There is no $C^1$ function $f$ mapping an open interval in $\mathbb{R}$ onto open ball in $\mathbb{R}^2$ 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? 
 A: 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.
