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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?

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    Have a look at this answer: http://math.stackexchange.com/questions/69915/why-isnt-mathbbr2-a-countable-union-of-ranges-of-curves/69970#699702011-10-05
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    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.2011-10-05
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    On the other hand, a *closed* set of measure zero is necessarily nowhere dense.2011-10-05
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    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.2011-10-05
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    @GeorgeLowther, thanks, this seems like something most understandable to me. If you have the time, would you mind expanding it into a full answer so I can accept? Maybe it's obvious, but why does a curve on a compact/finite interval have necessarily have finite length in $\mathbf{R}^2$?2011-10-06
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    @Pierre A curve on a finite interval need not have finite length in $\mathbb R^2$ (take, for example, a [space-filling curve](http://en.wikipedia.org/wiki/Space-filling_curve)). However, a $C^1$ curve $\gamma$ on an interval $[a,b]$ must, because its length can be computed directly as $\int_a^b|\dot{\gamma}(t)|\,dt$. See, for instance, [this Wikipedia snippet](http://en.wikipedia.org/wiki/Curve#Lengths_of_curves).2011-10-08
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    @GeorgeLowther Would you consider posting your argument as an answer to the [question](http://math.stackexchange.com/questions/69915/why-isnt-mathbbr2-a-countable-union-of-ranges-of-curves) 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).2011-10-08
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    @Nick: I just did, thanks.2011-10-09
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    @George Thank _you_ for sharing that trick!2011-10-09

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