Successor Set of Ordinary Transitive Set is Ordinary

Theorem

Let $x$ be a transitive set which is also ordinary.

Let $x^+$ denote the successor set of $x$:

$x^+ = x \cup \set x$

Then $x^+$ is also an ordinary set.

Proof

By definition of ordinary set:

$x \notin x$

Aiming for a contradiction, suppose $x^+ \in x^+$.

Then we have:

\(\ds x^+\)

\(=\)

\(\ds x \cup \set x\)

Definition of Successor Set

\(\ds \leadsto \ \ \)

\(\ds x^+\)

\(=\)

\(\ds x\)

Definition of Set Union

\(\, \ds \lor \, \)

\(\ds x^+\)

\(\in\)

\(\ds x\)

Let $x^+ \in x$

We have that $x$ is transitive.

Hence:

$x^+ \subseteq x$

Let $x^+ = x$

We have that: $x^+ = x \cup \set x$

Hence by definition of union of set of sets:

$x^+ \subseteq x$

Hence by Proof by Cases:

$x^+ \subseteq x$

But we have:

$x \in x^+$

and because $x^+ \subseteq x$ it follows that:

$x \in x$

But this contradicts our supposition that $x \notin x$.

Hence by Proof by Counterexample:

$x^+ \notin x^+$

$\blacksquare$

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 $3.2$