Why isn't $\sup(f(x)+g(x)) = \sup f(x) + \sup g(x)$?

Question

The claim that if $X$ is a non empty set and $f,g:X \to \mathbb{R}$ bounded functions then $$\sup_{x \in X} (f(x)+g(x)) = \sup_{x \in X} f(x) + \sup_{x \in X} g(x)$$ Is not true in general. But why is that? I thought that if $A$ and $B$ are bounded subsets of $\mathbb{R}$ and $A+B := \{a+b : a \in A, b \in B \}$ then: $\sup(A+B) = \sup (A) + \sup(B)$

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Here is my proof (Could someone check if it is correct ?)

Let $a \in A, b \in B$ then $a \leq \sup(A)$ and $b \leq \sup(B)$ so $a+b \leq \sup(A) + \sup(B)$. Since $a$ and $b$ were arbitrary then $\sup(A) + \sup(B)$ is an upper bound of $A+B$ so that $\sup(A+B) \leq \sup(A) + \sup (B)$.

Let $\epsilon >0$, there exists $a \in A$ and $b \in B$ such that: \begin{align*} \sup(A) - \dfrac{\epsilon}{2} \leq a \leq \sup(A) \\ \sup(B) - \dfrac{\epsilon}{2} \leq b \leq \sup(B) \end{align*}

So $\sup (A) + \sup(B) - \epsilon < a+b \leq \sup(A+B) \Rightarrow \sup (A)+ \sup (B) \leq \sup(A+B) + \epsilon$ and since $\epsilon$ was arbitrary then $\sup(A)+\sup(B) \leq \sup(A+B)$ and therefore: $$ \sup(A+B) = \sup (A) + \sup (B)$$

But if $X = [-1,1]$ and $f(x)=x$, $g(x)=-x$ then:

$$\sup_{x \in [-1,1]} (f(x)+g(x)) = 0$$ and $$\sup_{x \in [-1,1]} f(x) + \sup_{x \in [-1,1]} g(x) = 2$$

Why is $\sup (A+B) = \sup (A) + \sup (B)$ not true in general?

Answer 1

The problem here is that $\sup_{x \in X} f(x) + g(x)$ is not equal to $\sup_{(x,y) \in X \times X} f(x) + g(y)$, which would correspond to $\sup A + B$ if you define $A := \{f(x) \mid x \in X\}$ and $B := \{g(x) \mid x \in X\}$, because you are only considering the "diagonal" elements of $A + B$ in a way and not all the elements.
As Martin R commented earlier but deleted, consider $f := \sin =: -g$ to see what happens.

Answer 2

Here $A+B$ denotes the Minkowski sum, e.g. $$ A+B = \{a+b : a\in A, b\in B\}, $$ then yes, $\sup(A+B) = \sup(A)+\sup(B)$. The argument in the question proves this fact. However, $f(x)+g(x)$ is not the Minkowski sum of two sets. Rather, $f+g$ is a new function, defined by $$ (f+g)(x) = \{ (x,f(x)+g(x)) : x \in \operatorname{dom}(f) \cap \operatorname{dom}(g)\}, $$ where $\operatorname{dom}(f)$ denotes the domain of $f$. In other words, $f+g$ is a set of ordered pairs, where the first coordinate ranges over the common domain of $f$ and $g$. This is not a Minkowski sum.

A counterexample to the claim that $\sup(f(x)+g(x)) = \sup(f(x)) + \sup(g(x))$ can be obtained by taking $$ f(x) = \begin{cases} 1 & \text{if $x\in \mathbb{Q}$, and} \\ 0 & \text{otherwise}, \end{cases} \qquad\text{and}\qquad\ g(x) = \begin{cases} 0 & \text{if $x\in \mathbb{Q}$, and} \\ 1 & \text{otherwise}. \end{cases} $$ Here $$ \sup(f(x)+g(x)) = \sup(f(x)) = \sup(g(x)) = 1. $$