# Sup of two ordinals in a limit

I am not sure if this is a dumb question, but if I have a limit ordinal $$\xi$$, given any two $$\alpha,\beta <\xi$$, is it possible to find an ordinal $$\gamma$$ such that $$\alpha,\beta\leq\gamma<\xi$$?

Thanks!

Ordinals are totally ordered by set membership (at least if we are working with von Neumann ordinals). Consider set union of $$\alpha$$ and $$\beta$$. Then,
• If $$\alpha \in \beta$$, $$\alpha \cup \beta = \beta < \xi$$.
• If $$\beta \in \alpha$$, $$\alpha \cup \beta = \alpha < \xi$$.
• If $$\alpha = \beta$$, $$\alpha \cup \beta = \alpha < \xi$$.
So, $$\alpha,\beta \leq \alpha \cup \beta < \xi$$. $$\alpha \cup \beta$$ works as $$\gamma$$ in your question.
Another way to think about it: consider $$\gamma=\max(\alpha,\beta)<\xi$$. This ordinal satisfies, by definition, $$\gamma>\alpha,\beta$$ and so it works.
(Note that your property holds true for any set with an order; the fact that $$\xi$$ is a limit ordinal was not used).