O_n$ ($X$ is $T_3$) and use a standard theorem on decreasing sequences of compact sets, to prove that the $B_n$ form a countable local base.

Answer 2

The answer is no. Consider $X=[-1,1]$ with the usual topology. Then, for each $n\in\mathbb{N}$, we define the following sets:

$B_n:=(-\frac{1}{n},1)$ if $n$ is even,

$B_n:=(-1,\frac{1}{n})$ otherwise.

Then for any $x\in X$, we define $B_n(x):=(x+B_n)\cap X$.

If we argue at $x=0$, it is clear that $\cap B_n=\{0\}$, but this family is not a local base of $0$ because $B_n$ is not included in $(-\frac{1}{2},\frac{1}{2})$ for any $n$.