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 ${y \in E : | f_{i} (x - y) | < ϵ (1 \leq i \leq n)}$ where $f_{i}$ are members of $E^{*}$ and $ϵ > 0$ .

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 $A \in U_{x} ⟹ x \in A$ . 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 $τ$ on $X$ so that:

(1) Each member of $U_{x}$ is in $τ$ ; 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 $B \in τ$ 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 $τ_{0}$ , generated by all the sets $U = {U : U \in U_{x} (x \in X)}$ that is, members of $τ_{0}$ are arbitrary unions of sets of the norm $V_{1} \cap \dots \cap V_{n}$ for $V_{i} \in U$ . Condition (1) holds, and condition (2) will hold exactly when, given $V_{i} \in U$ each containing $x$ , there is $A \in U_{x}$ with $A \subseteq V_{1} \cap \dots \cap V_{n}$ . In particular:

(2a) Given $A_{1}, \dots, A_{n} \in U_{x}$ there is $A \in U_{x}$ with $A \subseteq A_{1} \cap \dots \cap A_{n}$ ;
(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 $V_{i} \in U$ 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 $A \subseteq A_{1} \cap \dots \cap A_{n} \subseteq V_{1} \cap \dots \cap V_{n}$ .

Consider now $τ_{1} = {A \subseteq X : x \in A ⟹ \exists B \in U_{x}, B \subseteq A} .$ Then $τ_{1}$ contains $\emptyset$ and $X$ , and is obviously closed under unions. Condition (2a) implies that $τ_{1}$ is closed under finite intersections, that is, that $τ_{1}$ is a topology. If $A \in τ$ and $x \in A$ then by (2), there is $C \in U_{x}$ with $C \subseteq A$ ; thus $A \in τ_{1}$ . Hence $τ_{0} \subseteq τ \subseteq τ_{1}$ .

Finally, given $A \in τ_{1}$ , if $x \in A$ then there is $B \in U_{x} \subseteq U$ with $B \subseteq A$ , and so $A$ is the union of members of $U$ , so $A \in τ_{0}$ . We conclude that our putative topology $τ$ is unique, and equals $τ_{0}$ (or $τ_{1}$ ). Further, we have shown that $τ$ 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.

Neighbourhood base

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