In Topology for analysis by A. Wilansky the theorem 11.1.2 states:
Let $(X,\mathcal{U})$ be a uniform space, let $\mathcal{B}$ be [uniformity] base for $\mathcal{U}$ and let $x \in X$. Then $\{U(x) : U \in \mathcal{B} \}$ is a [topology generated by uniformity $\mathcal{U}$] base at $x$.
I don't see how this can be possible as Wilansky and most treatments of topology I'm aware requires the base to consist of open sets. And this is not always the case for $\{U(x) : U \in \mathcal{B} \}$: for example, take $X =\mathbb{R} $ and let $\mathcal{U}$ be an uniformity generated by Eucledean distance $d(x,y) = |x-y|$, let $\mathcal{B} = \mathcal{U}$ and $x = 0$. Then, as $\mathcal{U}$ is upwards closed as a filter over $X^2$, the set $\{U(0) : U \in \mathcal{U} \}$ clearly contains sets like $[-1,1]$, which are not open.
In his proof Wilansky shows that for all $U(x)$ there is an open $G$ such that $x \in G \subset U(x)$. And also that for every neighborhood $N$ of $x$, there is a connector $U \in \mathcal{B}$ such that $U(x) \subset N$. I don't have any problems with this results, I just don't understand why all $U(x)$ must be open?
So, either I don't understand some definitions, or there is a mistake in a formulation of the theorem. If the later is the case, then what would be the correct formulation?
I will be grateful for any help.
They don’t have to be open: a neighbourhood base can even consist of compact neighbourhoods only, say.
The only requirements for a neighbourhood base at $x$ are : all the sets in it must be neighbourhoods of $x$ ( they are: Wilansky shows there is an open $G$ with $x \in G \subseteq B(x)$) and every other (open) neighbourhood of $x$ contains one of the $B(x)$ as a subset ( and he shows that too).