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We know that there are no $C^1$ functions which map open $E\subset \mathbb{R}^2$ INTO $\mathbb{R}$. (Actually we can drop the $C^1$ requirement and just use continuity). A nice related question is that There is no $C^1$ function $f$ mapping an open interval in $\mathbb{R}$ ONTO open ball in $\mathbb{R}^2$. But I can't prove it, any suggestion?

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is it ONTO or INTO? – Mercy King Aug 6 '12 at 1:44
Dear Mercy, the second part is Onto. – KWO Aug 6 '12 at 10:12
sard? sard? sard? – roy smith Oct 9 '15 at 19:51
up vote 3 down vote accepted

For the first question: if $f: E \to \mathbb R$ is continuous and one-to-one, let $B$ be an open ball with $\overline{B} \subset E$. If $x \in \partial B$, then since $x$ is a limit point of $B$ and of $E \backslash \overline{B}$, $f(x)$ must be in $\partial f(B)$. But $B$ is connected so $f(B)$ is connected (i.e. an interval in $\mathbb R$), and an interval has only two boundary points.

For the second question: a $C^1$ function $f$ is locally Lipschitz, and therefore for $E \subseteq \mathbb R$ the Hausdorff dimension of $f(E)$ is no greater than the Hausdorff dimension of $E$, and thus at most $1$.

EDIT: Here's a somewhat less sophisticated version. Consider a closed bounded interval $J$ on which your $C^1$ function $f$ is defined. Then $|f'|$ is bounded on $J$, say be $K$, and for any $x,y \in J$ we have $|f(x) - f(y)| \le K |x - y|$. Let $J$ have length $L$. For any positive integer $n$, we can split up $J$ into $n$ subintervals $J_j$ of length $L/n$, and the images of these under $f$ are $n$ sets of diameter at most $KL/n$. But for any $r > 0$ and positive integer $m$, a square of side $r$ in ${\mathbb R}^2$ contains $m^2$ points (forming a regular grid) whose distance from each other is at least $r/m$. If $n$ is sufficiently large, we can take $m$ so $KL/n < r/m$ but $m^2 > n$, and thus $f(J)$ can't contain any square. Therefore $f(J)$ can't contain any nonempty open set.

To complete the proof, you can use the Baire Category Theorem: if $I$ is an open interval, $f(I)$ is the union of countably many $f(J)$ for closed bounded intervals $J$, but
$f(J)$ is closed and nowhere dense, so the union of these can't contain any nonempty open set.

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Dear Sir, I have solved the first question, the second question is the one which I get stuck. Thanks for your answer but I am not sophisticated enough to understand Hausdorff dimension. Is there a more elementary suggestion? – KWO Aug 5 '12 at 23:55
Thanks sir, I understand it now. Thanks for your help. – KWO Aug 6 '12 at 10:02

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