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

This is a crosspost from MSE.

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 $$\gamma:[0,1]\to[0,1]^2$$.

I came across (something similar to) this question while thinking about this one.

• I answered the original polyomino question (which corresponds to arbitrary rectifiable curves). The smooth curve version is a bit harder and I don't immediately see how to handle it, so it remains open at the moment :-) Commented May 24, 2022 at 4:37
• Cool! I'll try to understand the answer in detail when I have time Commented May 24, 2022 at 12:08
• $A$ being open and by a compactness argument, couldn't we chose a finite set of points that $\gamma$ needs to go through for $\gamma+A$ to contain a closed ball of radius larger than $\epsilon$? Apologies, I know I'm missing something, but what is it? Commented Apr 20, 2023 at 11:40
• @YaakovBaruch maybe no curve of length $\leq1$ can pass through all the points in that finite set Commented Apr 20, 2023 at 12:52
• Ah, that length requierement slipped beteeen my eyes. Thank you. Commented Apr 20, 2023 at 15:10

The answer is no: if $$\varepsilon$$ is small enough, then for every open $$A \subset [0,1]^2$$ of measure at least $$1/100$$, there exists a smooth curve $$\gamma$$ of length $$\leq 1$$ such that $$\gamma+A$$ contains a $$\varepsilon$$-ball. The idea is to randomly shift by translates first at extremely small scales to cover a lot of balls (or squares) at that scale, then at less small scales to cover a lot of balls (or squares) at this larger scale, and so forth until one is able to cover a macroscopic ball or square.

First some reductions. By enlarging $$\varepsilon$$ a little bit, we can assume it is a negative power of two, and replace "$$\varepsilon$$-ball" by "dyadic $$\varepsilon$$-square". By inner regularity, $$A$$ contains a finite union of balls of measure at least $$1/200$$, hence contains a finite union $$A'$$ of closed dyadic squares of measure at least $$1/400$$. If $$N$$ is large enough, we can view $$A'$$ as a union of dyadic $$2^{-N}$$-squares, and assume that the $$2^{-N}$$-neighbourhood of $$A'$$ lies in $$A$$. It will then suffice to find a piecewise polygonal path $$\gamma'$$ of length at most $$1/2$$ such that $$\gamma'+A'$$ contains a dyadic $$\varepsilon$$-square, since one can smooth $$\gamma'$$ out at a scale much less than $$2^{-N}$$ to find a smooth curve $$\gamma$$ of length at most $$1$$ such that $$\gamma+A \subset \gamma'+A'$$.

Now introduce the hyperdyadic scales $$\varepsilon_n := \varepsilon^{2^n}$$ and integers $$J_n := \lfloor \log^{100} \frac{1}{\varepsilon_n} \rfloor$$. Clearly there is $$n_0$$ such that $$\varepsilon_{n_0+1} \leq 2^{-N}$$, hence $$A'$$ is now a union of dyadic $$\varepsilon_{n_0+1}$$-squares. We will choose, for each $$n=0,\dots,n_0$$, a set $$H_n$$ of $$J_n$$ shifts in the ball $$B(0, 10 \varepsilon_n)$$. By an easy induction on $$n_0$$, it is possible to find a piecewise polygonal path $$\gamma'$$ passing through the sumset $$H_0 + H_1 + \dots + H_{n_0}$$ (which is a finite set of cardinality at most $$J_0 \dots J_{n_0}$$) of length at most $$\sum_{n=0}^{n_0} 20 \varepsilon_n J_0 \dots J_n$$ which one can calculate to be less than $$1/2$$ if $$\varepsilon$$ is small enough (the double exponential decay of the $$\varepsilon_n$$ beats the quadratic exponential growth of the $$J_0 \dots J_n$$). Hence it will suffice to locate $$H_0,\dots,H_{n_0}$$ such that $$H_0 + H_1 + \dots + H_{n_0} + A'$$ contains a dyadic $$\varepsilon$$-square.

