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For each $x \in \mathbb{R}^2$, let $N_x$ be a null set of the unit circle $S^1$ (with respect to $1$-dimensional Lebesgue measure).

Further given that for each $u \in S^1$, we have that $u \not \in N_x$ for almost all $x \in \mathbb{R}^2$.

Does there necessarily exist a unit speed differentiable curve $\gamma : (-\varepsilon, \varepsilon) \rightarrow \mathbb{R}^2$ such that:

  • $\gamma'(t) \not \in N_{\gamma(t)}$ for all $t \in (-\varepsilon,\varepsilon)$?
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  • $\begingroup$ I have put a bounty on your very interesting question, I hope you don't mind! $\endgroup$
    – Didier
    Sep 20 at 15:53
  • $\begingroup$ @Didier I do not mind at all, thank you for showing an interest :) $\endgroup$ Sep 20 at 17:33

2 Answers 2

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No.

Here is an example where such curve does not exist.

Let $$N_x = \Bigl\{(u_1,u_2) \in S^1: u_1 = (\pi+q_1)x_1 + q_2\\ \text{or } u_2 = (\pi+q_1)x_2 + q_2;\; q_1,q_2\in \mathbb{Q}\Bigr\}. \tag{1}$$ Then $N_x$ is countable and for each $u\in S^1$ we have $u\in N_x$ only if $x_1$ or $x_2$ takes one of countably many values. Hence, both $N_x$ and $\{x: u\in N_x\}$ are null sets.

Yet no such $\gamma$ exists. In order to do this, we first show the following lemma.

Lemma 1. Let $g\colon (a,b) \to \mathbb{R}$ be a differentiable function satisfying $\forall t \in (a,b)\;\; \dot g(t) \notin \mathbb{Q}$. Then $g$ is affine, i.e. $g(t) = \alpha + \beta t$ for some $\alpha,\beta\in\mathbb{R}$.

Proof. $g$ is continuous because it is differentiable. Consider the set $$S = \left\{\frac{g(c_2) - g(c_1)}{c_2 - c_1}: c_1,c_2 \in (a,b), c_1 < c_2\right\}. \tag{2}$$ As an image of a connected set under a continuous map, $S$ is a connected subset of $\mathbb{R}$. On the other hand, from Mean value theorem $S\cap \mathbb{Q} = \emptyset$. Therefore, $S$ consists of a single point. If we denote that point by $\beta$, $t\mapsto g(t)- \beta t$ is a constant function (from the definition of $S$). $\;\square$

Now, assume to the contrary that such curve $\gamma$ exists. For any $q\in\mathbb{Q}, i \in \{0,1\}$ we can apply the lemma to the function $g_q: t \mapsto \gamma_i(t) - (\pi + q)\int_{0}^t \gamma_i(t') dt'$ to see that the function $\dot g_q: t \mapsto \dot\gamma_i(t) - (\pi + q_1) \gamma_i(t)$ is constant. Since $\gamma_i(t) = g_0(t) - g_1(t)$, we see that $\gamma_i$ is constant, contradicting the assumption of unit speed of $\gamma$.

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  • $\begingroup$ I am going to look at your answer in detail later for the bounty (I don't really have the time right now), but as for now, +1 $\endgroup$
    – Didier
    Sep 21 at 11:28
  • $\begingroup$ You've earned it, great answer! $\endgroup$
    – Didier
    Sep 22 at 10:13
  • $\begingroup$ This answers the question perfectly for me, and generalizes to the case when each $N_x$ is only assumed to be countable. Cheers. $\endgroup$ Sep 22 at 14:24
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Step 1 : We will define $N_o$ where $o=(0,0)$ is the origin in $ \mathbb{R}^2$ : $$N_o = \bigg\{ \bigg(\cos\ t,\sin\ t \bigg) \in \mathbb{S}^1\bigg| t=\frac{2\pi}{2^i}k,\ i,\ k\in \mathbb{Z}_+\bigg\} $$ where $\mathbb{Z}_+$ is a set of nonnegative integers.

For convenience, we call a real number of type $\frac{2\pi}{2^i}k$ a rational type. If not, we call it an irrational type.

Assume that $p$ is a point in $\mathbb{R}$ and we define its Euclidean distance $r= |p|$. If $r$ is rational type, then $R_r$ is $r$-rotation map on $\mathbb{S}^1$ with a clockwise direction. If $r$ is irrational, then $R_r$ is $r$-rotation map with a counter-clockwise direction. We define $T_r$ to be a $r$-rotation map on $\mathbb{S}^1$ with a counter-clockwise direction for any $r>0$.

