Set Difference and Intersection form Partition/Corollary 2

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

Then $$\left\{{T, \complement_S \left({T}\right)}\right\}$$ is a partition of $$S$$.

Proof
First we note that:


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

from the definition of proper subset.

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

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

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.