# Equivalence of Definitions of Topology Generated by Synthetic Basis

## Theorem

Let $S$ be a set.

Let $\BB$ be a synthetic basis on $S$.

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

### Definition 1

The **topology on $S$ generated by $\BB$** is defined as:

- $\tau = \set{\bigcup \AA: \AA \subseteq \BB}$

That is, the set of all unions of sets from $\BB$.

### Definition 2

The **topology on $S$ generated by $\BB$** is defined as:

- $\tau = \set {U \subseteq S: U = \bigcup \set {B \in \BB: B \subseteq U}}$

### Definition 3

The **topology on $S$ generated by $\BB$** is defined as:

- $\tau = \set {U \subseteq S: \forall x \in U: \exists B \in \BB: x \in B \subseteq U}$

## Proof

### Definition 1 iff Definition 2

Trivially, the reverse implication holds, as $\set {B \in \BB: B \subseteq U} \subseteq \BB$.

We now show that the forward implication holds.

Suppose that $U \in \tau$. Then, by definition:

- $\ds \exists \AA \subseteq \BB: U = \bigcup \AA$

By Union is Smallest Superset: General Result, we have:

- $\forall B \in \AA: B \subseteq U$

By the definition of a subset, it follows that:

- $\AA \subseteq \set {B \in \BB: B \subseteq U}$

From Union of Subset of Family is Subset of Union of Family:

- $\ds U = \bigcup \AA \subseteq \bigcup \set {B \in \BB: B \subseteq U}$

By Union is Smallest Superset: General Result:

- $\ds \bigcup \set {B \in \BB: B \subseteq U} \subseteq U$

By definition of set equality:

- $\ds U = \bigcup \set {B \in \BB: B \subseteq U}$

$\Box$

### Definition 1 iff Definition 3

From Set is Subset of Union: General Result, the forward implication directly follows.

We now show that the reverse implication holds.

By hypothesis, we have that:

- $\ds U \subseteq \bigcup \set {B \in \BB: B \subseteq U}$

By Union is Smallest Superset: General Result:

- $\ds \bigcup \set {B \in \BB: B \subseteq U} \subseteq U$

By definition of set equality:

- $\ds U = \bigcup \set {B \in \BB: B \subseteq U}$

The result follows.

$\blacksquare$