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

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.

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 $$0$$.

• Well, that was a clever trick :) Thanks a lot for the answer sir. – onurcanbektas Jul 16 '18 at 13:44

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.

• Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :) – onurcanbektas Jul 16 '18 at 13:52
• If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets. – Aweygan Jul 16 '18 at 14:02
• By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ? – onurcanbektas Jul 16 '18 at 15:50
• Since the distance between centers is at least $\frac{A}{n}$, we have $$\frac{A}{n}(n^2-1)\leq\sum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|\leq M\sum_{i=2}^{n^2-1}x_i-x_{i-1}\leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction. – Aweygan Jul 16 '18 at 15:59
• You're welcome, glad to help! – Aweygan Jul 16 '18 at 17:08