Then we define a null set at each point $p$ : $$ N_p = R_r(N_o ) $$

Step 2 : Assume that there is a unit speed curve $c$ s.t. a continuous tangent vector $c'(t)$ avoids the given null sets $N_{c(t)}$ for $0\leq t\leq \epsilon$. We will prove that there is no such curve $c$.

By a construction of the null set $N_{c(t)}$ we have uniquely $v(t)\in \mathbb{S}^1 - N_o$ s.t. $$R_{|c(t)|}(v(t) )=c'(t) = T_{|c(t)|} (w(t)) $$

Define $$A = \bigg\{ v(t) \in N_o\bigg| |c(t)| \ {\rm is\ rational} \bigg\},\ B =\bigg\{v(t)\in N_o\bigg| |c(t)|\ {\rm is\ irrational}\bigg\}$$ and $C=\{w(t)\in N_o \}$.

Clearly, $B$ is in $C$. For $v(t)=(\cos\ \phi,\sin\ \phi)\in A$, we have $$c'(t)=\bigg(\cos (\phi -|c(t)| ),\sin (\phi -|c(t)| ) \bigg) $$ and define $$ A'=\bigg\{ \bigg(\cos\ (\phi -2|c(t)|),\sin\ (\phi -2 |c(t) | ) \bigg) \bigg\}$$ Then $A'\cup B=C$ is a connected compact arc in $\mathbb{S}^1$.

Hence for some $t$, $ \phi -2|c(t)| $ should has a rational type which contradicts to the fact that $\phi$ is irrational and $|c(t)|$ is rational. Hence we conclude that $C$ is a point set.

Case 1 - $|c(t)|$ is const : Hence $c$ goes along a circular arc s.t. the number of its tangent vectors is infinite, which is impossible.

Case 2 - $|c(t)|$ is not const : Then $A =\{ \theta_A\},\ B=\{\theta_B\}$ are point sets.

Assume that $0<|c(t_1)| -| c(t_2)|$ is small. If rational type $r$ and irrational type $ r'$ are arbitrarily close to $|c(t_1)|$, then the vectors $e^{i(\theta_A-r)},\ e^{i(\theta_B+r)}$ are arbitrarily close to the tangent vector $c'(t_1)$. Hence $\theta_A-\theta_B =2|c(t_1)| $ up to $\pi$. Hence we have $2|c(t_1)|=2|c(t_2)|$ up to $\pi$, which is impossible. Hence we complete the proof.

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  • $\begingroup$ I had a hard time understanding the whole answer but it seems a great answer to me, although I gave the bounty to Fiktor who was first to answer $\endgroup$
    – Didier
    Sep 22 at 10:12
  • $\begingroup$ @Didier : Basic idea is to construct $N_p$ exactly for all $p\in \mathbb{R}^2$. First we construct $N_o$ where $o$ is the origin s.t. $N_o$ is dense in $\mathbb{S}^1$ and have a measure $0$. For $p\in \mathbb{R}^2$, we define $N_p$ to be $|p|$-rotation with a counter-clockwise or clockwise direction (resp. $|p|$ is irrational or rational) where $|p|$ is Euclidean magnitude of $p$. $\endgroup$
    – HK Lee
    Sep 22 at 10:44
  • $\begingroup$ Yes I got it. As I said, your answer is great $\endgroup$
    – Didier
    Sep 22 at 10:46
  • $\begingroup$ Why would $c'(t)$ be continuous? The question was about "unit speed differentiable curve" not "unit speed continuously differentiable curve", right? $\endgroup$
    – Fiktor
    Sep 22 at 16:06
  • $\begingroup$ Actually, I don't think the solution as currently written is correct even if we assume $c'(t)$ is continuous. The counterexample is $c(t) = (r(t) \cos(\theta(t)), r(t) \sin(\theta(t)))$ with $r,\theta$ computed by differential equations $r'(t) = \cos(r(t)+\varphi - \theta(t)), \theta'(t) = \sin(r(t)+\varphi-\theta)/r(t)$ for some $\varphi$. From counterclockwise condition rational $\varphi$ are disallowed, from clockwise condition for countably many $r$ (i.e. when $r$ is rational) countably many $\varphi$ are disallowed. Uncountably many remain. Here $A$ is not a point set. $\endgroup$
    – Fiktor
    Sep 22 at 16:59

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