A proof I'm reading relies on the following fact: Let $(J,\leq)$ be a well-ordered set, $A \subseteq J$ a subset, and suppose that there exists $i \in J$ such that $a < i$ for all $a \in A$. Then there exists $k < i$ such that $a \leq k$ for all $a \in A$.
I can't figure out how to prove this. $A$ must have a least upper bound because $J$ is well-ordered, but what stops this least upper bound from being $i$ itself?
This is indeed false without additional assumptions. For instance, $A$ could be empty and $i$ could be the least element of $J$. Or, $J$ could be $\mathbb{N}\cup\{\infty\}$, $A$ could be $\mathbb{N}$, and $i$ could be $\infty$.