Why is this the basis for the indiscrete topology?

Indiscrete topology only has $\{\varnothing, X\}$ where $X$ is the entire space.

I know this is probably a dumb question. Why is the basis (according to wikipedia) for this topology just $\{X\}$ instead of $\{\varnothing, X\}$?

I understand that $\{X\}$ satisfies the definition. Suppose $\mathcal{B} = \{X\}$

1. Then $\mathcal{B}$ covers $X$ - Check!

2. For all $B_1, B_2 \in \mathcal{B}, \ldots$ - Trivially satsifed since we don't have $B_1, B_2$

But how does this basis generate $\varnothing$?

1 Answer

If $\mathcal{B}$ is a basis, then for each open set $U$ there is a subset $\mathcal{A}\subseteq\mathcal{B}$ such that $U$ is the union of all of the elements of $\mathcal{A}$. In this example, for $U=\emptyset$, you can just take $\mathcal{A}=\emptyset$. The union of all the elements of $\mathcal{A}$ is the union of no sets, which is just the empty set.

By the way, you should say a basis, not the basis, since it's not the only basis for this topology: $\{\emptyset,X\}$ is also a basis.