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.