Relative Complement inverts Subsets/Proof 2
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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
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 |
$\blacksquare$