# Relative Complement inverts Subsets

## Theorem

Let $S$ be a set.

Let $A \subseteq S, B \subseteq S$ be subsets of $S$.

Then:

$A \subseteq B \iff \relcomp S B \subseteq \relcomp S A$

where $\complement_S$ denotes the complement relative to $S$.

## Proof

 $\displaystyle A$ $\subseteq$ $\displaystyle B$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle A \cap B$ $=$ $\displaystyle A$ Intersection with Subset is Subset‎ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \relcomp S {A \cap B}$ $=$ $\displaystyle \relcomp S A$ Relative Complement of Relative Complement $\displaystyle \leadstoandfrom \ \$ $\displaystyle \relcomp S A \cup \relcomp S B$ $=$ $\displaystyle \relcomp S A$ De Morgan's Laws: Complement of Intersection $\displaystyle \leadstoandfrom \ \$ $\displaystyle \relcomp S B$ $\subseteq$ $\displaystyle \relcomp S A$ Union with Superset is Superset

$\blacksquare$

## Also known as

This result can be referred to by saying that the subset operation is inclusion-inverting.