Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis

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Theorem

The following definitions of the concept of Topology Generated by Synthetic Sub-Basis are equivalent:

Definition 1

Define:

$\displaystyle \mathcal B = \left\{{\bigcap \mathcal F: \mathcal F \subseteq \mathcal S, \, \mathcal F \text{ is finite}}\right\}$

That is, $\mathcal B$ is the set of all finite intersections of sets in $\mathcal S$.

Note that $\mathcal F$ is allowed to be empty in the above definition.


The topology generated by $\mathcal S$, denoted $\tau \left({\mathcal S}\right)$, is defined as:

$\displaystyle \tau \left({\mathcal S}\right) = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\}$

Definition 2

The topology generated by $\mathcal S$, denoted $\tau \left({\mathcal S}\right)$, is defined as the unique topology on $X$ that satisfies the following axioms:

$\left({1}\right): \quad \mathcal S \subseteq \tau \left({\mathcal S}\right)$
$\left({2}\right): \quad$ For any topology $\mathcal T$ on $X$, the implication $\mathcal S \subseteq \mathcal T \implies \tau \left({\mathcal S}\right) \subseteq \mathcal T$ holds.

That is, $\tau \left({\mathcal S}\right)$ is the coarsest topology on $X$ for which every element of $\mathcal S$ is open.


Proof

Let $X$ be a set.

Let $\mathcal S \subseteq \mathcal P \left({X}\right)$ be a synthetic sub-basis on $X$.

Let $\mathcal B$ be the synthetic basis on $X$ generated by the synthetic sub-basis $\mathcal S$.

Let $\tau \left({\mathcal S}\right)$ be the topology on $X$ generated by the synthetic basis $\mathcal B$.


We now show that:

$\left({1}\right): \quad$ $\mathcal S \subseteq \tau \left({\mathcal S}\right)$.
$\left({2}\right): \quad$ For any topology $\mathcal T$ on $X$, the implication $\mathcal S \subseteq \mathcal T \implies \tau \left({\mathcal S}\right) \subseteq \mathcal T$ holds.


We have that:

$\mathcal S \subseteq \mathcal B \subseteq \tau \left({\mathcal S}\right)$

Hence by transitivity of $\subseteq$:

$\mathcal S \subseteq \tau \left({\mathcal S}\right)$


Suppose that $\mathcal T$ is a topology on $X$ such that $\mathcal S \subseteq \mathcal T$.

From General Intersection Property of Topological Space:

$\mathcal B \subseteq \mathcal T$

By open set axiom $\left({1}\right)$:

$\tau \left({\mathcal S}\right) \subseteq \mathcal T$


It follows that $\tau \left({\mathcal S}\right)$ is the unique topology on $X$ satisfying conditions $\left({1}\right)$ and $\left({2}\right)$.

$\blacksquare$