Relative Complement Partition

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 $\complement_S \left({T}\right) = S \setminus T$.

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.

$\blacksquare$

Sources

• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 6$: Example $6.4$