Consider the set C of real continuous functions defined on [0,1] such that f(0)=f(1). For each f∈C, we may consider the set A(f)={a∈[0,1]:f(x+a)=f(x) for some x∈[0,1−a]}.
Then ⋂f∈CA(f)={1,…,1n,…,0}
It is easy to show that 1n∈A(f) for each f∈C by partitioning [0,1] in n parts of equal length and using intermediate value theorem for the auxiliary function given by g(x)=f(x+1n)−f(x). Thus we have ⋂f∈CA(f)⊃{1,…,1n,…,0}.
How to prove the converse?
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