Does there exist any continuous curve $f:[0,1]\to\mathbb{C}$ for which $f(0)=0,f(1)=1$ and for which there is no pair of points $p,q\in f([0,1])$ such that $q-p=0.75$?
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$$f(x)=x + i \sin(2 \pi x ) $$
Milo Brandt
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If you want a visualization: Consider that we make curves $f$ and $f+0.75$ simultaneously; the latter must pass "above" $1$ (so as not to block $f$) and the former must avoid the latter. So we go up to make $f+0.75$ go over $1$ and dive so that $f$ avoids $f+0.75$. I think similar constructions probably prove that there do not have to points where $f(x)-f(y)=c$ for any $c\neq 0$. – Milo Brandt Apr 24 '15 at 01:22
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Nice construction, Meelo. Elegant too. – Berrick Caleb Fillmore Apr 24 '15 at 01:27