For ordinals, if $\beta \leq \alpha < \beta + \gamma$ then $\alpha = \beta + \delta$ for some $\delta < \gamma$

Question

I need to show, for ordinals $\alpha, \beta, \gamma, \delta$, that:

if $\beta \leq \alpha < \beta + \gamma$ then $\alpha = \beta + \delta$ for some $\delta < \gamma$.

I understand this should follow by well-orderedness of ordinals, but I'm not sure of how to phrase it precisely. (I actually need to show it for specific ordinals, but I feel it is true in this general form)

Answer 1

By well-orderedness, there exists a minimal $\delta$ with $\alpha\le \beta+\delta$.

• If $\delta=0$, we see that $\beta\le \alpha\le \beta$, hence $\alpha=\beta+0$, as desired.

• If $\delta=\epsilon+1$, the from $\beta+\epsilon<\alpha\le(\beta+\epsilon)+1$, we conclude $\alpha=\beta+\epsilon+1$, as desired.

• If $\delta$ is a limit ordinal, then $\beta+\delta$ is the limit of all $\beta+\epsilon$, $\epsilon<\delta$. As $\alpha >\beta+\epsilon$ for all $\epsilon<\delta$, we have $\alpha\ge \beta+\delta$ and once again $\alpha=\beta+\delta$.

Of course, once we have $\alpha=\beta+\delta$, it is clear from $\alpha<\beta+\gamma$ that $\delta<\gamma$.