Synthetic Basis and Analytic Basis are Compatible
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 $\paren {\text O 3 }$: Underlying Set is Element of Topology states that $S \in \tau$.
Therefore, by the definition of an analytic basis:
- $\ds \exists \SS \subseteq \BB: S = \bigcup \SS$
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 $\paren {\text O 2 }$: Pairwise Intersection of Open Sets states that $A \cap B \in \tau$.
Therefore, by the definition of an analytic basis:
- $\ds \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$
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 $\paren {\text O 1 }$: Union of Open Sets for a topology to conclude that:
- $\ds \forall U \in \tau': \exists \AA \subseteq \BB \subseteq \tau: U = \bigcup \AA \in \tau$
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:
- $\ds \BB = \set {\bigcup \set B: \set B \subseteq \BB} \subseteq \tau$
By definition, $\BB$ is an analytic basis for $\tau$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.2$: Bases: Proposition $3.2.4$