Set Difference and Intersection form Partition/Corollary 2

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

Then:
 * $\left\{{T, \complement_S \left({T}\right)}\right\}$

is a partition of $S$.

Proof
First we note that:


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

from the definition of proper subset.

From the definition of relative complement, we have $\complement_S \left({T}\right) = S \setminus T$.

It follows that $T \ne S \iff \complement_S \left({T}\right) \ne \varnothing$ from Set Difference with Self is Empty Set.

From Intersection with Relative Complement, $T \cap \complement_S \left({T}\right) = \varnothing$, that is, $T$ and $\complement_S \left({T}\right)$ are disjoint.

From Union with Relative Complement, $T \cup \complement_S \left({T}\right) = S$, that is, the union of $T$ and $\complement_S \left({T}\right)$ forms the whole set $S$.

Thus all the conditions for a partition are satisfied.