Can we find for any given $\varepsilon>0$ an open subset $A\subseteq[0,1]^2$ with measure $>\frac{1}{100}$ such that, for any smooth curve $\gamma:[0,1]\to\mathbb{R}^2$ of length $1$, the set $\gamma+A=\{\gamma(t)+a;t\in[0,1],a\in A\}$ does not contain any balls of radius $\varepsilon$?

I wouldn't mind changing the $\frac{1}{100}$ for any other positive constant. Also, I ask about smooth curves but it may make more sense to consider in general $1$-Lipschitz functions $\gamma:[0,1]\to[0,1]^2$.

For context, a positive answer to this question could be useful for this other question. But the question is also interesting in itself, of course.

  • $\begingroup$ If $A$ is open, it contains an open ball of radius $x$, now if we take $\gamma(t) = \sin\frac{20 \pi t}{x}$, $\gamma +A$ fills unit square (as moving even just this ball along $\gamma$ fills it). I guess something is wrong with this reasoning, but I can't find what. $\endgroup$
    – mihaild
    May 8 at 8:11
  • $\begingroup$ @mihaild I don't understand very well, your curve $\gamma$ is $[0,1]\to\mathbb{R}$, not $[0,1]\to\mathbb{R}^2$, and I don't think it has length $1$ $\endgroup$
    – Saúl RM
    May 8 at 8:20
  • $\begingroup$ Thanks, that's what I missed: the restriction on length. $\endgroup$
    – mihaild
    May 8 at 8:25
  • $\begingroup$ Btw, I am not sure why the "multivariable calculus" tag is more appropiate than the "recreational mathematics" one on this question. This is not part of any mathematics curriculum that I know of $\endgroup$
    – Saúl RM
    May 8 at 8:31
  • $\begingroup$ Would you be interested in the case when $\gamma$ has length, say, 10? $\endgroup$
    – Del
    May 9 at 20:26


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