Complement Relative to Subset is Subset of Complement Relative to Superset
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Theorem
Let $A, B, C$ be sets such that $A \subseteq B \subseteq C$.
Then:
- $\relcomp B A \subseteq \relcomp C A$
Proof
| \(\ds x\) | \(\in\) | \(\ds \relcomp B A\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds B\) | Definition of Relative Complement | ||||||||||
| \(\, \ds \land \, \) | \(\ds x\) | \(\notin\) | \(\ds A\) | Definition of Relative Complement | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds C\) | Definition of Subset: $B \subseteq C$ | ||||||||||
| \(\, \ds \land \, \) | \(\ds x\) | \(\notin\) | \(\ds A\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \relcomp C A\) | Definition of Relative Complement | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \relcomp B A\) | \(\subseteq\) | \(\ds \relcomp C A\) | Definition of Subset |
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $2 \ \text{(a)}$