Minimal Infinite Successor Set is Ordinal

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Theorem

Let $\omega$ denote the minimal infinite successor set.


Then $\omega$ is an ordinal.


Proof 1

The minimal infinite successor set is a set of ordinals by definition.

From the corollary of ordinals are well-ordered, it is seen that $\left({\omega, \Epsilon \! \restriction_\omega}\right)$ is a strictly well-ordered set.

It is to be shown by induction on minimal infinite successor set that $\forall n \in \omega: \omega_n = n$


Basis for the Induction

\(\displaystyle \omega_\varnothing\) \(=\) \(\displaystyle \left\{ {x \in \omega: x \in \varnothing}\right\}\) Definition of initial segment
\(\displaystyle \) \(=\) \(\displaystyle \varnothing\) Definition of empty set


Induction Hypothesis

Suppose that $\omega_n = n$ for some $n \in \omega$.


Induction Step

\(\displaystyle \omega_{n^+}\) \(=\) \(\displaystyle \left\{ {x \in \omega: x \in n^+}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {x \in \omega: x \in n \lor x \in \left\{ {n}\right\} }\right\}\) Definition of successor set
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {x \in \omega: x \in n \lor x = n}\right\}\) Definition of singleton
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {x \in \omega: x \in n}\right\} \cup \left\{ {x \in \omega: x = n}\right\}\) Definition of set union
\(\displaystyle \) \(=\) \(\displaystyle \omega_n \cup \left\{ {n}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle n \cup \left\{ {n}\right\}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle n^+\) Definition of successor set


And so $\omega$ is an ordinal.

$\blacksquare$


Proof 2

Let $K_I$ denote the set of all nonlimit ordinals.

Let $\operatorname{On}$ denote the set of all ordinals.

Let $a \in \omega$.

It follows that $a^+ \subseteq K_I$, so $a \in K_I$.

Thus:

$\omega \subseteq K_I \subseteq \operatorname{On}$


We now must prove that $\omega$ is a transitive set, at which point it will satisfy the Alternative Definition of Ordinal.

Let $x \in y$ and $y \in \omega$.

Then:

$y \in \operatorname{On} \land y^+ \subseteq K_I$

Because $y$ is an ordinal, it is transitive.

Therefore:

$x \subseteq y$

and:

$x^+ \subseteq y^+ \subseteq K_I$

Therefore, $x^+ \subseteq K_I$.

Applying the definition of Minimal Infinite Successor Set:

$x \in \omega$

so $\omega$ is transitive.

$\blacksquare$


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