Confused about this part topology part in Pederson Analysis Now.

Question

A bit confused about the first part of Proposition 1.2.12. So I know that $\tau(\varphi) := \inf \{\tau : \varphi \subset \tau \}$ since the collection of all topologies on $X$ under inclusion is a complete lattice. I think I understand the general definitions of a subbasis and basis given. But I don't really know what he means when he states $\tau(\varphi)$ consists of exactly those sets that are unions of sets, each of which is a finite intersection of sets from $\varphi$, together with $\emptyset, X$. I thought initially we are adding the union of every sub collection of $\varphi$ and interseection of every finite sub collection of $\varphi$ and adding those with the emptyset and $X$, but that doesn't seem what he is saying. Sorry if this is a stupid question, I am not good with English.

Answer 1

What we are doing with the definition of $\tau(\varphi)$ is first defining it as the smallest topology which contains $\varphi$, and then discussing explicit description of what $\tau(\varphi)$ is.

Consider a group $G$. A subset $A\subseteq G$ generates a subgroup which we can define as intersection of all subgroups of $G$ containing $A$. But we can also give explicit description of the subgroup generated by $A$: we first add all the inverses $A^{-1} = \{x^{-1}: x\in A\}$, and then all possible products $(A\cup A^{-1})^n = \{x_1...x_n : x_i\in A\cup A^{-1}\}$. Note how we first added inverses and then products. Would anything change if we first added all the products and then inverses? Of course, this wouldn't generate a subgroup in general. However if we were to take all possible combinations of "take product" and "take inverse" and took union of sets thus obtained, we would again arrive at the subgroup generated by $A$. But the special property of the process "take inverses then take all the products" lies in its simplicity.

Here also, $\tau(\varphi)$ could be generated using an ad hoc method of just taking all possibilities. And chosing just one of those possibilities, like add empty set, $X$, all unions, then finite intersections, would not work. And the right way to generate $\tau(\varphi)$ would be to first take $\varphi_0 = \varphi\cup\{X, \emptyset\}$, then $\varphi_1 = \{U_1\cap ...\cap U_n : U_i\in \varphi_0\}$ and finally $\tau(\varphi) = \varphi_2 = \{\bigcup_{i\in I} U_i : U_i\in \varphi_1\}$. Note that the equality $\tau(\varphi) = \varphi_2$ needs to be proven by checking that $\varphi_2$ is a topology i.e. that taking any more operations on $\varphi_2$ (finite intersections, arbitrary unions) gives us back an element of $\varphi_2$. Where a basis is special is that we can skip step of defining $\varphi_1$, so it's simpler and desirable (many natural generating sets of topologies i.e. subbases are actually bases; exceptions are e.g. compact-open topology).

Finally, the operations $\tau(\varphi)$ and subgroup generated by $A$ are examples of so called closure operators. They also include any other generating process in an algebraic structure, as well as taking smallest $\sigma$-algebra (important in measure theory). In fact there's a very natural example of a closure operator in a topological space, simply called the closure of a subset $S\subseteq X$ and denoted $\overline{S}$. It also has a more explicit description as $\overline{S} = \{x\in X : U\cap S\neq \emptyset, x\in U\in \tau\}$ though maybe not as nice. In fact it's a good exercise to prove it.