Let $\Omega$ be a set of ordinals such that for all $\alpha \in \Omega: |\alpha| < \kappa$, for a fixed infinite cardinal $\kappa$. Let $\alpha, \beta \in \Omega$ and denote $\gamma=\text{inf}\lbrace \mu \in \Omega \mid \alpha + \mu \geq \beta \rbrace$ and $\delta=\text{inf}\lbrace \mu \in \Omega \mid \beta + \mu \geq \alpha \rbrace$. I'm trying to prove that $\alpha +\gamma=\beta + \delta$. By symmetry, it suffices to show that $\alpha + \gamma \leq \beta + \delta$. Suppose $\beta + \delta < \alpha + \gamma$, then $\alpha \leq \beta + \delta < \alpha + \gamma$. Intuitively, this must somehow imply there exists a $\theta \in \Omega$, with $\theta < \gamma$, such that $\beta + \delta=\alpha + \theta$. How does one find this $\theta$?
EDIT:
The answer goes as follows. Since the ordinals are totally ordered, we can assume w.l.o.g. that $\alpha \geq \beta$, i.e. $\gamma =0$. Therefore, it remains to show that $\alpha < \beta + \delta$ is impossible. So assume $\alpha < \beta + \delta$. If there exists an ordinal $\mu \in \Omega$ such that $\alpha \leq \beta + \mu$, then we obtain a contradiction with the minimality of $\delta$. Suppose for all $\mu < \delta: \beta + \mu < \alpha$. Then, $\beta + \delta \leq \beta + \text{sup}_{\mu < \delta}(\mu) \leq \alpha$, a contradiction.