Infinite Ordinal can be expressed Uniquely as Sum of Limit Ordinal plus Finite Ordinal

From ProofWiki
Jump to navigation Jump to search


Let $x$ be an ordinal.

Suppose $x$ satisfies $\omega \subseteq x$.

Then $x$ has a unique representation as $\left({y + z}\right)$ where $y$ is a limit ordinal and $z$ is a finite ordinal.


Take $K_{II}$ to be the set of all limit ordinals.

Then set $y = \bigcup \left\{{w \in K_{II}: w \le x}\right\}$

The set $\left\{{w \in K_{II}: w \le x}\right\}$ 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 $x = \left({y + z}\right)$

Assume $\omega \le z$.

Then, again by Ordinal Subtraction when Possible is Unique:

$z = \left({\omega + w}\right)$

and so:

\(\displaystyle x\) \(=\) \(\displaystyle \left({y + \left({\omega + w}\right)}\right)\) Equality is Transitive
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({y + \omega}\right) + w}\right)\) Ordinal Addition is Associative

But $y + \omega$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Addition:

$\varnothing < \omega \implies y < y + \omega$

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:

$\left({y + w}\right) = \left({y + z}\right)$

By Ordinal Addition is Left Cancellable, we have $w = z$.

Thus $z$ is unique.

To prove uniqueness for $y$, suppose that $x = \left({y + u}\right)$ and $x = \left({w + z}\right)$.

Assume further, WLOG, that $y \le w$.


\(\displaystyle y \le w\) \(\implies\) \(\displaystyle \exists n: w = \left({y + n}\right)\) Ordinal Subtraction when Possible is Unique
\(\displaystyle \) \(\implies\) \(\displaystyle \left({y + u}\right) = \left({\left({y + n}\right) + z}\right)\) Substitutivity of Equality
\(\displaystyle \) \(\implies\) \(\displaystyle \left({y + u}\right) = \left({y + \left({n + z}\right)}\right)\) Ordinal Addition is Associative
\(\displaystyle \) \(\implies\) \(\displaystyle u = \left({n + z}\right)\) Ordinal Addition is Left Cancellable
\(\displaystyle \) \(\implies\) \(\displaystyle \left({n + z}\right) \in \omega\) by the fact that $u \in \omega$
\(\displaystyle \) \(\implies\) \(\displaystyle n \in \omega\) Ordinal is Less than Sum
\(\displaystyle \) \(\implies\) \(\displaystyle n \notin K_{II}\) by the fact that $\omega$ is the smallest limit ordinal
\(\displaystyle \) \(\implies\) \(\displaystyle n = \varnothing \lor \exists m: n = m^+\) Definition of Limit Ordinal

Assume that $\exists m: n = m^+$.

\(\displaystyle n = m^+\) \(\implies\) \(\displaystyle w = \left({y + m^+}\right)\) Definition of $n$
\(\displaystyle \) \(\implies\) \(\displaystyle w = \left({y + m}\right)^+\) Definition of Ordinal Addition
\(\displaystyle \) \(\implies\) \(\displaystyle w \notin K_{II}\) Definition of Limit Ordinal

This is clearly a contradiction, so $n = \varnothing$ and $w = y$.