# Union Distributes over Intersection/General Result/Proof

## 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:

$\ds S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

## Proof

### Union Subset of Intersection

Let $\ds x \in S \cup \bigcap \mathbb T$.

We need to show that:

$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

and then by definition of subset we will have shown that:

$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.

So, we have that $\ds x \in S \cup \bigcap \mathbb T$.

By definition of set union, $x \in S$ or $\ds x \in \bigcap \mathbb T$.

So there are two cases to consider:

$(1): \quad$ Suppose $x \in S$.

Then by definition of set union, $\forall X \in \mathbb T: x \in S \cup X$.

So:

$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

$(2): \quad$ Suppose $\ds x \in \bigcap \mathbb T$.

Then by definition of set intersection:

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

So by definition of set union:

$\forall X \in \mathbb T: x \in S \cup X$

So:

$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

In both cases we see that:

$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

so by Proof by Cases, we have that:

$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

$\Box$

### Intersection Subset of Union

Let $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.

We need to show that:

$\ds x \in S \cup \bigcap \mathbb T$

and then by definition of subset we will have shown that:

$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$.

So, we have that:

$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

By definition of set intersection:

$(A): \quad \forall X \in \mathbb T: x \in S \cup X$

There are two cases to consider:

$(1): \quad \forall X \in \mathbb T: x \in X$

Then by definition of set intersection:

$\ds x \in \bigcap_{X \mathop \in \mathbb T} X$

and so by definition of set union:

$\ds x \in S \cup \bigcap \mathbb T$

$(2): \quad \exists X \in \mathbb T: x \notin X$

From $(A)$ we have that $x \in S \cup X$.

But as $x \notin X$ it follows that $x \in S$.

Then by definition of set union:

$\ds x \in S \cup \bigcap \mathbb T$

In both cases we see that:

$\ds x \in S \cup \bigcap \mathbb T$

so by Proof by Cases, we have that:

$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$

$\Box$

So we have that:

$S \cup \ds \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

and

$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$

and so by definition of set equality:

$S \cup \ds \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

$\blacksquare$