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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.

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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 $0$.

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

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    $\begingroup$ 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 :) $\endgroup$ – onurcanbektas Jul 16 '18 at 13:52
  • $\begingroup$ 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. $\endgroup$ – Aweygan Jul 16 '18 at 14:02
  • $\begingroup$ 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 ? $\endgroup$ – onurcanbektas Jul 16 '18 at 15:50
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    $\begingroup$ 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. $\endgroup$ – Aweygan Jul 16 '18 at 15:59
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    $\begingroup$ You're welcome, glad to help! $\endgroup$ – Aweygan Jul 16 '18 at 17:08

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