# Equivalence of Definitions of Synthetic Basis

## Theorem

Let $S$ be a set.

The following definitions of the concept of Synthetic Basis are equivalent:

### Definition 1

A synthetic basis on $S$ is a subset $\mathcal B \subseteq \mathcal P \left({S}\right)$ of the power set of $S$ such that:

 $(B1)$ $:$ $\mathcal B$ is a cover for $S$ $(B2)$ $:$ $\displaystyle \forall U, V \in \mathcal B:$ $\exists \mathcal A \subseteq \mathcal B: U \cap V = \bigcup \mathcal A$

That is, the intersection of any pair of elements of $\mathcal B$ is a union of sets of $\mathcal B$.

### Definition 2

A synthetic basis on $S$ is a subset $\mathcal B \subseteq \mathcal P \left({S}\right)$ of the power set of $S$ such that:

$\mathcal B$ is a cover for $S$
$\forall U, V \in \mathcal B: \forall x \in U \cap V: \exists W \in \mathcal B: x \in W \subseteq U \cap V$

## Proof

### 1 implies 2

Let $U, V \in \mathcal B$.

Let $x \in U \cap V$.

$\displaystyle \exists \mathcal A \subseteq \mathcal B: U \cap V = \bigcup \mathcal A$

By definition of union, $\exists W \in\mathcal A : x \in W$.

By Set is Subset of Union: General Result, $W \subset U \cap V$.

Therefore:

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

$\Box$

### 2 implies 1

Let $U, V \in \mathcal B$.

Define the set:

$\displaystyle \mathcal A = \left\{{W \in \mathcal B: W \subseteq U \cap V}\right\} \subseteq \mathcal B$
$\displaystyle \bigcup \mathcal A \subseteq U \cap V$
$\displaystyle \forall x \in U \cap V: \exists W \in \mathcal A: x \in W$

Thus $\displaystyle U \cap V \subseteq \bigcup \mathcal A$

By definition of set equality:

$\displaystyle U \cap V = \bigcup \mathcal A$

$\blacksquare$