Successor Set of Transitive Set is Transitive
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Theorem
Let $S$ be a transitive set.
Then its successor set $S^+ = S \cup \set S$ is also transitive.
Proof
Recall that $S$ is transitive if and only if:
- $x \in S \implies x \subseteq S$
Hence:
\(\ds \forall x \in S^+: \, \) | \(\ds x\) | \(\in\) | \(\ds S\) | |||||||||||
\(\, \ds \lor \, \) | \(\ds x\) | \(=\) | \(\ds S\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\subseteq\) | \(\ds S\) | Definition of Transitive Set: $x \in S \implies x \subseteq S$ | ||||||||||
\(\, \ds \lor \, \) | \(\ds x\) | \(\subseteq\) | \(\ds S\) | Set is Subset of Itself: $x = S \implies x \subseteq S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\subseteq\) | \(\ds S\) | Proof by Cases: both $x \in S$ and $x = S$ lead to $x \subseteq S$ |
Then:
\(\ds \forall S: \, \) | \(\ds S\) | \(\subseteq\) | \(\ds S^+\) | Set is Subset of Union | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\subseteq\) | \(\ds S^+\) |
Thus we have:
- $x \in S \implies x \subseteq S$
Hence the result.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.1$