Infinite Ordinal can be expressed Uniquely as Sum of Limit Ordinal plus Finite Ordinal
Theorem
Let $x$ be an ordinal.
Suppose $x$ satisfies $\omega \subseteq x$.
Then $x$ has a unique representation as $\paren {y + z}$ where $y$ is a limit ordinal and $z$ is a finite ordinal.
Proof
Take $K_{II}$ to be the set of all limit ordinals.
Then set $y = \bigcup \set {w \in K_{II}: w \le x}$
The set $\set {w \in K_{II}: w \le x}$ is non-empty because $\omega \subseteq x$.
By Union of Ordinals is Least Upper Bound, $y \in K_{II}$ and $y \le x$.
By Ordinal Subtraction when Possible is Unique, there is a unique $z$ such that $x = \paren {y + z}$
Assume $\omega \le z$.
Then, again by Ordinal Subtraction when Possible is Unique:
- $z = \paren {\omega + w}$
and so:
\(\ds x\) | \(=\) | \(\ds \paren {y + \paren {\omega + w} }\) | Equality is Transitive | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {y + \omega} + w}\) | Ordinal Addition is Associative |
But $y + \omega$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Addition:
- $\O < \omega \implies y < y + \omega$
This contradicts the fact that $y$ is the largest limit ordinal smaller than $x$.
Therefore, $z \in \omega$.
Thus we have proven that such a selection of $y$ and $z$ exists.
Suppose $z$ and $w$ both satisfy:
- $\paren {y + w} = \paren {y + z}$
By Ordinal Addition is Left Cancellable, we have $w = z$.
Thus $z$ is unique.
It remains to prove uniqueness for $y$.
Suppose that $x = \paren {y + u}$ and $x = \paren {w + z}$.
Without loss of generality, assume further that $y \le w$.
Then:
\(\ds y\) | \(\le\) | \(\ds w\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists n: \, \) | \(\ds w\) | \(=\) | \(\ds \paren {y + n}\) | Ordinal Subtraction when Possible is Unique | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {y + u}\) | \(=\) | \(\ds \paren {\paren {y + n} + z}\) | Substitutivity of Equality | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {y + u}\) | \(=\) | \(\ds \paren {y + \paren {n + z} }\) | Ordinal Addition is Associative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds u\) | \(=\) | \(\ds \paren {n + z}\) | Ordinal Addition is Left Cancellable | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {n + z}\) | \(\in\) | \(\ds \omega\) | as $u \in \omega$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(\in\) | \(\ds \omega\) | Ordinal is Less than Sum | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(\notin\) | \(\ds K_{II}\) | as $\omega$ is the smallest limit ordinal | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(=\) | \(\ds \O\) | Definition of Limit Ordinal | ||||||||||
\(\ds \exists m: \, \) | \(\, \ds \lor \, \) | \(\ds n\) | \(=\) | \(\ds m^+\) |
Aiming for a contradiction, suppose that $\exists m: n = m^+$.
\(\ds n\) | \(=\) | \(\ds m^+\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(=\) | \(\ds \paren {y + m^+}\) | Definition of $n$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(=\) | \(\ds \paren {y + m}^+\) | Definition of Ordinal Addition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(\notin\) | \(\ds K_{II}\) | Definition of Limit Ordinal |
This is clearly a contradiction.
Hence:
- $n = \O$
and so:
- $w = y$
The result follows.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.13$