If I have a subbase $B$ of a topology $(X,T)$ (a subset of the power set of $X$ where the union of all elements in $B$ is $X$), then my topology notes state the collection of unions of finite intersections of elements of $B$ form a topology.
Here is my attempt at a proof:
1) By taking the union of every singleton of $B$ (considering a singleton as a finite intersection with itself), I get $X$, and by taking the empty union I get the empty set.
2) The union of a union of finite intersection of elements of $B$ is again a union of finite intersections of elements of $B$
3) The finite intersection of a union of finite intersection of elements of $B$ is again a finite intersection of elements for $B$ because…
Now for 3), I am a bit confused. Suppose I have two sets $A,C$ that are each unions of finite intersections of elements of $B$. How can I show that their intersection is again a union of finite intersection of elements of $B$ ? Hints and insights appreciated.
Best Answer
For the third, use $$ \bigcap_{i=1}^n \bigcup_{\alpha\in I_i}A^i_\alpha = \bigcup_{(\alpha_1,\ldots, \alpha_n)\in I_1\times\ldots\times I_n} \bigcap_{i=1}^n A_{\alpha_i}^i$$ (And check that I wrangled the sets right. Also remember $n=2$ suffices.)
Alternatively, you can prove the set of all finite intersections form a base, and that the arbitrary unions of a base are a topology.
Finally, observe this is the smallest topology containing the subbase (i.e. the intersection of all topologies containing it).