Subset Relation is Transitive/Proof 1
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Theorem
The subset relation is transitive:
- $\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$
Proof
| \(\ds \) | \(\) | \(\ds \paren {R \subseteq S} \land \paren {S \subseteq T}\) | ||||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \paren {x \in R \implies x \in S} \land \paren {x \in S \implies x \in T}\) | Definition of Subset | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \paren {x \in R \implies x \in T}\) | Hypothetical Syllogism | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds R \subseteq T\) | Definition of Subset |
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Theorem $1.1 \ \text{(b)}$
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering