Image of $C^1$ function does not contain open set Let $f:\mathbb{R}\rightarrow\mathbb{R}^2$ be a $C^1$ function. Prove that the image of $f$ contains no open set of $\mathbb{R}^2$.
So say $f(x)=(g(x),h(x))$. Since $f$ is $C^1$, we have that $g'(x),h'(x)$ both exist and are continuous functions in $x$. To show that $f$ contains no open set of $\mathbb{R}^2$, it suffices to show that $f$ does not contain any open ball in $\mathbb{R}^2$. Suppose, for contradiction, that it contains the ball centered at $(a,b)$ with radius $r$. How can I continue?
 A: Let $F:\mathbb{R}^2\rightarrow\mathbb{R}^2$ be given by $F(x,y)=f(x)$. Then every point of $\mathbb{R}^2$ is a critical point of $F$ (that is $\det J(F)(x)=0$ for every $x\in\mathbb{R}^2$, where $J(F)(x)$ denotes the Jacobian of $F$ at $x$). By Sard's theorem, $f(\mathbb{R})=F(\mathbb{R}^2)$ has zero measure. Since non-empty open sets have positive measure, $f(\mathbb{R})$ cannot contain any open set.
A: Here is a more prosaic approach that relies on $f$ being Lipschitz on compact intervals.
Choose an interval $[t_0,t_1]$. Since $f$ is $C^1$, it is uniformly lipschitz on this interval with some rank $L$.
Let $D = f([t_0,t_1])$.
We can use $L$ to find an upper bound on the measure of $D$.
Let $R(\tau_1,\tau_0) = \{ x | \|x-f(\tau_0) \| \le L |\tau_1-\tau_0| \} = \overline{B}(f(\tau_0), L |\tau_1-\tau_0|)$. We have $f([\tau_0,\tau_1]) \subset R(\tau_1,\tau_0)$, and we can estimate $m R(\tau_1,\tau_0) \le L^2 |\tau_1-\tau_0|^2$.
Hence, for any $n$, we can split the interval $[t_0,t_1]$ into $n$ parts to get $D \subset \cup_{k=0}^{n-1} R(t_0+k\frac{t_1-t_0}{n}, t_0+(k+1)\frac{t_1-t_0}{n})$, and so $mD \le n L^2 \frac{1}{n^2} = \frac{L^2}{n}$. Since $n$ is arbitrary, we have $mD = 0$.
Since $\mathbb{R} = \cup_{m} [-m,m]$, we see that $m f(\mathbb{R}) = 0$.
Since $m B(x,\epsilon) >0$ for any $\epsilon >0$, it follows that $f(\mathbb{R})$ can contain no open ball.
