# Union from Synthetic Basis is Topology/Proof 1

## Theorem

Let $\BB$ be a synthetic basis on a set $X$.

Let $\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$.

Then $\tau$ is a topology on $X$.

## Proof

We proceed to verify the open set axioms for $\tau$ to be a topology on $S$.

### Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

Let $\AA \subseteq \tau$.

It is to be shown that:

$\ds \bigcup \AA \in \tau$

Define:

$\ds \AA' = \bigcup_{U \mathop \in \AA} \set {B \in \BB: B \subseteq U}$

By Union is Smallest Superset: Family of Sets, it follows that $\AA' \subseteq \BB$.

$\ds \bigcup \AA = \bigcup_{U \mathop \in \AA} \bigcup \set {B \in \BB: B \subseteq U} = \bigcup \AA' \in \tau$

$\Box$

### Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets

Let $U, V \in \tau$.

It is to be shown that:

$U \cap V \in \tau$

Define:

$\OO = \set {A \cap B: A, B \in \BB, \, A \subseteq U, \, B \subseteq V}$

By the definition of a synthetic basis:

$\forall A, B \in \BB: A \cap B \in \tau$

Hence, by the definition of a subset, it follows that $\OO \subseteq \tau$.

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets, which has been verified above for $\tau$, we have:

$\ds \bigcup \OO \in \tau$

Since set intersection preserves subsets, we have:

$\ds \forall W \in \OO: \exists A, B \in \BB: W = A \cap B \subseteq U \cap V$

By Union is Smallest Superset: General Result, it follows that:

$\ds \bigcup \OO \subseteq U \cap V$

By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:

$\ds \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \OO$

That is, by the definition of a subset:

$\ds U \cap V \subseteq \bigcup \OO$

By definition of set equality:

$\ds U \cap V = \bigcup \OO \in \tau$

$\Box$

### Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

By the definition of a synthetic basis, $\BB$ is a cover for $S$.

By Equivalent Conditions for Cover by Collection of Subsets, it follows that:

$\ds S = \bigcup \BB \in \tau$

$\Box$

All the open set axioms are fulfilled, and the result follows.

$\blacksquare$