# Set Difference Union Intersection

## Theorem

$S = \paren {S \setminus T} \cup \paren {S \cap T}$

## Proof 1

 $\ds \paren {S \setminus T} \cup \paren {S \cap T}$ $=$ $\ds \paren {\paren {S \setminus T} \cup S} \cap \paren {\paren {S \setminus T} \cup T}$ Union Distributes over Intersection $\ds$ $=$ $\ds S \cap \paren {\paren {S \setminus T} \cup T}$ Set Difference Union First Set is First Set $\ds$ $=$ $\ds S \cap \paren {S \cup T}$ Set Difference Union Second Set is Union $\ds$ $=$ $\ds S$ Intersection Absorbs Union

$\blacksquare$

## Proof 2

 $\ds \paren {S \setminus T} \cup \paren {S \cap T}$ $=$ $\ds S \setminus \paren {T \setminus T}$ Set Difference with Set Difference is Union of Set Difference with Intersection $\ds$ $=$ $\ds S \setminus \O$ Set Difference with Self is Empty Set $\ds$ $=$ $\ds S$ Set Difference with Empty Set is Self

$\blacksquare$

## Proof 3

$S \setminus T \subseteq S$
$S \cap T \subseteq S$

Hence from Union is Smallest Superset:

$\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$

Let $s \in S$.

Either:

$s \in T$, in which case $s \in S \cap T$ by definition of set intersection

or

$s \notin T$, in which case $s \in S \setminus T$ by definition of set difference.

That is, by definition of set union:

$s \in \paren {S \setminus T} \cup \paren {S \cap T}$

and so by definition of subset:

$S \subseteq \paren {S \setminus T} \cup \paren {S \cap T}$

Hence the result by definition of set equality.

$\blacksquare$