Relative Complement inverts Subsets

From ProofWiki
Jump to navigation Jump to search

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.


Sources