Problem in Supremum of cosets .

Question

Let $X$ be a nonempty set and let $f\colon X\rightarrow\mathbb{R}$ have bounded range in $\mathbb{R}$. If $a\in\mathbb{R}$, show that $$\sup(a+f(x))=a+\sup(f(x)).$$ I know above condition but facing it hard to prove.

Answer 1

Intuitively, it means that, if you move graph of $f$ vertically, then all bounds also move.

More general, if $A\subseteq\mathbb{R}$ is bounded above, then $a+\sup A=\sup(a+A)$:

First, $a+b\le a+\sup A$ for all $b\in A$. Therefore $\sup(a+A)\le a+\sup A$.

Now, supose not equality. Then, $\sup(a+A)-a<\sup A$. Thus, there is $b\in A$ such that $\sup(a+A)-a<b\le\sup A$. But $b=a+b-a\le\sup(a+A)-a<b$. This contradiction shows that inequality is not possible.

Then, in particular, if $A=\{f(x)\}$, the result follows.