Possible Duplicate:
Horizontal chord of length $\frac{1}{2}$ in the graph of a continuous function.
Consider the set $C$ of real continuous functions defined on $[0,1]$ such that $f(0) = f(1)$. For each $f \in C$, we may consider the set $A(f) = \{a \in [0,1] : f(x+a) = f(x) \mbox{ for some } x \in [0,1-a]\}$.
Then $$\bigcap\limits_{f \in C} A(f) = \left\{1, \ldots, \frac {1}{n}, \ldots, 0\right\}$$
It is easy to show that $\displaystyle\frac 1n \in A(f)$ for each $f \in C$ by partitioning $[0,1]$ in $n$ parts of equal length and using intermediate value theorem for the auxiliary function given by $\displaystyle g(x) = f\left(x+\frac {1}{n}\right) - f(x)$. Thus we have $\displaystyle\bigcap\limits_{f \in C} A(f) \supset \left\{1, \ldots, \frac 1n, \ldots, 0\right\}$.
How to prove the converse?
