# Prove subbasis generates topology

I have a problem with this exercise: Prove that $S = \{[a\, +\infty), a\in \mathbb{R} \}$ generates a topology.

The topology generated is the collection of all unions of finite intersections. Now, because $[a, \infty)\cap[b, \infty)$ is never empty, no finite intersection will be empty, then the union of finite intersections will not be empty. This would lead me to think that it is not a problem to generate $\mathbb{R}$, but where is the empty set? If the union of finite intersections is never empty there is no empty set.

According to the exercise, $S$ is a subbasis, but how can the empty set be generated? My guess would be that it generate the lower limit topology because the intervals are of the form $[a,\infty)$.

• The empty union (the union of the subfamily $\emptyset \subseteq S$) is $\emptyset$. And the topology is not the lower limit one, we cannot even separate two distinct points or singletons are not closed. So $T_1$and $T_2$ fail. – Henno Brandsma May 9 '17 at 19:11

## 1 Answer

While doing union operation if you choose your index set to be an empty set then you end up with the empty set. While doing the intersection if you pick your index from an empty set then you end up with the whole set.

A bit confusing and philosophical but to give you some sense.

The identity element of multiplication is 1. Multiplying nothing with nothing you will end up with identity whic is 1.

The identity element of addition is 0. Adding nothing to nothing will end up with identity 0

So if you intersect any set,say A, with the whole set,say X, you get A. X is identity for the intersection.

And same way, an empty set is identity for union.