Subset Relation is Transitive/Proof 1

Theorem

The subset relation is transitive:

$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$

$\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 $

$\leadsto$

$\ds \paren {x \in R \implies x \in T}$

$\ds $

$\leadsto$

$\ds R \subseteq T$

Definition of Subset

$\blacksquare$

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