In every set of ordinals there is a maximal

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Given $B \not= \emptyset$ a finite set of ordinals, prove that there is a maximal element in $B$ with respect to $\in$ , meaning $\exists A \in B \forall x \in B (x=A or x \in A)$

I think this is true because of the well-order of ordinals, but i don't know how to prove it, because for instance $B = \{ \emptyset, \omega\}$ so obviously $\omega$ is the maximal element in $B$ but $\omega$ is not finite, and every proof i thought of works only for finite ordinals.

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Being an ordinal isn't particularly important:

Theorem: If $X$ is a finite, totally ordered set, then $X$ has a maximum.

This isn't too hard to prove by induction.

Since the ordinals are totally ordered by $\in$, $B$ (with the ordering given by $\in$) is a finite totally ordered set, so the theorem applies.

Incidentally, for this particular problem, there is a set-theoretic calculation of the maximum too: $\max(B) = \cup B$ (of course, you'll have to prove that $\cup B \in B$.