Union from Synthetic Basis is Topology/Proof 1

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

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

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

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

Proof
Note that $U \in \vartheta$ iff:


 * $U = \displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

Let us now verify the three conditions for $\vartheta$ to be a topology on $X$.

Proof of $\left({1}\right)$
Let $\mathcal A \subseteq \vartheta$.

Then:
 * $\displaystyle \bigcup \mathcal{A} = \bigcup_{U \mathop \in \mathcal{A}} U = \bigcup_{U \mathop \in \mathcal{A}} \bigcup \left\{{B \in \mathcal{B}: B \subseteq U}\right\} \in \vartheta$

Proof of $\left({2}\right)$
Suppose that $U, V \in \vartheta$.

Define:
 * $\mathcal B_U = \left\{{B \in \mathcal B: B \subseteq U}\right\}$
 * $\mathcal B_V = \left\{{B \in \mathcal B: B \subseteq V}\right\}$

By the distributivity of intersection over union, we have:
 * $\displaystyle U \cap V = \bigcup_{A \mathop \in \mathcal B_U} A \cap \bigcup_{B \mathop \in \mathcal B_V} B = \bigcup_{A \mathop \in \mathcal B_U} \bigcup_{B \mathop \in \mathcal B_V} \left({A \cap B}\right)$

By the definition of a synthetic basis, $\forall A, B \in \mathcal B: A \cap B \in \vartheta$.

Hence $U \cap V \in \vartheta$ by condition $\left({1}\right)$ for a topology, which is proven above.

Proof of $\left({3}\right)$
By the definition of a synthetic basis, $X \in \vartheta$.

Also see

 * Generated Topology