Non-surjectivity of a function

Question

Let $f:[0,1]\to [0,1]^2$ a $C^1$ function.

Show that $f$ is not surjective. Any help would be appreciable.

Answer 1

Say $I=[0,1]$. In fact a Lipschitz function from $I$ to $I^2$ cannot be surjective. People prove this by showing that the area of $f(I)$ must be zero. There are some subtleties there - one can give a much more elementary argument:

Divide $I^2$ into a grid of $N=n^2$ squares, each of size $1/n\times 1/n$. Say $G_n$ is the collection of center points of those squares. So $G_n\subset I^2$, $G_n$ contains $N$ points, and $$|p-q|\ge 1/n\quad(p,q\in G_n, p\ne q).$$

Divide $I$ into $N-1$ intervals $I_1,\dots,I_{N-1}$, each of length $1/(N-1)$. Suppose $f:I\to I^2$ is surjective. A pigeonhole argument shows that there exist two points in $G_n$ that are in the image of the same $I_j$. (To be explicit about the pigeonholes: $G_n\subset\bigcup_{i=1}^{N-1}f(I_i)$, while $G_n$ contains $N$ points...)

So there exist $s,t\in I$ with $|s-t|\le1/(n^2-1)$ and $|f(s)-f(t)|\ge1/n$. Since $$\frac{1/n}{1/(n^2-1)}\to\infty$$this shows that $f$ is not Lipschitz.