Tumbleweed

$S$ is the set of all numbers of form $\frac{1}{n}$ , where $n$ is a positive integer. Hence the answer is $2520 \times (\frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{10}) = 7381$ .

First let's assume $c=\frac{1}{n}$ for some integer $n \geq 2$ (case $n=1$ is trivial) and let $f$ be any function satisfying the problem description. Let $g(x) = f(x+c)-f(x)$ be defined for $x \in [0, 1-c]$ .

Note that $g(0) + g(c) + g(2c) \ldots + g((n-1)c) = f(1) - f(0) = 0$ (by telescoping). Hence there are two values $a, b$ such that $g(a) \geq 0$ and $g(b) \leq$ (otherwise the sum above would be strictly negative or strictly positive). As $g$ is continuous, Darboux property tells us that it is equal to zero somewhere.

Now, let's assume $c$ is not of form $\frac{1}{n}$ . Let $n$ be the largest integer such that $cn < 1$ . Let $r = 1 - cn < c$ .

I will define $f$ in such a way that $f(x+c) > f(x)$ for all $x \in [0, 1-c]$ . To define the value of $f$ for $x$ I will express $x=ac+b$ where $a$ is the largest integer that $ac \leq x$ , hence $0 \leq b < c$ .

Let $f(ac + b) = a - n \frac{b}{r}$ , if $b \leq r$ and $f(ac+b) = a - n + (n+1) \frac{b-r}{c-r}$ otherwise. The only non-trivial issues to check are:

1) $f$ is continuous where $b=0$ . To check this, note that $f(ac) = a$ regardless if we use $b=0$ in the first definition or $b=c$ in the second definition.

2) $f$ is continuous where $b=r$ . To check this note that $f(ac + r) = a - n$ regardless if we use $b=r$ in the first definition or in the second definition.

3) $f(x+c) \geq f(x)$ for all $x$ . To check this, note that if $a$ is increased by 1 and $b$ does not change, the value of $f$ increases by 1.

The function defined by Paul Walter is a little difficult to visualize. I am including below the graph of the function $f(x)$ when $c=0.3$