The verifiable sets of a basis, plus the whole space, form a sub-basis

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statements, with full knowledge of the source text and of the intended meaning. It was not produced by a blind, independent auditor, so it must not be mistaken for independent testimony; reviewers should compare the Lean code against the source themselves.

Setting (common to all statements of this mission). $\Omega$ is an arbitrary type (it may be empty), and a statement is a subset $s \subseteq \Omega$. $D$ is an arbitrary experimental domain on $\Omega$, i.e. a family $\mathcal D$ of subsets of $\Omega$ such that

• $\Omega \in \mathcal D$ and $\emptyset \in \mathcal D$;
• $s \cap t \in \mathcal D$ whenever $s, t \in \mathcal D$;
• $\bigcup_{n \in \mathbb N} f_n \in \mathcal D$ whenever $f_n \in \mathcal D$ for all $n \in \mathbb N$;
• there is a countable family $B_0 \subseteq \mathcal D$ such that every member of $\mathcal D$ belongs to $\langle B_0 \rangle$, the smallest family containing $B_0$, $\Omega$ and $\emptyset$ and closed under binary intersections and countable unions.

The theoretical domain $\bar{\mathcal D}$ is the smallest family of subsets of $\Omega$ that contains $\mathcal D$ and $\Omega$ and is closed under complement, binary intersection and countable union. A possibility is a set $x \in \bar{\mathcal D}$ with $x \neq \emptyset$ such that for every $t \in \bar{\mathcal D}$ either $x \subseteq t$ or $x \cap t = \emptyset$. $X$ is the type of possibilities (two possibilities are equal exactly when their underlying sets are equal). For $s \subseteq \Omega$ the verifiable set is $U(s) = \{x \in X : x \cap s \neq \emptyset\}$. The space $X$ carries the topology $\mathcal T$ generated by $\{U(s) : s \in \mathcal D\}$, i.e. the coarsest topology on $X$ in which every $U(s)$ with $s \in \mathcal D$ is open.

Statement. For every type $\Omega$, every experimental domain $D$ on $\Omega$ and every basis $B$ of $\mathcal D$ ($B\subseteq\mathcal D$, $\mathcal D\subseteq\langle B\rangle$; $B$ need not be countable), the topology on $X$ generated by the family $\{U(b): b\in B\}\cup\{X\}$ is equal to $\mathcal T$, the topology generated by $\{U(s): s\in\mathcal D\}$.