Subset Relation is Transitive
Theorem
The subset relation is transitive:
- $\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$
Proof 1
| \(\ds \) | \(\) | \(\ds \paren {R \subseteq S} \land \paren {S \subseteq T}\) | ||||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \paren {x \in R \implies x \in S} \land \paren {x \in S \implies x \in T}\) | 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 |
$\blacksquare$
Proof 2
Let $V$ be a basic universe.
By definition of basic universe, $R$, $S$ and $T$ are all elements of $V$.
By the Axiom of Transitivity, $R$, $S$ and $T$ are all classes.
We are given that $R \subseteq S$ and $S \subseteq T$.
Hence by Subclass of Subclass is Subclass, $R$ is a subclass of $T$.
By Subclass of Set is Set, it follows that $R$ is a subset of $T$.
Hence the result.
$\blacksquare$
Sources
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