# 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 1

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

$\blacksquare$

## Proof 2

### Sufficient Condition

Let $A \subseteq B$.

Then:

 $\ds x$ $\in$ $\ds \relcomp S B$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds S \setminus B$ Definition of Relative Complement $\ds \leadsto \ \$ $\ds x$ $\notin$ $\ds B$ Definition of Set Difference $\ds \leadsto \ \$ $\ds x$ $\notin$ $\ds A$ Definition of Subset $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds S \setminus A$ Definition of Set Difference $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \relcomp S A$ Definition of Relative Complement $\ds \leadsto \ \$ $\ds \relcomp S B$ $\subseteq$ $\ds \relcomp S A$ Definition of Subset

$\Box$

### Necessary Condition

 $\ds \relcomp S B$ $\subseteq$ $\ds \relcomp S A$ $\ds \leadsto \ \$ $\ds \relcomp S {\relcomp S A}$ $\subseteq$ $\ds \relcomp S {\relcomp S B}$ from Sufficient Condition $\ds \leadsto \ \$ $\ds A$ $\subseteq$ $\ds B$ Relative Complement of Relative Complement

$\blacksquare$

## Also known as

This result can be referred to by saying that:

the subset operation is inclusion-inverting
the relative complement operation, considered as a mapping on the ordered structure $\struct {S, \subseteq}$, is decreasing