Spivak Calculus ed.3 chapter 3 problem 27 (a) The question is:
a) Suppose f (x)= x + 1. Are there any functions g such that f $\circ$ g = g $\circ$ f?
Then the answer provided states:
The condition f $\circ$ g = g $\circ$ f means that g(x) + 1 = g(x + 1) for all x. There are many such g. In fact, g can be defined arbitrarily for 0$\leq$x$<$1, and its values for other x determined from this equation.
I understand the first part of the answer, it is obvious that  g(x) + 1 = g(x + 1). I don't understand however why the interval 0$\leq$x$<$1 is special versus any other interval. Also, what are these many such functions?  I would appreciate any help with this, as I am using the textbook to self-study.
 A: 
I don't understand however why the interval 0$\leq$x$<$1 is special versus any other interval

Because of this $\color{red}1$ in $\,g(x+\color{red}1)=g(x)+1\,$ which, after repeated application, gives: $$g(x) = \lfloor x \rfloor + g\left(\{x\}\right)$$
In other words, the values of $g\left(\{x\}\right)$ univocally define $g(x)$ on the entire real axis. But the fractional part $\{x\} \in [0,1)\,$, so it is enough to define $g$ on $[0,1)$.
For a trivial example, you can define $g$ to be identically $0$ on $[0,1)$, which then gives $g(x) = \lfloor x \rfloor\,$, which obviously satisfies $g(x+1)=g(x)+1\,$.

[ EDIT ]   On a second reading and depending on where the accent lies, the question could also be interpreted as why $[0,1)$ is "special" versus some other interval of length $1$, for example $[13,14)\,$.

The answer to that is that $[0,1)$ is not special in that sense. $g$ could be arbitrarily defined on any interval $[a, a+1)$ for some $a \in \mathbb{R}\,$, then extended to the entire $\mathbb{R}$ using the same idea as above.
