Consider some ordered set $(X,\leq)$ and bounded subsets $A,B\subseteq X$. It is to show that "$\sup(A\cup B)=\sup\{\sup(A),\sup(B)\}$, provided the relevant suprema exist". I have no problem in showing the statement, but what's the smallest number of $\sup$'s we have to assume existence of?
Does existence of any of the involved $\sup$'s imply existence of another one?
In $\mathbb R$, since $\sup(A\cup B)\geq \sup(A)$ and $\sup(A\cup B)\geq \sup(B)$, the existence of $\sup(A\cup B)$ implies the existence of the other two.
Outside $\mathbb R$, we can't even be sure of that, since taking $B=\{2\}$ and $A=\{q| q^2<2\}$ both as subsets of $\mathbb Q$ means that $\sup(A\cup B)=2$ however $\sup(A)$ does not exist.
Also, no other implications like this (where one supremum implies another) exist, since you can have $A=\{0\}$ and $B=\mathbb R$, in which case $\sup(A)$ exists, but $\sup(A\cup B)$ does not.