Union from Synthetic Basis is Topology/Proof 2
Theorem
Let $\BB$ be a synthetic basis on a set $X$.
Let $\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$.
Then $\tau$ is a topology on $X$.
Proof
We use Equivalence of Definitions of Topology Generated by Synthetic Basis.
We proceed to verify the open set axioms for $\tau$ to be a topology on $X$.
Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets
Let $\AA \subseteq \tau$.
It is to be shown that:
- $\ds \bigcup \AA \in \tau$
By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:
- $\ds \forall x \in \bigcup \AA: \exists U \in \AA: \exists B \in \BB: x \in B \subseteq U \subseteq \bigcup \AA$
By the transitivity of $\subseteq$, the result follows.
$\Box$
Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets
Let $U, V \in \tau$.
It is to be shown that:
- $U \cap V \in \tau$
By the definition of a synthetic basis, we have that:
- $\forall A, B \in \BB: A \cap B \in \tau$
Therefore, it follows from Set is Subset of Union: General Result and Set Intersection Preserves Subsets that:
- $\ds \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: \exists C \in \BB: x \in C \subseteq A \cap B \subseteq U \cap V$
By the transitivity of $\subseteq$, the result follows.
$\Box$
Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology
By the definition of a synthetic basis, we have that:
- $\forall x \in X: \exists B \in \BB: x \in B \subseteq X$
$\Box$
All the open set axioms are fulfilled, and the result follows.
$\blacksquare$