How to prove that if $f:\mathbb{R} \to \mathbb{R}^2$ is of class $C^1$, then $f$ is not onto.

Question

In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that

If $f:\mathbb{R} \to \mathbb{R}^2$ is of class $C^1$, show that $f$ does not carry $\mathbb{R}$ onto $\mathbb{R}^2$. In fact show that $f(\mathbb{R})$ contains no open subset of $R^2$.

I have started with assuming that $f(\mathbb{R})$ contains an open set $U$ of $\mathbb{R}^2$, and by the continuity of $f$, I have argued that $f^{-1}(U)$ is open in $\mathbb{R}$; however, after this point, I am stuck, so I would appreciate any hint or help.

Answer 1

Consider the map$$\begin{array}{rccc}F\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&f(x).\end{array}$$Then $F$ is of class $C^1$. Therefore, since $\mathbb{R}\times\{0\}$ has measure $0$, $F\bigl(\mathbb{R}\times\{0\}\bigr)$ has measure $0$ too (here, I am using the fact that functions of class $C^1$ map sets of measure $0$ into sets of measure $0$, which is lemma 18.1 in Munkres' textbook). But $F\bigl(\mathbb{R}\times\{0\}\bigr)=f(\mathbb{R})$ and a subset of $\mathbb{R}^2$ which contains a non-empty open subset cannot have measure $0$.

Answer 2

Here's another approach.

For each $N\in\mathbb N$, the restriction $f_N$ of $f$ to $[-N,N]$ is Lipschitz. Use this to show that $f_N([-N,N])$ does not contain a rectangle, hence has empty interior. Then since $f(\mathbb R)=\cup_Nf([-N,N])$, the Baire category theorem yields the result.