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 $( y + z )$ 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 $y = \bigcup \{ w \in K_{II} : w \le x \}$

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

Assume $\omega \le z$. Then, again by Ordinal Subtraction when Possible is Unique, $z = ( \omega + w )$.

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


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

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.

$z$ must be unique because if $z$ and $w$ both satisfy the hypothesis, then $( y + w ) = ( y + z )$, so by Ordinal Addition is Left Cancellable, we have $w = z$.

To prove uniqueness for $y$, suppose that $x = ( y + u )$ and $x = ( w + z )$. Assume further, without loss of generality that $y \le w$.

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

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