Factorization of Limit Ordinals
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Theorem
Let $x$ be a limit ordinal.
Then:
- $x = \paren {\omega \times y}$ for some $y \in \On$
where $\omega$ is the minimally inductive set.
Proof
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 $z = 0$.
Therefore:
\(\ds x\) | \(=\) | \(\ds \paren {\omega \times y} + z\) | Division Theorem for Ordinals | |||||||||||
\(\ds \) | \(=\) | \(\ds \omega \times y\) | Ordinal Addition by Zero |
$\blacksquare$