Union from Synthetic Basis is Topology/Proof 2

Proof
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 $X$.

Let $\AA \subseteq \tau$.

It is to be shown that:
 * $\ds \bigcup \AA \in \tau$

By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:
 * $\ds \forall x \in \bigcup \AA: \exists U \in \AA: \exists B \in \BB: x \in B \subseteq U \subseteq \bigcup \AA$

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

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

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

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.

Also see

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