Basis Condition for Coarser Topology/Corollary 1
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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 $\BB_1$ 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:
- $\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
From Basis Condition for Coarser Topology, $\tau_1$ is coarser than $\tau_2$.
$\blacksquare$