# Intersection Distributes over Union/General Result

## Theorem

Let $S$ and $T$ be sets.

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T$ be a subset of $\powerset T$.

Then:

$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

## Proof

### Intersection Subset of Union

Let $\ds x \in S \cap \bigcup \mathbb T$.

We need to show that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and then by definition of subset we will have shown that $\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.

So, we have that $\ds x \in S \cap \bigcup \mathbb T$.

By definition of set intersection, $x \in S$ and $\ds x \in \bigcup \mathbb T$.

From $\ds x \in \bigcup \mathbb T$ we know that:

$\exists X \in \mathbb T: x \in X$

and so:

$\ds \exists X \in \mathbb T: x \in S \cap X$

So by definition of set union:

$\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

So:

$\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

$\Box$

### Union Subset of Intersection

Let $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.

We need to show that $\ds x \in S \cap \bigcup \mathbb T$ and then by definition of subset we will have shown that $\ds \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$.

So, we have that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.

By definition of set union:

$\exists X \in \mathbb T: x \in S \cap X$

By definition of set intersection, we have that $x \in S$ and $x \in X$.

By definition of set union:

$\ds x \in \bigcup \mathbb T$

So by definition of set intersection, we have that:

$\ds x \in S \cap \bigcup \mathbb T$

So:

$\ds \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$

$\Box$

So we have that:

$\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

and:

$\ds \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$

and so by definition of set equality:

$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

$\blacksquare$