Empty Intersection iff Subset of Complement/Proof 2
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Corollary to Intersection with Complement is Empty iff Subset
- $S \cap T = \O \iff S \subseteq \relcomp {} T$
Proof
From Intersection with Complement is Empty iff Subset
- $S \subseteq T \iff S \cap \relcomp {} T = \O$
Then we have:
\(\ds \) | \(\) | \(\ds S \nsubseteq \relcomp {} T\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \neg \paren {\forall x \in S: x \in \relcomp {} T}\) | Definition of Subset | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \exists x \in S: x \notin \relcomp {} T\) | Denial of Universality | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \exists x \in S: x \in T\) | Definition of Set Complement | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds x \in S \cap T\) | Definition of Set Intersection | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds S \cap T \ne \O\) | Definition of Disjoint Sets |
Thus:
\(\ds \) | \(\) | \(\ds S \cap T = \O\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \forall x \in S: x \in \relcomp {} T\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds S \subseteq \relcomp {} T\) |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $6$