Set Difference and Intersection form Partition

Theorem

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

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

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

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 $T$.

Proof

We have from Set Difference Intersection Second Set is Empty Set that:

$\left({S \setminus T}\right) \cap T = \varnothing$

and hence immediately from Intersection with Empty Set:

$\left({S \setminus T}\right) \cap \left({S \cap T}\right) = \varnothing$

So we have that $S \setminus T$ and $S \cap T$ are disjoint.

Next we note that from Set Difference Union Intersection we have that:

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

That is all we need to assert that $S \setminus T$ and $S \cap T$ form a partition of $S$.

$\blacksquare$

Proof of Corollary

From above, we have that:

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

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

Finally we note from Set Difference Disjoint with Reverse that $\left({S \setminus T}\right) \cap \left({T \setminus S}\right) = \varnothing$.

So:

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

and the result follows.

$\blacksquare$