Basis Condition for Coarser Topology

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 $\BB_1$ and $\BB_2$ satisfy:
 * $\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$

then $\tau_1$ is coarser than $\tau_2$.

Proof
Let $\BB_1$ and $\BB_2$ satisfy:
 * $\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$

Let $U$ be any element of $\BB_1$.

Then there exists $\AA_2 \subseteq \BB_2$ such that $U = \bigcup \AA_2$.

By definition of the topology generated by $\BB_2$, it follows that $U \in \tau_2$.

Since $U$ was arbitrary, it follows that $\BB_1 \subseteq \tau_2$.

Let $W$ be any element of $\tau_1$.

By definition of the topology generated by $\BB_1$:
 * there exists $\AA_1 \subseteq \BB_1$ such that $W = \bigcup \AA_1$.

Since $\BB_1 \subseteq \tau_2$, it follows that $\AA_1 \subseteq \tau_2$.

By the open set axiom $(O1)$, then $W \in \tau_2$.

Since $W$ was arbitrary, it follows that $\tau_1 \subseteq \tau_2$.

The result follows.