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 $t \in (-\varepsilon,\varepsilon)$?
  • $\begingroup$ 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



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 $\gamma$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.