# Equivalent Definitions of Synthetic Basis

## Theorem

Let $X$ be a set.

Then for a subset $\BB \subseteq \powerset X$ of the power set of $X$, the following are equivalent:

$(1): \quad$ $\displaystyle \forall A, B \in \BB: \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$.
$(2): \quad$ $\displaystyle \forall A, B \in \BB: \forall x \in A \cap B: \exists W \in \BB: x \in W \subseteq A \cap B$.

## Proof

### $(1)$ implies $(2)$

Suppose that $A, B \in \BB$.

$\displaystyle \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$

By Set is Subset of Union: General Result, we have:

$\displaystyle \forall W \in \AA: W \subseteq \bigcup \AA = A \cap B$

Therefore:

$\displaystyle \forall x \in A \cap B: \exists W \in \AA \subseteq \BB: x \in W \subseteq A \cap B$

$\Box$

### $(2)$ implies $(1)$

Suppose that $A, B \in \BB$.

Define the set:

$\displaystyle \AA = \set {W \in \BB: W \subseteq A \cap B} \subseteq \BB$

By Union is Smallest Superset: General Result, we have:

$\displaystyle \bigcup \AA \subseteq A \cap B$

By hypothesis, we have:

$\displaystyle \forall x \in A \cap B: \exists W \in \AA: x \in W$

Hence, by the definition of a subset and the definition of set union:

$\displaystyle A \cap B \subseteq \bigcup \AA$

Therefore, by definition of set equality:

$\displaystyle A \cap B = \bigcup \AA$

$\blacksquare$