Given any two real numbers $x,y$ we can compute $$\max(x,y)=\frac{|\,a-b\,|+a+b}{2}$$ using "nice" functions and operations (adding/substracting, absolute value, halving). Inductively we can calculate $\max(x_1,\dots,x_n)$ for any finite collection $x_1,\dots,x_n$ of real numbers.
Now, given a (non-empty) set $A\subseteq\mathbb{R}$ is there a formula to compute $\sup(A)$ in general?
The first problem that I see is that the existence of suprema is an axiom and thus is maybe "out of reach" for operations defined using every other axiom but this one. This already makes me suspicious of a "nice/elementary" formula/algorithm taking a set and gibing its supremum back. In a possible answer please state why your formula is "elemental".
My first attempt is to use something like $\sup(A)=\Vert \mathbf{1}_A\Vert_\infty$ where $\mathbf{1}_A(x)=x$ is the identity restricted to $A$. I see many problems with this: first, while $$\Vert\mathbf{1}_A\Vert_p = \left(\,\int_A |\,\mathbf{1}_A\,|^p\,\right)^{1/p} $$
is "nice", $\sup(A)=\Vert\mathbf{1}_A\Vert_\infty=\lim_p\Vert\mathbf{1}_A\Vert_p$ is taking the limit and I would be willing to admit limits as "elemental" I feel like there is a "computability gap" (computable meaning "an explicit calculation exists"). Also, the value of the integral does depend on the "endpoints" of $A$ i.e if $A$ is an interval
$$\sup A=\lim_p\left(\,\int_A\Vert\mathbf{1}_A\Vert^p\, \right)^{1/p}=\left(\,\frac{\sup(A)^{p+1}-\inf(A)^{p+1}}{p+1}\,\right)^{1/p}$$
You also have to asume that $A$ is lebesgue measurable and while it may not depend precisely on $\sup$ and $\inf$ it certainly needs a lot of info of $A$. The formula for the $\max$ didn't need anything about the numbers: you have the numbers, you have the maximum, no previous checking.
Also, an integral is kind of like a limit and maybe we will be suspicious.
Sorry for the lengthy question.
Thanks!