Showing $\{B \in \mathcal B: x \in B\}$ is filterbasis of neighbourhood filter $\mathcal U(x)$ $\forall x \in X$

Question

Let $(X,\mathcal T)$ be a topological space and $\mathcal B$ a basis of $\mathcal T$. How can I show that for every $x \in X$ the system $\{B \in \mathcal B: x \in B\}$ is a filterbasis of the neighbourhood filter $\mathcal U(x)$?

Answer 1

This is true just by applying definitions:

Suppose that $U \in \mathcal{U}(x)$. This means that $U$ is a neighbourhood of $x$ so there exists an open set $O_x$ such that $x \in O_x \subseteq U$.

As $\mathcal{B}$ is a base, ($O_x$ is a union of base elements so) there exists a $B \in \mathcal{B}$ such that $x \in B \subseteq O_x$.

But then $B \in \mathcal{B}_x$ by definition. And $B \subseteq U$. So any member of $\mathcal{U}(x)$ contains a member of $\mathcal{B}_x$ which is the definition of the latter being a filter base for the former.

Done.