Union from Synthetic Basis is Topology/Proof 1

Theorem
Let $\mathcal B$ be a synthetic basis for a set $A$.

Let $\vartheta = \left\{{U \in \mathcal P \left({A}\right): U \mbox{ is a union of sets from } \mathcal B}\right\}$.

Then $\vartheta$ is a topology for $A$.

$\vartheta$ is called the topology arising from the basis $\mathcal B$.

Proof

 * $(1): \quad$ From the definition of synthetic basis, $A \in \vartheta$. It is understood that we are allowed to take the union of no sets from $\mathcal B$, so $\varnothing \in \vartheta$.


 * $(2): \quad$ A union of unions of sets from $\mathcal B$ is again a union of sets from $\mathcal B$.


 * $(3): \quad$ Let $\displaystyle U = \bigcup_{i \in I} B_{i1}, V = \bigcup_{j \in J} B_{j2}$ for some indexing sets $I, J$ where all the $B_{i1} \in \mathcal B, B_{j2} \in \mathcal B$.

Then:
 * $\displaystyle U \cap V = \bigcup_{\left({i, j}\right) \in I \times J \left({B_{i1} \cap B_{j2}}\right)}$

which is a union of sets from $\mathcal B$.