# 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