Does it follow that $\sup\emptyset = 0$ if the domain is $\mathbb{R}_{\geq 0} \cup \{\infty\}$? I am currently working on a paper where I am taking $\min$s of $\max$s over sets of non-negative real numbers including positive infinity, i.e. my equations look something like $\min\max\{x_1,\ldots,x_n\}$ where $x_i \in \mathbb{R}_{\geq 0} \cup \{\infty\}$.
However, for technical reasons, I want it that $\min\emptyset = \infty$ and $\max\emptyset = 0$, so my idea is to use $\inf$ and $\sup$ instead of $\min$ and $\max$ (since for finite sets $\min$ is the same as $\inf$ and $\sup$ the same as $\max$).
So my question is: Does it follow that $\sup\emptyset = 0$ if the domain is $\mathbb{R}_{\geq 0} \cup \{\infty\}$?

Yes: Every number is an upper bound of the empty set, and $0$ is the least such number in your domain.
(By contrast, $\max\emptyset$ doesn't exist by the usual definitions. If you really wanted to define $\max\emptyset$, you would have to extend the definition of $\max$ in a way that risks being misleading.)
If your concern is that the equation $\sup\emptyset=0$ tacitly depends on the domain, you could clarify by abbreviating $[0,\infty]=\mathbb R_{\geq0}\cup\{\infty\}$ and then writing $\sup_{[0,\infty]}\emptyset=0$. This would distinguish the equation from the more familiar result $\sup_{[-\infty,\infty]}\emptyset=-\infty$. For intuitive explanations of the latter, see Infimum and supremum of the empty set and related questions.
