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. 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. A union of unions of sets from $$\mathcal B$$ is again a union of sets from $$\mathcal B$$.


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

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

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