Union Distributes over Intersection/General Result/Proof

Union Subset of Intersection
Let $\displaystyle x \in S \cup \bigcap \mathbb T$.

We need to show that:
 * $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

and then by definition of subset we will have shown that:
 * $\displaystyle S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.

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

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

So there are two cases to consider:

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

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

So:
 * $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

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

Then by definition of set intersection:
 * $\displaystyle \forall X \in \mathbb T: x \in X$

So by definition of set union:
 * $\displaystyle \forall X \in \mathbb T: x \in S \cup X$

So:
 * $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

In both cases we see that:
 * $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

so by Proof by Cases, we have that:
 * $\displaystyle S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

Intersection Subset of Union
Let $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.

We need to show that:
 * $\displaystyle x \in S \cup \bigcap \mathbb T$

and then by definition of subset we will have shown that:
 * $\displaystyle \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$.

So, we have that:
 * $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

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:
 * $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} X$

and so by definition of set union:
 * $\displaystyle x \in S \cup \bigcap \mathbb T$


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

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:
 * $\displaystyle x \in S \cup \bigcap \mathbb T$

In both cases we see that:
 * $\displaystyle x \in S \cup \bigcap \mathbb T$

so by Proof by Cases, we have that:
 * $\displaystyle \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$

So we have that:
 * $S \cup \displaystyle \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

and
 * $\displaystyle \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$

and so by definition of set equality:
 * $S \cup \displaystyle \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$