I know that this question had been answered before here but I am asking to please check a method used by me which resulted in the wrong conclusion.
Let $X$ be a nonempty set, let $f$ and $g$ be defined on $X$ and have bounded range in $\mathbb R$.
Prove that $\sup \{f(x)+g(x):\space x\in X\}\leq \sup \{f(x):\space x\in X\}+\sup \{g(x):\space x\in X\}$
My approach:
Let $u=\sup \{f(x):\space x\in X\}$ and $v=\sup \{g(x):\space x\in X\}$
$$u\geq f(x)\space\forall\space x\in X$$
$$v\geq g(x)\space\forall\space x\in X$$
$$\therefore u+v\geq f(x)+g(x)\space\forall\space x\in X$$
Thus we can imply that $u+v$ is an upper bound of $f(x)+g(x)$
Let $w$ be another upper bound for $f(x)+g(x)$
$$\therefore w>f(x)+g(x)\space \forall \space x\in X$$
$$w-g(x)>f(x)\space \forall \space x\in X$$
$\therefore w-g(x)$ is an upper bound for $\{f(x):\space x\in X\}$
$$\therefore w-g(x)>u\space \forall \space x\in X$$
$$w-u>g(x)\space \forall \space x\in X$$
Thus we can imply that $w-u$ is an upper bound for $\{g(x):\space x\in X\}$
$$\therefore w-u>v\Rightarrow w>u+v$$
Now since $w$ is arbitrary, we can imply that any upper bound for $\{f(x)+g(x):\space x\in X\}$ would be greater than $u+v$
Thus we can imply that $u+v=\sup \{f(x)+g(x):\space x\in X\}$
$\therefore \sup \{f(x)+g(x):\space x\in X\}=\sup \{f(x):\space x\in X\}+\sup \{g(x):\space x\in X\}$ which clearly is not correct.
I know that all this effort was irrelevant and unnecessary for this question but please help me find the mistake in this solution
THANKS
Best Answer
First, you already know that the proof was complete once you reached $u+v$ is an upper bound of $\{f(x)+g(x):x \in X\}.$
The problem in your attempt to prove the converse is concluding $w -g(x)$ is an upper bound for $\{f(x):x \in X\}.$ For a fixed $x,$ you have $w-g(x)\geq f(x)$ but you may not have $w-g(x)\geq f(y)$ for some $y \neq x.$ Therefore you cannot say $w-g(x)$ is an upper bound of $\{f(y):y \in X\}$.