Topology Defined by Basis

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Let $S$ be a set.

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

Suppose that

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


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


We have to prove Open Set Axioms $(\text O1)-(\text O3)$:

$(\text O1): \quad$ The union of an arbitrary subset of $\tau$ is an element of $\tau$.

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 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 $\tau$

$\bigcup \FF \in \tau$


$(\text O2): \quad$ 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 $\GG_A$ 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:

$\exists D \in \GG_A: x \in D$


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

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

Then by $(\text B1)$:

$\exists U \in \BB: x \in U \subseteq D \cap E$

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 $\tau$:

$A \cap B \in \tau$


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

By $(\text B2)$ 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.