If $x$ is a set of ordinals, then $\bigcup x=\alpha$ is an ordinal and the $\text{sup}(x)$. Why is $x\subseteq \alpha+1$ rather than $\alpha$?
$\alpha+1$ is from notes by J. Roitman. I think it should be $\subseteq\alpha$.
After all as the least upper bound, every element of $x$ is $\leq\alpha$. And as $\alpha$ and the members of $x$ are ordinals, they are transitive. So aren't the members of members of $x$, i.e., $\bigcup x$, members of $\alpha$ making $x$ a subset of $\alpha$?
Thanks
You've shown that every element of an element of $x$ is also an element of $\alpha$, but you want to prove that every element of $x$ is also an element of $\alpha$. This is false. Consider the case of a singleton set $x = \{\alpha\}$. Then $\bigcup x = \alpha$, but $x \nsubseteq \alpha$ because $\alpha \notin \alpha$.