The domain of the function $f$ is the set of real numbers. For every number $a$, if $a<x<a+100$, then $a \leq f(x) \leq a+100$. Determine the function $f$!
The condition intuitively means that $f$ maps every open interval of length 100 into the same closed interval of length 100. This is clearly satisfied by the function $f(x)=x$, since in this case the image of any set is identical to the set itself.

We will show that there are no other solutions to the problem. If $g(x) \neq x$, then there exists some $x_{1}$ such that $g\left(x_{1}\right) \neq x_{1}$. The proof now hinges on the fact that different numbers can be "separated" by an open interval of length 100, meaning there is an open interval of length 100 that contains $x_{1}$ but not $g\left(x_{1}\right)$—and $g\left(x_{1}\right)$ is different from the endpoints of the interval. The diagram shows the cases where $x_{1}>g\left(x_{1}\right)$ and $x_{1}<g\left(x_{1}\right)$.
![](https://cdn.mathpix.com/cropped/2024_05_02_43c1b81d6f8c26343d28g-1.jpg?height=232&width=1020&top_left_y=472&top_left_x=542)

In this case, the condition does not hold for such intervals, so if there is an $x_{1}$ such that $g(x)_{1} \neq x_{1}$, then $g$ cannot be a solution.