# Is there an open subset $A$ of $[0,1]^2$ with measure $>\frac{1}{100}$ that satisfies this property?

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.

• 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. May 8 at 8:11
• @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 $1$ May 8 at 8:20
• Thanks, that's what I missed: the restriction on length. May 8 at 8:25
• 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 May 8 at 8:31
• Would you be interested in the case when $\gamma$ has length, say, 10?
– Del
May 9 at 20:26