# Neighbourhood base

Posted on 31st January 2019

Reading a paper with my office mate, we ended up having a discussion about the notion of an "open neighbourhood base" in a topological space. For example, I might informally say that the weak topology on a Banach space $$E$$ has, around a point $$x$$, an open neighbourhood base is given by the sets $\{y\in E : |f_i(x-y)|<\epsilon \ (1\leq i\leq n) \}$ where $$f_i$$ are members of $$E^*$$ and $$\epsilon > 0$$.

This raises a natural question:

Suppose we have a set $$X$$ and for each $$x\in X$$ we have specified a collection $$U_x$$ of subsets of $$X$$, such that $$A\in U_x \implies x\in A$$. When is there a topology on $$X$$ such that $$U_x$$ are the "basic open neighbourhoods" of $$x$$?

What precisely do we mean by this? I think this is a little unclear, so let's be precise. We seek a topology $$\tau$$ on $$X$$ so that:

• (1) Each member of $$U_x$$ is in $$\tau$$; and
• (2) Each $$U_x$$ should be a filter-base for the collection of all neighbourhoods of $$x$$. That is, if $$x \in B \subseteq A \subseteq X$$ with $$B\in\tau$$ then there is $$C\in U_x$$ with $$C\subseteq A$$. Clearly it suffices to consider the cases when $$A=B$$.

Consider the topology, say $$\tau_0$$, generated by all the sets $$\newcommand{\mc}{\mathcal} \mc U = \{U : U\in U_x \ (x\in X) \}$$ that is, members of $$\tau_0$$ are arbitrary unions of sets of the norm $$V_1\cap\cdots\cap V_n$$ for $$V_i\in\mc U$$. Condition (1) holds, and condition (2) will hold exactly when, given $$V_i\in\mc U$$ each containing $$x$$, there is $$A\in U_x$$ with $$A\subseteq V_1\cap\cdots\cap V_n$$. In particular:

• (2a) Given $$A_1,\cdots,A_n\in U_x$$ there is $$A\in U_x$$ with $$A\subseteq A_1\cap\cdots\cap A_n$$;
• (2b) Given $$B\in U_y$$ with $$x\in B$$, there is $$A\in U_x$$ with $$A\subseteq B$$;

These two conditions then imply (2), for given $$V_i\in\mc U$$ each containing $$x$$, by (2b) we can find $$A_i\in U_x$$ with $$A_i\subseteq V_i$$, and then by (2a) there is $$A\in U_x$$ with $$A\subseteq A_1\cap\cdots\cap A_n \subseteq V_1\cap\cdots\cap V_n$$.

Consider now $\tau_1 = \{ A\subseteq X : x\in A\implies\exists\,B\in U_x, \ B\subseteq A\}.$ Then $$\tau_1$$ contains $$\emptyset$$ and $$X$$, and is obviously closed under unions. Condition (2a) implies that $$\tau_1$$ is closed under finite intersections, that is, that $$\tau_1$$ is a topology. If $$A\in\tau$$ and $$x\in A$$ then by (2), there is $$C\in U_x$$ with $$C\subseteq A$$; thus $$A\in \tau_1$$. Hence $$\tau_0 \subseteq \tau \subseteq \tau_1$$.

Finally, given $$A\in\tau_1$$, if $$x\in A$$ then there is $$B\in U_x\subseteq\mc U$$ with $$B\subseteq A$$, and so $$A$$ is the union of members of $$\mc U$$, so $$A\in\tau_0$$. We conclude that our putative topology $$\tau$$ is unique, and equals $$\tau_0$$ (or $$\tau_1$$). Further, we have shown that $$\tau$$ exists exactly when (2a) and (2b) hold (which are conditions on the families $$U_x$$).

Let us return to the weak topology. (2a) is clear, but to show (2b) is a little exercise in picking epsilons.