Basis Condition for Coarser Topology/Corollary 2
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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:
- $\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
From Basis Condition for Coarser Topology:
- $\tau_1$ is coarser than $\tau_2$
$\blacksquare$