Range of curve on a compact interval is nowhere dense 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?
 A: It seems to me that the $C^1$ condition is almost a red herring, in that it is much more restrictive than weaker conditions that are more immediately linked to what you want to show. For example, suppose the curve is locally linear at each of its points. Then, given any point $x$ on the curve, all sufficiently small disks centered at $x$ (FYI, "there exist arbitrarily small disks centered at $x$" would suffice) have the property that the portion of the curve within that disk stays within a rectangular strip centered on a diameter of the disk and whose width relative to the diameter of the disk can be made arbitrarily small (FYI, "strip width less than disk diameter" would suffice, even if this ratio approaches $1$ as the disk diameter approaches $0$). That is, for all sufficiently small disks centered at $x$, the curve does not wander all around the disk, but instead the curve stays uniformly close to a diameter of the disk. It immediately follows that every neighborhood of $x$ contains a sub-neighborhood disjoint from the curve. (Think about how easy it is to find a sub-neighborhood of $\{(x,y) \in {\mathbb R}^{2}: x^2 + y^2 < 1\}$ that is disjoint from $\{(x,y) \in {\mathbb R}^{2}: |y| < 10^{-6}\}.)$ Finally, depending on what your definition of "nowhere dense" is, you might need to finish up by showing the property I introduced, which could be called "locally nowhere dense at each of its points", implies the set is globally nowhere dense.
A: 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.
A: If the curve is defined as the continuous image of I= [0,1] (instead of (0,1) ), then, if the image contained an open set, you then have a bijection between compact and Hausdorff, so that you have an embedding of I into its image. Then, if the image contained a subset S of measure >0 , you would get a relatively open subset of (0,1) to map into a relatively-open subset of S, which violates invariance of domain.
