Set Difference and Intersection form Partition

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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:

$S \setminus T \ne \varnothing$
$T \setminus S \ne \varnothing$
$S \cap T \ne \varnothing$

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 Intersection with Second Set is Empty Set:

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

and hence immediately from Intersection with Empty Set:

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

So $S \setminus T$ 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 $S$.

$\blacksquare$