Equivalence of Definitions of Topology Generated by Synthetic Basis

Theorem
Let $S$ be a set.

Let $\mathcal B$ be a synthetic basis on $S$.

Let $\tau$ be the topology on $S$ generated by the synthetic basis $\mathcal B$:
 * $\displaystyle \tau = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\}$

Then:


 * $\displaystyle ({1}): \quad \forall U \subseteq S: U \in \tau \iff U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$


 * $({2}): \quad \forall U \subseteq S: U \in \tau \iff \forall x \in U: \exists B \in \mathcal B: x \in B \subseteq U$

Proof of $({1})$
Trivially, the "$\impliedby$" implication holds, as $\left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq \mathcal B$.

We now show that the "$\implies$" implication holds.

Suppose that $U \in \tau$. Then, by definition:
 * $\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$

By Union Smallest: General Result, we have::
 * $\displaystyle \forall B \in \mathcal A: B \subseteq U$

By the definition of a subset, it follows that:
 * $\displaystyle \mathcal A \subseteq \left\{{B \in \mathcal B: B \subseteq U}\right\}$

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

By Union Smallest: General Result, we have:
 * $\displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq U$

Hence, by Equality of Sets, it follows that:
 * $\displaystyle U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

as claimed.