For ordinals $\delta$ and $\beta$, if $\delta\in$ or $=\beta$, then there exists $\gamma\in$ or $=\beta$ such that $\beta=\delta+\gamma$

Question

Prove that for ordinals $\delta$ and $\beta$, if $\delta \in$ or $= \beta$ then there exists an ordinal $\gamma \in$ or $= \beta$ such that $\beta = \delta + \gamma$.

I tried using transfinite induction on $\delta$

Base case seems simple, since you can let $\gamma = \beta$, but I'm stuck on the proof for both the successor ordinal and limit ordinal cases.

My class has not yet defined ordinal subtraction, so I can't use that in the proof.

Any help or hints would be appreciated!

Answer 1

(Note that you can write $\delta\le\beta$ instead of the clumsy $\delta\in$ or $=\beta$.)

HINT: Fix $\delta$ and prove it by induction on $\beta$ for all $\beta\ge\delta$.

• Suppose that you know that $\delta+\xi=\beta$; what must $\gamma$ be in order that $\delta+\gamma=\beta+1$?
• Suppose that $\beta$ is a limit ordinal, and for each $\rho$ such that $\delta\le\rho<\beta$ you know that there is an ordinal $\xi_\rho$ such that $\delta+\xi_\rho=\rho$; what is the natural candidate for an ordinal $\gamma$ such that $\delta+\gamma=\beta$?