We will establish the more general claim that for any $$0 \leq n \leq n_0+1$$, we can select $$H_n,\dots,H_{n_0}$$ such that $$H_n + \dots + H_{n_0} + A'$$ contains a union of dyadic $$\varepsilon_{n}$$-squares in $$[0,1]^2$$, of total measure at least $$\frac{1}{400} - 1000^{-n-1}$$; setting $$n=0$$ will give the claim.

We prove this claim by downward induction on $$n$$. When $$n=n_0+1$$ the claim follows since $$A'$$ itself is a union of dyadic $$\varepsilon_{n_0+1}$$ squares in $$[0,1]^2$$ of measure at least $$\frac{1}{400}$$. Now assume inductively that $$0 \leq n \leq n_0$$ and the claim has already been proven for $$n+1$$, thus we have already located $$H_{n+1},\dots,H_{n_0}$$ such that $$H_{n+1} + \dots + H_{n_0} + A'$$ contains a union $$B$$ of dyadic $$\varepsilon_{n+1}$$-squares in $$[0,1]^2$$, of total measure at least $$\frac{1}{400} - 1000^{-n-2}$$.

We cover $$B$$ by dyadic $$\varepsilon_n$$-squares. If we let $$B'$$ be the union of all the dyadic $$\varepsilon_n$$-squares in which $$B$$ has relative density at least $$1000^{-n-2}$$, then $$B \backslash B'$$ has measure at most $$1000^{-n-2}$$, hence $$B'$$ (which contains $$B \cap B'$$) has measure at least $$\frac{1}{400} - 2 \times 1000^{-n-2}$$. We now use the probabilistic method, picking $$H_n$$ to be $$J_n$$ random elements of $$B(0,10\varepsilon_n)$$. A standard union bound calculation (discretising each $$\varepsilon_n$$-square at scale $$\frac{1}{10} \varepsilon_{n+1}$$, say into a lattice of cardinality $$O( (\varepsilon_n/\varepsilon_{n+1})^2 )$$) shows that each $$\varepsilon_n$$-square in $$B'$$ will lie in $$H_n+B$$ with probability at least $$1 - O( (\varepsilon_n/\varepsilon_{n+1})^2 ( 1 - 1000^{-n-3} )^{J_n} )$$ which by the choice of parameters can be seen to be at least $$1 - 1000^{-n-2}$$. Thus, by linearity of expectation (first moment method), one can choose $$H_n$$ so that $$H_n+B$$ covers a subcollection of $$\varepsilon_n$$-squares in $$B'$$ of measure at least $$\frac{1}{400} - 1000^{-n-1}$$, closing the induction.

• Thanks! Nice argument! (In case someone else misinterprets this, $\log^{100}\frac{1}{\varepsilon_n}$ does not mean iterating the $\log$ function $100$ times) Commented Apr 20, 2023 at 20:49
• @SaúlRM I have never seen the $\log^{100}$ to denote the $100$-fold iterated logarithm; I have sometimes seen $\log_{100}$ to denote it. Commented Apr 22, 2023 at 18:51
• @mathworker21 It is common in complex dynamics to use superscript $^n$ for the n-fold iteration. For example $\exp^2(z)=\exp(\exp(z))$. Woudn't $\log_{100}$ be the log base 100? Commented Apr 22, 2023 at 19:20
• @D.S.Lipham "Woudn't $\log_{100}$ be the log base 100?" It is, for any given paper, until I see the notation section :) . Don't shoot the messenger. Commented Apr 22, 2023 at 19:32
• @TerryTao My brain has slowed over the years so it's going to take me a bit more time to finish understanding every detail, but I'm pretty sure that "closed dyadic squares of measure at most 1/400" is a typo and you meant "at least". Or am I totally lost? (And I can't correct it by editing and rely on reviewers to prevent me from embarrassing myself because of limitations of the StackExchange software framework.) Commented May 11, 2023 at 21:13