Factorization of Limit Ordinals

From ProofWiki
Jump to navigation Jump to search


Let $x$ be a limit ordinal.


$x = \paren {\omega \times y}$ for some $y \in \On$

where $\omega$ is the minimally inductive set.


By the Division Theorem for Ordinals:

$x = \paren {\omega \times y} + z$

for some unique $y$ and $z \in \omega$.

Aiming for a contradiction, suppose $z \ne 0$.

Because $z \in \omega$, $z$ is not a limit ordinal.

Therefore, by the definition of limit ordinal:

$z = w^+$

for some $w \in \omega$.

But this means that:

\(\ds x\) \(=\) \(\ds \paren {\omega \times y} + w^+\) Division Theorem for Ordinals
\(\ds \) \(=\) \(\ds \paren {\paren {\omega \times y} + w}^+\) Definition of Ordinal Addition

This means that $x$ is the successor of some ordinal.

Hence $x$ cannot be a limit ordinal.

But this contradicts the assumption that $x$ is a limit ordinal.

It follows that $z = 0$.


\(\ds x\) \(=\) \(\ds \paren {\omega \times y} + z\) Division Theorem for Ordinals
\(\ds \) \(=\) \(\ds \omega \times y\) Ordinal Addition by Zero