Union from Synthetic Basis is Topology/Proof 1

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Theorem

Let $\BB$ be a synthetic basis on a set $X$.

Let $\displaystyle \tau = \set {\bigcup \AA: \AA \subseteq \BB}$.


Then $\tau$ is a topology on $X$.


Proof

We proceed to verify the open set axioms for $\tau$ to be a topology on $S$.


$(\text O 1):$ Union of Open Sets

Let $\AA \subseteq \tau$.

It is to be shown that:

$\displaystyle \bigcup \AA \in \tau$


Define:

$\displaystyle \AA' = \bigcup_{U \mathop \in \AA} \set {B \in \BB: B \subseteq U}$


By Union is Smallest Superset: Family of Sets, it follows that $\AA' \subseteq \BB$.

Hence, by Equivalence of Definitions of Topology Generated by Synthetic Basis and General Distributivity of Set Union:

$\displaystyle \bigcup \AA = \bigcup_{U \mathop \in \AA} \bigcup \set {B \in \BB: B \subseteq U} = \bigcup \AA' \in \tau$

$\Box$


$(\text O 2):$ Pairwise Intersection of Open Sets

Let $U, V \in \tau$.

It is to be shown that:

$U \cap V \in \tau$


Define:

$\OO = \set {A \cap B: A, B \in \BB, \, A \subseteq U, \, B \subseteq V}$

By the definition of a synthetic basis:

$\forall A, B \in \BB: A \cap B \in \tau$

Hence, by the definition of a subset, it follows that $\OO \subseteq \tau$.

By open set axiom $(1)$, which has been verified above for $\tau$, we have:

$\displaystyle \bigcup \OO \in \tau$


Since set intersection preserves subsets, we have:

$\displaystyle \forall W \in \OO: \exists A, B \in \BB: W = A \cap B \subseteq U \cap V$

By Union is Smallest Superset: General Result, it follows that:

$\displaystyle \bigcup \OO \subseteq U \cap V$


By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:

$\displaystyle \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \OO$

That is, by the definition of a subset:

$\displaystyle U \cap V \subseteq \bigcup \OO$


By definition of set equality:

$\displaystyle U \cap V = \bigcup \OO \in \tau$

$\Box$


$(\text O 3):$ Set Itself

By the definition of a synthetic basis, $\BB$ is a cover for $S$.

By Equivalent Conditions for Cover by Collection of Subsets, it follows that:

$\displaystyle S = \bigcup \BB \in \tau$

$\blacksquare$


Also see


Sources