Topology Defined by Basis

Theorem
Let $S$ be a set.

Let $\mathcal B$ be a set of subsets of $S$.

Suppose that
 * $(B1): \quad \forall A_1, A_2 \in \mathcal B: \forall x \in A_1 \cap A_2: \exists A \in \mathcal B: x \in A \subseteq A_1 \cap A_2$
 * $(B2): \quad \forall x \in X: \exists A \in \mathcal B: x \in A$
 * $\tau = \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}$

Then:
 * $T = \left({S, \tau}\right)$ is a topological space
 * $\mathcal B$ is a basis of $T$.

Proof
We have to prove Open Set Axioms $(O1)-(O3)$:
 * $(O1): \quad$ The union of an arbitrary subset of $\tau$ is an element of $\tau$.

Let $\mathcal F \subseteq \tau$.

Define by definition of $\tau$ a family $\left({\mathcal G_A}\right)_{A \in \mathcal F}$ such that
 * $\forall A \in \mathcal F: A = \bigcup \mathcal G_A \land \mathcal G_A \subseteq \mathcal B$.

By General Distributivity of Set Union:
 * $\displaystyle \bigcup \bigcup_{A \in \mathcal F} \mathcal G_A = \bigcup_{A \in \mathcal F} \bigcup \mathcal G_A = \bigcup \mathcal F$

By Union of Subsets is Subset/Family of Sets:
 * $\displaystyle \bigcup_{A \in \mathcal F} \mathcal G_A \subseteq \mathcal B$

Thus by definition of $\tau$
 * $\bigcup \mathcal F \in \tau$


 * $(O2): \quad$ The intersection of any two elements of $\tau$ is an element of $\tau$.

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

By definition of $\tau$ there exist subsets $\mathcal G_A$ and $\mathcal G_B$ of $\mathcal B$ such that:
 * $A = \bigcup \mathcal G_A$ and $B = \bigcup \mathcal G_B$ and $\mathcal G_A, \mathcal G_B \subseteq \mathcal B$

Set $\mathcal G_C = \left\{{C \in \mathcal B: C \subseteq A \cap B}\right\}$

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

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

Let $x \in A \cap B$.

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

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

Analogically:
 * $\exists E \in \mathcal G_B: x \in E$

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

Then by $(B1)$:
 * $\exists U \in \mathcal B: x \in U \subseteq D \cap E$

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 $\mathcal G_C$:
 * $U \in \mathcal G_C$

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

This ends the proof of inclusion.

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

By definition of subset:
 * $\mathcal G_C \subseteq \mathcal B$

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


 * $(O3): \quad S$ is an element of $\tau$.

By $(B2)$ and definition of union:
 * $\bigcup \mathcal B = X$

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

It remains to prove that $\mathcal B$ 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 $\mathcal G \subseteq \mathcal B$ such that:
 * $ U = \bigcup \mathcal G$

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

By definition of subset: $A \in \mathcal B$

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

Thus the result by definition of basis.