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:

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 $x = \paren {w + z}$.

, assume further that $y \le w$.

Then:

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

This is clearly a contradiction.

Hence:
 * $n = \O$

and so:
 * $w = y$

The result follows.