Union from Synthetic Basis is Topology/Proof 1

Theorem
Let $\mathcal B$ be a synthetic basis on a set $S$.

Let $\displaystyle \tau = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\}$.

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

$\tau$ is called the topology arising from, or generated by, the basis $\mathcal B$.

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

$\left({O1}\right):$ Union of Open Sets
Let $\mathcal A \subseteq \tau$.

It is to be shown that:
 * $\displaystyle \bigcup \mathcal A \in \tau$

Define:
 * $\displaystyle \mathcal A' = \bigcup_{U \mathop \in \mathcal A} \left\{{B \in \mathcal B: B \subseteq U}\right\}$

By Union is Smallest Superset: Family of Sets, it follows that $\mathcal A' \subseteq \mathcal B$.

Hence, by Equivalent Definitions of Topology Generated by Synthetic Basis and Union Distributes over Union: General Result:
 * $\displaystyle \bigcup \mathcal A = \bigcup_{U \mathop \in \mathcal A} \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} = \bigcup \mathcal A' \in \tau$

$\left({O2}\right):$ Pairwise Intersection of Open Sets
Let $U, V \in \tau$.

It is to be shown that:
 * $U \cap V \in \tau$

Define:
 * $\mathcal O = \left\{{A \cap B: A, B \in \mathcal B, \, A \subseteq U, \, B \subseteq V}\right\}$

By the definition of a synthetic basis:
 * $\forall A, B \in \mathcal B: A \cap B \in \tau$

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

By open set axiom $\left({1}\right)$, which has been verified above for $\tau$, we have:
 * $\displaystyle \bigcup \mathcal O \in \tau$

Since set intersection preserves subsets, we have:
 * $\displaystyle \forall W \in \mathcal O: \exists A, B \in \mathcal B: W = A \cap B \subseteq U \cap V$

By Union is Smallest Superset: General Result, it follows that:
 * $\displaystyle \bigcup \mathcal O \subseteq U \cap V$

By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:
 * $\displaystyle \forall x \in U \cap V: \exists A, B \in \mathcal B: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \mathcal O$

That is, by the definition of a subset:
 * $\displaystyle U \cap V \subseteq \bigcup \mathcal O$

By definition of set equality:
 * $\displaystyle U \cap V = \bigcup \mathcal O \in \tau$

$\left({O3}\right):$ Set Itself
By the definition of a synthetic basis, $\mathcal B$ is a cover for $S$.

By Equivalent Conditions for Cover by Collection of Subsets, it follows that:
 * $\displaystyle S = \bigcup \mathcal B \in \tau$

Also see

 * Topology Generated by Synthetic Basis
 * Generated Topology