# Union from Synthetic Basis is Topology

## Contents

## Theorem

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

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

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

$\tau$ is called the **topology arising from**, or **generated by**, **the basis $\BB$**.

## Proof 1

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

### $(\text O 1):$ Union of Open Sets

Let $\AA \subseteq \tau$.

It is to be shown that:

- $\displaystyle \bigcup \AA \in \tau$

Define:

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

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

Hence, by Equivalence of Definitions of Topology Generated by Synthetic Basis and General Distributivity of Set Union:

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

$\Box$

### $(\text O 2):$ 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 $(1)$, which has been verified above for $\tau$, we have:

- $\displaystyle \bigcup \OO \in \tau$

Since set intersection preserves subsets, we have:

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

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

- $\displaystyle \bigcup \OO \subseteq U \cap V$

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

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

That is, by the definition of a subset:

- $\displaystyle U \cap V \subseteq \bigcup \OO$

By definition of set equality:

- $\displaystyle U \cap V = \bigcup \OO \in \tau$

$\Box$

### $(\text O 3):$ Set Itself

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:

- $\displaystyle S = \bigcup \BB \in \tau$

$\blacksquare$

## Proof 2

We use Equivalence of Definitions of Topology Generated by Synthetic Basis.

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

### $(\text O 1):$ Union of Open Sets

Let $\AA \subseteq \tau$.

It is to be shown that:

- $\displaystyle \bigcup \AA \in \tau$

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

- $\displaystyle \forall x \in \bigcup \AA: \exists U \in \AA: \exists B \in \BB: x \in B \subseteq U \subseteq \bigcup \AA$

By the transitivity of $\subseteq$, the result follows.

$\Box$

### $(\text O 2):$ Pairwise Intersection of Open Sets

Let $U, V \in \tau$.

It is to be shown that:

- $U \cap V \in \tau$

By the definition of a synthetic basis, we have that:

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

Therefore, it follows from Set is Subset of Union: General Result and Set Intersection Preserves Subsets that:

- $\displaystyle \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: \exists C \in \BB: x \in C \subseteq A \cap B \subseteq U \cap V$

By the transitivity of $\subseteq$, the result follows.

$\Box$

### $(\text O 3):$ Set Itself

By the definition of a synthetic basis, we have that:

- $\forall x \in X: \exists B \in \BB: x \in B \subseteq X$

Hence the result.

$\blacksquare$

## Also see

- Definition:Topology Generated by Synthetic Basis
- Definition:Topology Generated by Synthetic Sub-Basis

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction