Set Difference and Intersection form Partition/Corollary 2

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

Then:
 * $\set {T, \relcomp S T}$

is a partition of $S$.

Proof
First we note that:


 * $\O \subsetneqq T \implies T \ne \O$
 * $T \subsetneqq S \implies T \ne S$

from the definition of proper subset.

From the definition of relative complement, we have:
 * $\relcomp S T = S \setminus T$.

It follows from Set Difference with Self is Empty Set that:
 * $T \ne S \iff \relcomp S T \ne \O$

From Intersection with Relative Complement is Empty:
 * $T \cap \relcomp S T = \O$

That is, $T$ and $\relcomp S T$ are disjoint.

From Union with Relative Complement:
 * $T \cup \relcomp S T = S$

That is, the union of $T$ and $\relcomp S T$ forms the whole set $S$.

Thus all the conditions for a partition are satisfied.