# 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 Basis Condition for Coarser Topology: $\tau_1$ is coarser than $\tau_2$

$\blacksquare$