# Neighbourhood bases, continued.

Posted on 26th February 2019

We saw before that on a set $$X$$ we can specify a (unique) topology by, for each $$x\in X$$, specifying a collection of sets $$U_x$$ which will satisfy that:

• Each $$V\in U_x$$ contains $$x$$ and will be open;
• Every open $$C \ni x$$ will be such that there is $$V\in U_x$$ with $$V\subseteq C$$,

if and only if we have the conditions that:

• Given $$A_1,\cdots,A_n\in U_x$$, there is $$A\in U_x$$ with $$A\subseteq A_1\cap\cdots\cap A_n$$;
• Given $$B\in U_y$$ with $$x\in B$$, there is $$A\in U_x$$ with $$A\subseteq B$$.

However, we might ask: what is the advantage of specifying the "basic open sets" about each point, rather than just specifying a base for the topology?

One advantage comes when you consider subspaces. Let $$Y\subseteq X$$ be a subset. There are then two natural ways to construct a topology on Y:

• Use the subspace topology; that is, the open subsets of $$Y$$ are exactly the sets $$U\cap Y$$ for $$U$$ open in $$X$$;
• For each $$y\in Y$$ let $$V_y = \{ A\cap Y : A\in U_y\}$$ and generate a topology using $$V_y$$ as the basic open neighbourhoods of $$y$$.

Do these give the same topology? We will see that the answer is yes. Firstly, it is clear that every member of $$V_y$$ is open in the subspace topology. Conversely, given $$V = U\cap Y$$ open in the subspace topology, for $$y\in V$$, there is $$A\in U_y$$ with $$A\subseteq U$$, so $$B=A\cap Y$$ is such that $$B\in V_y$$ with $$B\subseteq V$$. Thus $$V$$ is in the topology generated by the $$V_y$$.

In many situations, the sets $$V_y$$ will occur very naturally. For example, the topology induced by a metric has as basic open sets about $$x$$ the open balls centred at $$x$$. The above reasoning now immediately shows that the subspace topology on $$Y\subseteq X$$ will agree with the topology given by restricting to metric to $$Y$$ and then considering $$Y$$ as its own metric space.

This post was originally motivated by Fell's topology on a set of (equivalence classes) of unitary representations of a locally compact group $$G$$. A basic open set about a representation $$\pi$$ is $$W=W(\pi,\varphi_i,Q,\epsilon)$$ where $$\epsilon>0$$, $$Q\subseteq G$$ is compact, and $$(\varphi_i)_{i=1}^n$$ are functions of positive type associated to $$\pi$$, and $$\rho\in W$$ exactly when we can approximate each $$\varphi_i$$, up to $$\epsilon$$ error on $$Q$$, by sums of functions of positive type associated to $$\rho$$.

Clearly the first (intersection related) condition holds for this family. The second condition seems trickier: if $$\rho\in W$$, then we wish to find $$W' = W(\rho,\varphi_j',Q',\epsilon')$$ such that if $$\rho'\in W'$$ then $$\rho'\in W$$. Really this is a triangle inequality argument. $|\varphi_i(g) - \sum_k \phi_{i,k}(g)|<\epsilon \qquad (g\in Q)$ where each $$\phi_{i,k}$$ is associated to $$\rho$$. (This follows as $$\rho\in W$$). As $$Q$$ is compact, we can decrease $$\epsilon$$ slightly and still have a true inequality: this allows us some wiggle room. Now if each $$\phi_{i,k}$$ is very well approximated on $$Q$$ by sums of functions of positive type associated to $$\rho'$$, then the $$\varphi_i$$ can also be $$\epsilon$$ approximated by such functions. Thus $$\rho'\in W'(\rho,\phi_{i,k},Q,\epsilon')$$ does imply $$\rho'\in W$$, for some small $$\epsilon'>0$$.

So this topology interacts nicely with subspaces. In particular, the definition works well for subsets of the (proper class) collection of all representations.