Union from Synthetic Basis is Topology

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Theorem

Let $\mathcal B$ be a synthetic basis on a set $X$.

Let $\displaystyle \tau = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\}$.


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


$\tau$ is called the topology arising from, or generated by, the basis $\mathcal B$.


Proof 1

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


$(O1):$ Union of Open Sets

Let $\mathcal A \subseteq \tau$.

It is to be shown that:

$\displaystyle \bigcup \mathcal A \in \tau$


Define:

$\displaystyle \mathcal A' = \bigcup_{U \mathop \in \mathcal A} \set {B \in \mathcal B: B \subseteq U}$


By Union is Smallest Superset: Family of Sets, it follows that $\mathcal A' \subseteq \mathcal B$.

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

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

$\Box$


$(O2):$ Pairwise Intersection of Open Sets

Let $U, V \in \tau$.

It is to be shown that:

$U \cap V \in \tau$


Define:

$\mathcal O = \set {A \cap B: A, B \in \mathcal B, \, A \subseteq U, \, B \subseteq V}$

By the definition of a synthetic basis:

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

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

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

$\displaystyle \bigcup \mathcal O \in \tau$


Since set intersection preserves subsets, we have:

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

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

$\displaystyle \bigcup \mathcal O \subseteq U \cap V$


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

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

That is, by the definition of a subset:

$\displaystyle U \cap V \subseteq \bigcup \mathcal O$


By definition of set equality:

$\displaystyle U \cap V = \bigcup \mathcal O \in \tau$

$\Box$


$(O3):$ Set Itself

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

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

$\displaystyle S = \bigcup \mathcal B \in \tau$

$\blacksquare$


Proof 2

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 $X$.


$(O1):$ Union of Open Sets

Let $\mathcal A \subseteq \tau$.

It is to be shown that:

$\displaystyle \bigcup \mathcal A \in \tau$


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

$\displaystyle \forall x \in \bigcup \mathcal A: \exists U \in \mathcal A: \exists B \in \mathcal B: x \in B \subseteq U \subseteq \bigcup \mathcal A$

By the transitivity of $\subseteq$, the result follows.

$\Box$


$(O2):$ 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:

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

Therefore, it follows from Set is Subset of Union: General Result and Set Intersection Preserves Subsets that:

$\displaystyle \forall x \in U \cap V: \exists A, B \in \mathcal B: A \subseteq U, \, B \subseteq V: \exists C \in \mathcal B: x \in C \subseteq A \cap B \subseteq U \cap V$

By the transitivity of $\subseteq$, the result follows.

$\Box$


$(O3):$ Set Itself

By the definition of a synthetic basis, we have that:

$\forall x \in X: \exists B \in \mathcal B: x \in B \subseteq X$

Hence the result.

$\blacksquare$


Also see


Sources