Intersection Distributes over Union

Theorem
Set intersection is distributive over set union:


 * $$R \cap \left({S \cup T}\right) = \left({R \cap S}\right) \cup \left({R \cap T}\right)$$

General Result
Let $$S$$ and $$T$$ be sets.

Let $$\mathcal P \left({T}\right)$$ be the power set of $$T$$.

Let $$\mathbb T$$ be a subset of $$\mathcal P \left({T}\right)$$.

Then:
 * $$S \cap \bigcup \mathbb T = \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$

Proof
$$ $$ $$

Intersection Subset of Union
Let $$x \in S \cap \bigcup \mathbb T$$.

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

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

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

From $$x \in \bigcup \mathbb T$$ we know that:
 * $$\exists X \in \mathbb T: x \in X$$

and so:
 * $$\exists X \in \mathbb T: x \in S \cap X$$

So by definition of set union:
 * $$x \in \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$

So:
 * $$S \cap \bigcup \mathbb T \subseteq \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$

Union Subset of Intersection
Let $$x \in \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$.

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

So, we have that $$x \in \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$.

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:
 * $$x \in \bigcup \mathbb T$$

So by definition of set intersection, we have that:
 * $$x \in S \cap \bigcup \mathbb T$$

So:
 * $$\bigcup_{X \in \mathbb T} \left({S \cap X}\right) \subseteq S \cap \bigcup \mathbb T$$

So we have that:
 * $$S \cap \bigcup \mathbb T \subseteq \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$

and
 * $$\bigcup_{X \in \mathbb T} \left({S \cap X}\right) \subseteq S \cap \bigcup \mathbb T$$

and so by definition of Equality of Sets:
 * $$S \cap \bigcup \mathbb T = \bigcup_{X \in \mathbb T} \left({S \cap X}\right)$$

Also see

 * Union Distributes over Intersection