Proving that $\sup A = k - \inf B$ when $A,B$ are related by a map. Suppose that $A = \{ a_1,a_2,\dots,a_n \}$ and $B = \{ \psi(a_1),\psi(a_2), \dots, \psi(a_n) \}$ for some map $\psi$.
Suppose that there exists a $k$ such that for every $a \in A,$ $a = k - \psi(a),$ in other words, $\psi(a) = a - k$ is a translation.
I would like to show that $\sup A = \sup\{k - \psi(a)\} = k - \inf B$.
The inspiration for this question comes from Spivak's Calculus problem 13-21, where he uses $\sup \{L(g, P)\} = \sup \{ k - U(f,\psi(P))\} = k - \inf\{U(f,\psi(P))\}$.
My attempt goes something like this: we know every element $a$ can be written in the form $k - \psi(a)$, so the sets $\{ a : a \in A \}$ and $\{ k - \psi(a) : a \in A \}$ are clearly equal. If we want to maximize $k - \psi(a)$, we want $\psi(a)$ (over all $a \in A$) to be as small as possible since we want to subtract as small of a value as possible. This happens when we leave $k$ alone, and find the greatest lower bound of all $\psi(a)$. In other words, we take the greatest lower bound of the set of all $\psi(a)$ which is $\inf B$.
 A: Since $\psi$ is traslation, $\psi(a) \leq \psi(b)$ is equivalent to $a \leq b;$ if $a_k \geq a_i$ for all $i,$ then $\psi(a_k) \geq \psi(a_i)$ for all $i,$ showing that $\psi(a_k)$ is the supremum of $\psi(A) = B.$
A: I believe that you understood well the problem, meaning your idea is correct. Let us now work a little bit more in order to make the proof precise.
Let $A \subseteq \mathbb{R}$ be a finite set of real numbers; i.e. $A = \{a_1, \dots, a_n\}$ ($n \in \mathbb{N}$), where $a_i \in \mathbb{R}$, $I = 1, \dots, n$. Further, let us consider a map $\psi \colon A \to \mathbb{R}$ and set $B$ to be the image of $\psi$; i.e. $B = \{\psi(a) \in \mathbb{R} \colon a \in A\}$. By hypothesis, we know that there exists some real constant $k$ such that $a = k - \psi(a)$, for each $a \in A$.
First, note a few details about what we have. Our set $A$ is a finite set of real numbers. Hence, there exists a maximum and a minimum of $A$ (try to prove this by yourself and comeback if you have any doubt). Plus, we also know that the maximum and the minimum of $A$ are, respectively, the supremum and the infimum of $A$ (this is also a fact that you can easily check by yourself). In short, the set $A$ has a supremum and a infimum. By a completely analogous argument, we can conclude the same thing for the set $B$.
Now that we have assure that the supremum of $A$ and the infimum of $B$ exists, we must prove that
$$\sup A = k - \inf B.$$
This equality tell us that $k - \inf B$ is the supremum of $A$. So, we clearly must prove that $k - \inf B$ is the supremum of $A$. Why? Because the supremum of a set, when it exists, is unique.
Recall that $s \in \mathbb{R}$ is a supremum of a set $S$ if verify the following two conditions:

*

*$s$ is an upper bound of $S$; i.e. $\quad \forall x \in S \;\; (x \leq s)$;

*Any other upper bound of $S$ is greater than or equal to $s$; i.e. $\quad \forall x \in S \;\; (x \leq r) \implies s \leq r$.

We start by proving condition 1. Let $x \in A$. By hypothesis, we know that $x = k - \psi(x)$. It follows from the definition of infimum of a set, that $\inf B \leq \psi(x)$. Hence
$$\inf B \leq \psi(x) \implies -\psi(x) < -\inf B \implies k - \psi (x) < k - \inf B \implies x < k - \inf B.$$
Note that we just proved that, for each $x \in A$, we have that $x < k - \inf B$. This means that $k - \inf B$ is an upper bound of $A$, and this completes the proof of the condition 1.
For the condition 2., I will leave the details to be filled by you (come back later if any doubt arises). Let $r \in \mathbb{R}$ be another upper bound of $A$; i.e. such that $x \leq r$, for each $x \in A$. With this, let's prove that $k - \inf B \leq r$. The key observation that you must do (and you must prove it as well) is to answer the question "does it $\inf B \in B$?". This will suffice to show condition 2.
I hope this helps you to "formalise" your intuitive idea of the proof (which was indeed correct). Any further doubts, come back and ask us :)
