Union from Synthetic Basis is Topology/Proof 1

Theorem
Let $\mathcal B$ be a synthetic basis for a set $A$.

Let $\vartheta = \left\{{U \in \mathcal P \left({A}\right): U \mbox{ is a union of sets from } \mathcal B}\right\}$.

Then $\vartheta$ is a topology for $A$.

$\vartheta$ is called the topology arising from the basis $\mathcal B$.

Proof
Note that $U \subseteq A$ is in $\vartheta$ iff:


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

Let us now verify the three conditions for $\vartheta$ to be a topology on $A$.

Proof of $(1)$
From the definition of synthetic basis, $A \in \vartheta$.

It is understood that we are allowed to take the union of no sets from $\mathcal B$, so $\varnothing \in \vartheta$ by Union of Empty Set.

Proof of $(2)$
Suppose that $U_i \in \vartheta$ for all $i \in I$.

It is to be shown that:


 * $\displaystyle \bigcup_{i \mathop \in I} U_i \in \vartheta$

So let us consider the following collection of sets from $\mathcal B$:


 * $\mathbb B := \left\{{B \in \mathcal B: \exists i \in I: B \subseteq U_i}\right\}$

For each $B \in \mathbb B$, we have (for some $i \in I$):


 * $B \subseteq U_i \subseteq \displaystyle \bigcup_{i \mathop \in I} U_i$

whence by Union Smallest: General Result:


 * $\displaystyle \bigcup \mathbb B \subseteq \bigcup_{i \mathop \in I} U_i$

For the converse inclusion, let $x \in \displaystyle \bigcup_{i \mathop \in I} U_i$.

By definition of set union, there is some $i \in I$ such that $x \in U_i$.

That is, since $U_i \in \vartheta$, by the above observation:


 * $x \in \displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U_i}\right\}$

All the sets $B$ in this union clearly are also in $\mathbb B$; therefore we conclude:


 * $x \in \displaystyle \bigcup \mathbb B$

By definition of subset, this means:


 * $\displaystyle \bigcup_{i \mathop \in I} U_i \subseteq \bigcup \mathbb B$

and by Equality of Sets, we deduce:


 * $\displaystyle \bigcup_{i \mathop \in I} U_i = \bigcup \mathbb B$

Thus $\displaystyle \bigcup_{i \mathop \in I} U_i$ is in $\vartheta$.

Proof of $(3)$
For $U, V \in \vartheta$, it is to be shown that:


 * $U \cap V \in \vartheta$

To achieve this, define a collection $\mathbb B$ of sets from $\mathcal B$ by:


 * $\mathbb B := \left\{{B \in \mathcal B }\ \middle\vert \ { \exists C, D \in \mathcal B: C \subseteq U, D \subseteq V, B \subseteq C \cap D }\right\}$

For any $C, D$ with $C \subseteq U, D \subseteq V$ we have by Intersection Preserves Subsets that:


 * $C \cap D \subseteq U \cap V$

By Subset Relation is Transitive, it follows that for all $B \in \mathbb B$:


 * $B \subseteq U \cap V$

From Union Smallest: General Result, therefore:


 * $\displaystyle \bigcup \mathbb B \subseteq U \cap V$

For the converse inclusion, let $x \in U \cap V$.

Thus $x \in U$ and $x \in V$; as $U, V \in \vartheta$, this means:


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

Thus there are $C, D \in \mathcal B$ with $C \subseteq U$, $D \subseteq V$ and:


 * $x \in C \cap D$

Since $\mathcal B$ is a synthetic basis for $A$, it follows that $C \cap D \in \vartheta$, so that:


 * $C \cap D = \displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq C \cap D}\right\}$

Every $B$ in the above collection is seen to satisfy the condition to be in $\mathbb B$ as well, i.e. $B \in \mathbb B$.

Therefore we deduce:


 * $x \in \displaystyle \bigcup \mathbb B$

By definition of subset, this means:


 * $U \cap V \subseteq \displaystyle \bigcup \mathbb B$

and by Equality of Sets, we conclude:


 * $U \cap V = \displaystyle \bigcup \mathbb B$

establishing $U \cap V \in \vartheta$.

Having verified all three conditions, we conclude $\vartheta$ is a topology on $A$.

Also see

 * Generated Topology