Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Have a look at this answer: math.stackexchange.com/questions/69915/… –  Giuseppe Negro Oct 5 '11 at 17:50
    
Just for the protocol, nowhere density has little to do with measure. $\mathbb Q$ is measure zero (heck, it is even countable) but it is still dense; while you can produce unbounded fat Cantor sets with infinite measure which are still nowhere dense. –  Asaf Karagila Oct 5 '11 at 17:52
    
On the other hand, a closed set of measure zero is necessarily nowhere dense. –  Nate Eldredge Oct 5 '11 at 17:54
1  
No curve of finite length can be dense in any part of the plane. You can use the fact that an $N\times N$ grid in the unit square consists of $N^2$ points all distance at least $1/N$ apart. So a curve passing through them all must have length at least $N$. Let $N$ go to infinity. –  George Lowther Oct 5 '11 at 17:58
1  
@GeorgeLowther Would you consider posting your argument as an answer to the question linked to by the OP here as well? I really like your technique, and I think the asker of that question is looking for something along the lines of your argument (he just initiated a bounty for a simpler answer to his question). –  Nick Strehlke Oct 8 '11 at 18:45

3 Answers 3

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.

share|improve this answer
    
Minor correction - the length should be $\int |\gamma'(t)|dt$, shouldn't it? –  Thomas Andrews Mar 5 '13 at 17:46
    
@Thomas: Fixed, thanks. –  George Lowther May 14 '13 at 0:45

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.

share|improve this answer
    
There are space filling curves, those are not $C^1$ curves. Sure, this is not "the best bound" for nowhere dense curves, but it's one that is simple enough to explain to a second year math student. –  Asaf Karagila Oct 5 '11 at 19:00
    
I suspect that "continuous and with finitely many self-intersections" is sufficient, thanks to the Jordan curve theorem. On the other hand, relaxing either of these restrictions will allow the image to be dense. –  Henning Makholm Oct 5 '11 at 19:31

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.

share|improve this answer
    
But nobody said anything about injectivity of the curve. See also space filling curve on Wikipedia. –  t.b. Oct 5 '11 at 20:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.