TITLE: Range of curve on a compact interval is nowhere dense
QUESTION [7 upvotes]: I glanced through this question on why $\mathbf{R}^2$ is not of the first category. 
I understand how this would follow if the image of a curve on a compact/finite interval in $\mathbf{R}$ is nowhere dense in $\mathbf{R}^2$. I didn't understand any of the answers, since I haven't learned any measure theory. Also, I browsed through the referenced text, and this question appears before any measure theory is introduced. 
Is there a proof that the image of a $C^1$ curve on compact/finite (one or the other) interval is nowhere dense in $\mathbf{R}^2$ that only uses ideas from general topology, and not measure theory?

REPLY [15 votes]: A $C^1$ curve $\gamma\colon[0,1]\to\mathbb{R}^2$ must be nowhere dense by the following argument.
First, $\gamma$ has finite length $\int_0^1\vert\gamma^\prime(t)\vert\,dt$.
Next, for any positive integer $n$, consider the $(n+1)^2$ points $(i/n,j/n)$ for $0\le i,j\le n$. These all lie in the unit square $[0,1]^2$, and each pair of points is distance at least $1/n$ apart. So, a curve passing through them all has length at least $((n+1)^2-1)/n=n+2$. Letting $n$ go to infinity, we see that a finite length curve cannot map onto $[0,1]^2$.
So, the image of $\gamma$ cannot contain the unit square and, by scaling, it cannot contain any nonempty open subset of $\mathbb{R}^2$. As the image is closed, this means that it is nowhere dense.
Note: I also used this argument here.