Natural Number is Transitive Set

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Theorem

Let $n$ be a natural number.

Then $n$ is a transitive set.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$n$ is a transitive set.


Basis for the Induction

From Empty Class is Supercomplete, we have that $\O$ is supercomplete.

Thus by definition of supercomplete, $\O$ is a transitive class.

By the axiom of the empty set, $\O$ is a set.

Hence $\O$ is a transitive set.

Thus $\map P 0$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$k$ is a transitive set.


from which it is to be shown that:

$k^+$ is a transitive set.


Induction Step

This is the induction step:

\(\ds x\) \(\in\) \(\ds k^+\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds k \cup \set k\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds k\)
\(\, \ds \lor \, \) \(\ds x\) \(\in\) \(\ds k\) Definition of Transitive Set: from Induction Hypothesis

If $x \in k$ then, because $k$ is transitive, $x \subseteq k$.

If $x = k$, then, because $k^+ = k \cup \set k$, it follows that $x \subseteq k$ by definition of union of class.

In either case:

$x \subseteq k$

But we have $k \subseteq k^+$ by definition of union of class.

Hence:

$x \subseteq k^+$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: n$ is a transitive set.

$\blacksquare$


Sources