Relative Complement inverts Subsets/Proof 1

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

\(\ds A\)

\(\subseteq\)

\(\ds B\)

\(\ds \leadstoandfrom \ \ \)

\(\ds A \cap B\)

\(=\)

\(\ds A\)

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

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(k)}$