# The Unapologetic Mathematician

## Bases and Subbases

We’ve defined topologies by convergence of nets, by neighborhood systems, and by closure operators. In each case, we saw some additional hypothesis — sometimes more and sometimes less explicitly — to restrict which data actually corresponded to a topological space. That is, many neighborhood systems give rise to the same topology, which in turn induces only one of those neighborhood systems. Now let’s turn back to our original definition of topology and see how we can weaken it in a similar way.

Remember that we defined the closure of a set $A$ in a topological space $(X,\tau)$ as the smallest closed set containing $A$. To get at it, we took the intersection of all the closed sets containing $A$. And we knew that at least one such closed set existed because the whole space $X$ was closed. We’re going to do the exact same thing to come up with topologies.

So let’s take a collection $\sigma\subseteq P(X)$ of subsets of $X$. We want the smallest collection $\tau\subseteq\sigma$ of subsets of $X$ that contains $\sigma$ so that $\tau$ is a topology. To get at it, we consider all the topologies on $X$ that contain $\sigma$, and then take their intersection. As we saw back when we first defined topologies, this intersection will again be a topology, and it will be contained in any topology containing $\sigma$. And we know that we have at least one topology containing $\sigma$ because the discrete topology has all of $P(X)$ as its open sets.

Let’s see how we can build up the topology $\tau$ from $\sigma$ more directly. What is it that prevents $\sigma$ from being a topology itself? Well, it might not be closed under taking arbitrary unions and finite intersections. So let’s start with $\sigma$ and throw in all the unions of finite intersections of elements of $\sigma$. We’ll use the convention that the union of no subsets of $X$ is the empty set $\varnothing$, while the intersection of no subsets of $X$ is the entire set $X$. This means we at least have $\varnothing$ and $X$ as unions of finite intersections.

Now let’s consider the intersection of two such sets. That is, if we start with $\bigcup\limits_{a\in\mathcal{A}}\bigcap\limits_{i\in\mathcal{I}_a}U_{a,i}$ and $\bigcup\limits_{b\in\mathcal{B}}\bigcap\limits_{i\in\mathcal{I}_b}U_{b,i}$, then we get the intersection $\left(\bigcup\limits_{a\in A}\bigcap\limits_{i\in\mathcal{I}_a}U_{a,i}\right)\cap\left(\bigcup\limits_{b\in B}\bigcap\limits_{i\in\mathcal{I}_b}U_{b,i}\right)=\bigcup\limits_{a\in A}\bigcup\limits_{b\in B}\left(\bigcap\limits_{i\in\mathcal{I}_a}U_{a,i}\cap\bigcap\limits_{i\in\mathcal{I}_b}U_{b,i}\right)$

which is again a union of finite intersections. Similarly, if we take an arbitrary union of these unions of finite intersections, we get another union of finite intersections. And any topology containing the sets in $\sigma$ must contain these sets. So this is exactly the topology generated by $\sigma$. In this case, we call $\sigma$ a subbase for the topology it generates.

“Subbase”? What happened to “base”? Well, a base for a topology is sort of halfway in between a subbase and a topology. First of all, we require that the elements of $\sigma$ cover $X$. That is, every point in $X$ shows up in at least one of the sets in $\sigma$. We also require that if $S_1$ and $S_2$ are in $\sigma$ and $x\in S_1\cap S_2$ then there is some $S_3$ in $\sigma$ with $x\in S_3\subseteq S_1\cap S_2$. Thus we can write the intersection of any two elements of $\sigma$ as a union of other elements of $\sigma$. The covering property says that we can write the empty intersection as a union as well. And so we don’t need to take any intersections at all — only unions. That is, a base $\beta$ for a topology $\tau$ on a set $X$ is a collection of subsets of $X$ so that every subset in $\tau$ is a union of subsets in $\beta$. In particular, if we start with any collection of sets $\sigma$ and throw in all the finite intersections of subsets in $\sigma$ we get a base for the topology generated by $\sigma$.

Probably the nicest thing about defining a topology with a subbase is that the subbase is all we need to check continuity. More explicitly: let $\sigma\subseteq P(Y)$ be a subbase generating a topology $\tau_Y$ on a set $Y$, and let $(X,\tau_X)$ be any topological space. Then we have defined a function $f:X\rightarrow Y$ to be continuous if $f^{-1}(U)\in\tau_X$ for each $U\in\tau_Y$. What I’m asserting here is that we can weaken this to say that $U\in\sigma$ implies $f^{-1}(U)\in\tau_X$. For then any set in $\tau_Y$ is the union of finite intersections of sets in $\sigma$, and the preimage of such a set is then the union of the finite intersections of the preimages of the sets in $\sigma$. So if these are all open, so will be the preimage of every set in $\tau_Y$.

November 22, 2007