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:

$\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$

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

$\blacksquare$