Complement Union with Superset is Universe/Corollary
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Corollary to Complement Union with Superset is Universe
- $S \cup T = \mathbb U \iff \map \complement S \subseteq T$
where:
- $S \subseteq T$ denotes that $S$ is a subset of $T$
- $S \cup T$ denotes the union of $S$ and $T$
- $\complement$ denotes set complement
- $\mathbb U$ denotes the universal set.
Proof
Let $X = \map \complement S$.
Then:
\(\ds X \subseteq T\) | \(\iff\) | \(\ds \map \complement X \cup T = \mathbb U\) | Complement Union with Superset is Universe | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \complement S \subseteq T\) | \(\iff\) | \(\ds \map \complement {\map \complement S} \cup T = \mathbb U\) | substituting $X = \map \complement S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \complement S \subseteq T\) | \(\iff\) | \(\ds S \cup T = \mathbb U\) | Complement of Complement |
$\blacksquare$
Also see
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(a)}$
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(d)}$