If $\alpha<\beta$, there exists a ordinal $\gamma$ such that $\alpha + \gamma = \beta$. Why? I want to prove the following proposition.
proposition: Let $\alpha,\beta$ be ordinals. If $\alpha<\beta$, there exists a ordinal $\gamma$ such that $\alpha + \gamma = \beta$.
I am trying to prove this by  transfinite induction. However, I have some trouble with the case $\beta$ is a limit ordinal.(I know the definition of the sum of a ordinal and a limit ordinal)
please give me some ideas.
 A: The other answers are fine, but since you explicitly asked for the limit case in a proof by transfinite induction, let's see how that goes as well.
Assume $\beta=\sup_{\delta<\beta}\delta$. By induction hypothesis, for each $\delta<\beta$ you can pick an ordinal $\gamma_\delta$ such that $\alpha+\gamma_\delta=\delta$. Now let $\gamma:=\sup_{\delta<\beta} \gamma_\delta$. Then $$\alpha+\gamma=\alpha+\sup_{\delta<\beta} \gamma_\delta=\sup_{\delta<\beta}(\alpha+\gamma_\delta)=\sup_{\delta<\beta} \delta=\beta.$$
A: Let $\gamma$ be the smallest ordinal that satisfies $\alpha+\gamma\geq \beta$. Then you see $\alpha+\gamma=\beta$.
Otherwise, suppose $\alpha+\gamma>\beta$, it is obvious that $\gamma$ can not be $\eta+1$, so suppose it is a limit one. By definition of $\sup$, there is some $\eta<\gamma$ such that $\alpha+\eta\geq \beta$, which is a contradiction.
A: As sets, $\alpha$ is a subset of $\beta$. Consider $C=\beta\setminus\alpha$, the set-theoretic difference. Then $C$ is a well-ordered set, having the same order type as some ordinal $\gamma$. Then $\beta=\alpha+\gamma$.
