# A unit speed differentiable curve which avoids a null set of directions

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

## 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 $$\gamma$$.

• 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 Sep 21 at 11:28
• You've earned it, great answer! Sep 22 at 10:13
• This answers the question perfectly for me, and generalizes to the case when each $N_x$ is only assumed to be countable. Cheers. 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.

• 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 Sep 22 at 10:12
• @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 $p$. Sep 22 at 10:44
• Yes I got it. As I said, your answer is great Sep 22 at 10:46
• Why would $c'(t)$ be continuous? The question was about "unit speed differentiable curve" not "unit speed continuously differentiable curve", right? Sep 22 at 16:06
• 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. Sep 22 at 16:59