Basis Condition for Coarser Topology/Corollary 1

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:
 * $\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$

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

Proof
Let $\BB_1$ and $\BB_2$ satisfy:
 * $\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$

Let $U \in \BB_1$.

Let $\AA = \set{ V \in \BB_2 : V \subseteq U}$

From Union of Family of Sets is Smallest Superset:
 * $\bigcup \AA \subseteq V$

Let $x \in U$.

Then:
 * $\exists V_x \in \BB_2 : x \in V_x \subseteq U$

Thus:
 * $V_x \in \AA$

and
 * $x \in V_x \subseteq \bigcup \AA$

Since $x$ was arbitrary, it follows that:
 * $U \subseteq \bigcup \AA$

By the definition of set equality:
 * $U = \bigcup \AA$

Since $U$ was arbitrary, it follows that $\BB_1$ and $\BB_2$ satisfy:
 * $\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$

From Basis Condition for Coarser Topology, $\tau_1$ is coarser than $\tau_2$.