Set Complement inverts Subsets/Corollary
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Corollary to Set Complement inverts Subsets
Let $S$ and $T$ be sets.
Then:
- $S \subseteq \map \complement T \iff T \subseteq \map \complement S$
where:
- $S \subseteq \map \complement T$ denotes that $S$ is a subset of the set complement of $T$.
Proof
\(\ds S\) | \(\subseteq\) | \(\ds \map \complement T\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement {\map \complement T}\) | \(\subseteq\) | \(\ds \map \complement S\) | Set Complement inverts Subsets | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds T\) | \(\subseteq\) | \(\ds \map \complement S\) | Complement of Complement |
$\blacksquare$
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 $1 \ \text{(f)}$