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?

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.
