Union from Synthetic Basis is Topology/Proof 1

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

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

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

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

Proof
Note that $U \in \vartheta$ iff:


 * $U = \displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

Let us refer to the above statement by the symbol $\left({*}\right)$.

Proof of $\left({*}\right)$
Trivially, the condition stated is sufficient.

We show that the condition stated is necessary.

Suppose that $U \in \vartheta$. Then, by definition:
 * $\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$

By Union Smallest: General Result, we have::
 * $\displaystyle \forall B \in \mathcal A: B \subseteq U$

By the definition of a subset, it follows that:
 * $\displaystyle \mathcal A \subseteq \left\{{B \in \mathcal B: B \subseteq U}\right\}$

Therefore:
 * $\displaystyle U = \bigcup \mathcal A \subseteq \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

By Union Smallest: General Result, we have:
 * $\displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq U$

Hence, by Equality of Sets, it follows that:
 * $\displaystyle U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

as claimed.

We now proceed to verify the three conditions for $\vartheta$ to be a topology on $X$.

Proof of $\left({1}\right)$
Let $\mathcal A \subseteq \vartheta$.

It is to be shown that:
 * $\displaystyle \bigcup \mathcal A \in \vartheta$

Define:
 * $\displaystyle \mathcal A' = \bigcup_{U \mathop \in \mathcal A} \left\{{B \in \mathcal B: B \subseteq U}\right\}$

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

Hence, by $\left({*}\right)$ and Union of Unions: General Result:
 * $\displaystyle \bigcup \mathcal A = \bigcup_{U \mathop \in \mathcal A} \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} = \bigcup \mathcal A' \in \vartheta$

Proof of $\left({2}\right)$
Let $U, V \in \vartheta$.

It is to be shown that:
 * $U \cap V \in \vartheta$

Define:
 * $\Theta = \left\{{A \cap B: A, B \in \mathcal B, \, A \subseteq U, \, B \subseteq V}\right\}$

By the definition of a synthetic basis:
 * $\forall A, B \in \mathcal B: A \cap B \in \vartheta$

Hence, by the definition of a subset, it follows that $\Theta \subseteq \vartheta$.

By axiom $\left({1}\right)$ for a topology, which has been verified above, we have:
 * $\displaystyle \bigcup \Theta \in \vartheta$

Since intersection preserves subsets, we have:
 * $\displaystyle \forall W \in \Theta: \exists A, B \in \mathcal B: W = A \cap B \subseteq U \cap V$

By Union Smallest: General Result, it follows that:
 * $\displaystyle \bigcup \Theta \subseteq U \cap V$

By $\left({*}\right)$ and Subset of Union: General Result, we have:
 * $\displaystyle \forall x \in U \cap V: \exists A, B \in \mathcal B: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \Theta$

Therefore, by the definition of a subset, it follows that:
 * $\displaystyle U \cap V \subseteq \bigcup \Theta$

Hence, by Equality of Sets, we have:
 * $\displaystyle U \cap V = \bigcup \Theta \in \vartheta$

Proof of $\left({3}\right)$
By the definition of a synthetic basis, $X \in \vartheta$.

Also see

 * Generated Topology