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 $$\mathcal{B}$$.