amparvardi wrote: Let $f(x)$ be a continuous function defined on the closed interval $0 \leq x \leq 1$. Let $G(f)$ denote the graph of $f(x): G(f) = \{(x, y) \in \mathbb R^2 | 0 \leq$$ x \leq 1, y = f(x) \}$. Let $G_a(f)$ denote the graph of the translated function $f(x - a)$ (translated over a distance $a$), defined by $G_a(f) = \{(x, y) \in \mathbb R^2 | a \leq x \leq a + 1, y = f(x - a) \}$. Is it possible to find for every $a, \ 0 < a < 1$, a continuous function $f(x)$, defined on $0 \leq x \leq 1$, such that $f(0) = f(1) = 0$ and $G(f)$ and $G_a(f)$ are disjoint point sets ? No, it's not. Choose for example $a=\frac 12$ so that the two curves are $f(x)$ and $f_1(x)=f(x-\frac 12)$ For $x=\frac 12$, we get $f(x)-f_1(x)=f(\frac 12)$ and for $x=1$, we get $f(x)-f_1(x)=-f(\frac 12)$ And so, since continuous, $\exists u\in[\frac 12,1]$ such that $f(u)=f_1(u)$ In fact it's possible to show that we can find such $f(x)$ for any $a\in(0,1)$ except for $a=\frac 1k$ where $k$ is a positive integer $>1$