Suppose I have a set of ordinals $x$. Then $\cup x$ is an ordinal and is the supremum of $x$. Is is possible for $\cup x \in x$? In other words, can $x$ contain its supremum?
I'm asking this because I found that $\cup x \subseteq x$.
proof. Let $\gamma \in \cup x$. Then $\gamma \in \beta $ for some $\beta \in x$. By transitivity of the ordinals, $\gamma \in x$.
On
Yes, e.g. for $x=\{0,1\}$ whose union is $1 \in x$. This also happens for infinite successor ordinals, like $x=\{0,1,2,3,5,7,9,\ldots,\omega,\omega+1\}$ whose union is $\omega+1 \in x$ again.
Nothing special about ordinals, a supremum can be a maximum in any ordered set.
For all ordinals $\alpha, \beta$: $$\alpha \in \beta \iff \alpha \subsetneq \beta \iff \alpha < \beta$$ by their transitivity. It's how their ordering is defined.
If $x$ is a successor ordinal, then $\cup x=(x-1)\in x$. On the other hand, if $x$ is a limit ordinal, we get $\cup x=x\notin x$