Union from Synthetic Basis is Topology/Proof 2

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

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

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

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

Proof
We use Equivalent Definitions of Topology Generated by Synthetic Basis.

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

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

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

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

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

$({O2}):$ 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:
 * $\forall A, B \in \mathcal B: A \cap B \in \tau$

Therefore, it follows from Subset of Union: General Result and Set Intersection Preserves Subsets that:
 * $\displaystyle \forall x \in U \cap V: \exists A, B \in \mathcal B: A \subseteq U, \, B \subseteq V: \exists C \in \mathcal B: x \in C \subseteq A \cap B \subseteq U \cap V$

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

$({O3}):$ Set Itself
By the definition of a synthetic basis, we have that:
 * $\forall x \in X: \exists B \in \mathcal B: x \in B \subseteq X$

Hence the result.

Also see

 * Topology Generated by Synthetic Basis
 * Generated Topology