Synthetic Basis and Analytic Basis are Compatible

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S, \tau}$ be a topological space.


Then $\BB$ is an analytic basis for $\tau$ if and only if $\tau$ is the topology on $S$ generated by the synthetic basis $\BB$.


Proof

Necessary Condition

Suppose that $\BB$ is an analytic basis for $\tau$.

We proceed to check the axioms for $\BB$ to be a synthetic basis on $S$.


Open set axiom $(\text O 3)$ states that $S \in \tau$.

Therefore, by the definition of an analytic basis:

$\displaystyle \exists \SS \subseteq \BB: S = \bigcup \SS$

By Equivalent Conditions for Cover by Collection of Subsets, $\BB$ is a cover for $S$.

That is, axiom $(\text B 1)$ for a synthetic basis is satisfied by $\BB$.


Suppose that $A, B \in \BB$.

By the definition of an analytic basis, $\BB \subseteq \tau$.

Therefore $A, B \in \tau$.

Open set axiom $(\text O 2)$ states that $A \cap B \in \tau$.

Therefore, by the definition of an analytic basis:

$\displaystyle \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$

That is, axiom $(\text B 2)$ for a synthetic basis is satisfied by $\BB$.


Therefore, $\BB$ is a synthetic basis on $S$.


Let $\tau'$ be the topology on $S$ generated by the synthetic basis $\BB$.

It follows from the definition of an analytic basis that $\tau \subseteq \tau'$.

Since the subset relation is transitive, we can apply open set axiom $(\text O 1)$ for a topology to conclude that:

$\displaystyle \forall U \in \tau': \exists \AA \subseteq \BB \subseteq \tau: U = \bigcup \AA \in \tau$

That is, $\tau' \subseteq \tau$.

By definition of set equality:

$\tau = \tau'$

$\Box$


Sufficient Condition

Suppose that $\BB$ is a synthetic basis on $S$, and that $\tau$ is the topology on $S$ generated by $\BB$.

Then:

$\displaystyle \BB = \set {\bigcup \set B: \set B \subseteq \BB} \subseteq \tau$

By definition, $\BB$ is an analytic basis for $\tau$.

$\blacksquare$


Sources