Basis Condition for Coarser Topology/Corollary 2

Theorem
Let $S$ be a set.

Let $\BB_1$ and $\BB_2$ be two bases on $S$.

Let $\tau_1$ and $\tau_2$ be the topologies generated by $\BB_1$ and $\BB_2$ respectively.

If $\BB_1 \subseteq \BB_2$ then $\tau_1$ is coarser than $\tau_2$.

Proof
Let $\BB_1 \subseteq \BB_2$.

Let $U \in \BB_1$.

Let $\AA = \set{U}$.

Then:
 * $\AA \subseteq \BB_2$

and
 * $U = \bigcup \AA$

So:
 * $\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$

From Leigh.Samphier/Sandbox/Basis Condition for Coarser Topology: $\tau_1$ is coarser than $\tau_2$