# 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 $\SS \subseteq \TT$.

$\BB \subseteq \TT$
$\map \tau \SS \subseteq \TT$

$\blacksquare$