Topology Defined by Basis

Theorem
Let $S$ be a set.

Let $\BB$ be a set of subsets of $S$.

Suppose that
 * $(\text B1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$
 * $(\text B2): \quad \forall x \in X: \exists A \in \BB: x \in A$
 * $\tau = \set {\bigcup \GG: \GG \subseteq \BB}$

Then:
 * $T = \struct {S, \tau}$ is a topological space
 * $\BB$ is a basis of $T$.

Proof
We have to prove Open Set Axioms:

Let $\FF \subseteq \tau$.

Define by definition of $\tau$ a family $\family {\GG_A}_{A \mathop \in \FF}$ such that
 * $\forall A \in \FF: A = \bigcup \GG_A \land \GG_A \subseteq \BB$.

By General Self-Distributivity of Set Union:
 * $\ds \bigcup \bigcup_{A \in \FF} \GG_A = \bigcup_{A \in \FF} \bigcup \GG_A = \bigcup \FF$

By Union of Subsets is Subset/Family of Sets:
 * $\ds \bigcup_{A \in \FF} \GG_A \subseteq \BB$

Thus by definition of $\tau$
 * $\bigcup \FF \in \tau$

Let $A$ and $B$ be elements of $\tau$.

By definition of $\tau$ there exist subsets $\GG_A$ and $\GG_B$ of $\BB$ such that:
 * $A = \bigcup \GG_A$ and $B = \bigcup \GG_B$ and $\GG_A, \GG_B \subseteq \BB$

Set $\GG_C = \set {C \in \BB: C \subseteq A \cap B}$

By Union of Subsets is Subset:
 * $\bigcup \GG_C \subseteq A \cap B$

We will prove the inclusion: $A \cap B \subseteq \bigcup \GG_C$

Let $x \in A \cap B$.

Then by definition of intersection:
 * $v \in A$ and $x \in B$

Hence by definition of union:
 * $\exists D \in \GG_A: x \in D$

Analogically:
 * $\exists E \in \GG_B: x \in E$

By definition of subset $D, E \in \BB$ and by definition of intersection $x \in D \cap E$.

Then by $(\text B 1)$:
 * $\exists U \in \BB: x \in U \subseteq D \cap E$

By Set is Subset of Union/Set of Sets:
 * $D \subseteq A$ and $E \subseteq B$

Then by Set Intersection Preserves Subsets:
 * $D \cap E \subseteq A \cap B$

Hence by Subset Relation is Transitive:
 * $U \subseteq A \cap B$

Then by definition of $\GG_C$:
 * $U \in \GG_C$

Thus by definition of union:
 * $x \in \bigcup \GG_C$

This ends the proof of inclusion.

Then by definition of set equality:
 * $A \cap B = \bigcup \GG_C$

By definition of subset:
 * $\GG_C \subseteq \BB$

Thus by definition of $\tau$:
 * $A \cap B \in \tau$

By $(\text B 2)$ and definition of union:
 * $\bigcup \BB = X$

Because $\BB \subseteq \BB$ by definition of $\tau$:
 * $S \in \tau$

It remains to prove that $\BB$ is a basis of $T$.

Let $U$ be an open set of $S$.

Let $x$ be a point of $S$ such that $x \in U$

By definition of $\tau$ there exists $\GG \subseteq \BB$ such that:
 * $ U = \bigcup \GG$

By definition of union:
 * $\exists A \in \GG: x \in A$

By definition of subset: $A \in \BB$

Thus by Set is Subset of Union:
 * $x \in A \subseteq U$

Thus the result by definition of basis.