Set Difference and Intersection form Partition

Theorem

Let $S$ and $T$ be sets such that:

$S \setminus T \ne \O$

$S \cap T \ne \O$

where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.

Then $S \setminus T$ and $S \cap T$ form a partition of $S$.

Corollary 1

Let $S$ and $T$ be sets such that:

$S \setminus T \ne \O$

$T \setminus S \ne \O$

$S \cap T \ne \O$

Then $S \setminus T$, $T \setminus S$ and $S \cap T$ form a partition of $S \cup T$, the union of $S$ and $T$.

Corollary 2

Let $\O \subsetneqq T \subsetneqq S$.

Then:

$\set {T, \relcomp S T}$

is a partition of $S$.

Proof

From Set Difference and Intersection are Disjoint:

$S \setminus T$ and $S \cap T$ are disjoint.

Next from Set Difference Union Intersection:

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

Thus by definition $S \setminus T$ and $S \cap T$ form a partition of $S$.

$\blacksquare$