Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis
Theorem
The following definitions of the concept of Topology Generated by Synthetic Sub-Basis are equivalent:
Definition 1
Define:
- $\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \FF \text{ is finite} }$
That is, $\BB$ is the set of all finite intersections of sets in $\SS$.
Note that $\FF$ is allowed to be empty in the above definition.
The topology generated by $\SS$, denoted $\map \tau \SS$, is defined as:
- $\ds \map \tau \SS = \set {\bigcup \AA: \AA \subseteq \BB}$
Definition 2
The topology generated by $\SS$, denoted $\map \tau \SS$, is defined as the unique topology on $X$ that satisfies the following axioms:
- $(1): \quad \SS \subseteq \map \tau \SS$
- $(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ holds.
That is, $\map \tau \SS$ is the coarsest topology on $X$ for which every element of $\SS$ is open.
Proof
Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a synthetic sub-basis on $X$.
Let $\BB$ be the synthetic basis on $X$ generated by the synthetic sub-basis $\SS$.
Let $\map \tau \SS$ be the topology on $X$ generated by the synthetic basis $\BB$.
We now show that:
- $(1): \quad$ $\SS \subseteq \map \tau \SS$
- $(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ holds.
We have that:
- $\SS \subseteq \BB \subseteq \map \tau \SS$
Hence by transitivity of $\subseteq$:
- $\SS \subseteq \map \tau \SS$
Suppose that $\TT$ is a topology on $X$ such that $\SS \subseteq \TT$.
From General Intersection Property of Topological Space:
- $\BB \subseteq \TT$
By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:
- $\map \tau \SS \subseteq \TT$
It follows that $\map \tau \SS$ is the unique topology on $X$ satisfying conditions $(1)$ and $(2)$.
$\blacksquare$