## Are there jordan curves with non-zero area?

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### Are there jordan curves with non-zero area?

When one of the professors who specializes in a math field (I can't remember which one) related to this question was asked, he said "I don't know."

### Re: Are there jordan curves with non-zero area?

methinks no. The usual space-filling curves (peano curves) aren't Jordan curves, it seems like being a Jordan curve doesn't give one enough flexibility to have positive measure.

- scowdich
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### Re: Are there jordan curves with non-zero area?

I thought Jordan Curves were defined as non-overlapping loops, implying that they all have nonzero area. Have I missed something?

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### Re: Are there jordan curves with non-zero area?

scowdich wrote:I thought Jordan Curves were defined as non-overlapping loops, implying that they all have nonzero area. Have I missed something?

Yes; most loops have zero area (as opposed to the area contained in the loop, which of course has nonzero area).

What is the area of the boundary of a circle?

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

- scowdich
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### Re: Are there jordan curves with non-zero area?

Well, if you drew it with a really dull pencil...

### Re: Are there jordan curves with non-zero area?

I have little knowledge in this area so forgive my misunderstandings. Isn't a hilbert curve non-overlapping?

- jestingrabbit
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### Re: Are there jordan curves with non-zero area?

The hilbert curve meets itself on x=1/2 or y=1/2 or any of a whole bunch of other places.

A space filling curve must either meet itself or not be continuous. I forget how to prove it, but I'm pretty sure that if you take the symmetric difference between an open set and its closure then you have a set of measure zero. Every Jordan curve can be realised as such a set I believe.

A space filling curve must either meet itself or not be continuous. I forget how to prove it, but I'm pretty sure that if you take the symmetric difference between an open set and its closure then you have a set of measure zero. Every Jordan curve can be realised as such a set I believe.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

### Re: Are there jordan curves with non-zero area?

Apparently the answer is yes:

http://links.jstor.org/sici?sici=0002-9947(190301)4%3A1%3C107%3AAJCOPA%3E2.0.CO%3B2-T

http://links.jstor.org/sici?sici=0002-9947(190301)4%3A1%3C107%3AAJCOPA%3E2.0.CO%3B2-T

- jestingrabbit
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### Re: Are there jordan curves with non-zero area?

Holy crap. I think I can construct one now.

Start with a sequence {a

Then give A some width by making B=A*[0,1] (where * is cartesian product) and similarly construct B

To get the curve, make a series of hats for the B

If you do that right I believe you'll get something continuous with a positive exterior measure. Close the curve if you must.

Edit: this doesn't work.

Start with a sequence {a

_{i}} (index starting at 0) such that Sum( 2^{i}a_{i}) < 1. Then let {A_{i}} be a sequence of subsets of the interval [0,1], such that A_{0}=[0,1] and construct A_{i+1}from A_{i}by cutting out an inteval of length a_{i}from each inteval that makes A_{i}. Let A be the intersection of the A_{i}'s. This is the usual contruction of a Cantor discontinuum with positive measure.Then give A some width by making B=A*[0,1] (where * is cartesian product) and similarly construct B

_{i}'s.To get the curve, make a series of hats for the B

_{i}'s. See the attached really crappy illustration.If you do that right I believe you'll get something continuous with a positive exterior measure. Close the curve if you must.

Edit: this doesn't work.

Last edited by jestingrabbit on Tue Oct 16, 2007 11:04 am UTC, edited 2 times in total.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

### Re: Are there jordan curves with non-zero area?

@jestingrabbit:

As I recall, a Jordan curve in R

Your curve seems to be defined as a limit of Jordan curves f

Can you actually show that f

If that doesn't hold then your construction is not a curve at all, never mind an injective curve.

Looking at it I don't believe there's any way you can make that work.

Informally: your curves "jump" too far with each new step. The size of the jumps has to shrink to give a correctly defined limit curve.

Additional problem: the resulting set is a countable union of sets with measure zero in R

As I recall, a Jordan curve in R

^{2}is defined as a continuous injective map f: [0,1] -> R^{2}with f(0) = f(1).Your curve seems to be defined as a limit of Jordan curves f

_{1}, f_{2}, f_{3}, ...Can you actually show that f

_{n}(t) has a defined limit for all t in [0,1]?If that doesn't hold then your construction is not a curve at all, never mind an injective curve.

Looking at it I don't believe there's any way you can make that work.

Informally: your curves "jump" too far with each new step. The size of the jumps has to shrink to give a correctly defined limit curve.

Additional problem: the resulting set is a countable union of sets with measure zero in R

^{2}, so it has measure zero anyway.- jestingrabbit
- Factoids are just Datas that haven't grown up yet
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### Re: Are there jordan curves with non-zero area?

Informally, at the nth step you have 2

However, I think there's another problem. The part of the curve that intersects y=0 or y=1 has measure 0, so we only need to determine the area of the collection of vertical lines. At each step, a finite number of lines is added. So there are only a countable number of lines in total, and so the area of that curve is necessarily 0.

The idea must be to create a curve whose image contains no open sets, but which still has positive measure. If the image did contain an open set that would imply a homeomorphism between the line and a two dimensional space which is impossible (as those sets aren't homeomorphic).

Back to the drawing board.

Edit: didn't read your further problem bit before I posted.

^{n}straight bits that need to be replaced by U's. You can clearly do that without redefining any of the rest of the curve, and after you've put the U bit there, you wont ever need to do anything to it again. So, you could define f(t) as being f_{N}(t) where N is such that for all n>N, f_{N}(t)=f_{n}(t). To make f(0) = f(1) you can tack on part of a circle by reparametrizing the segment which intersects x=1.However, I think there's another problem. The part of the curve that intersects y=0 or y=1 has measure 0, so we only need to determine the area of the collection of vertical lines. At each step, a finite number of lines is added. So there are only a countable number of lines in total, and so the area of that curve is necessarily 0.

The idea must be to create a curve whose image contains no open sets, but which still has positive measure. If the image did contain an open set that would imply a homeomorphism between the line and a two dimensional space which is impossible (as those sets aren't homeomorphic).

Back to the drawing board.

Edit: didn't read your further problem bit before I posted.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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### Re: Are there jordan curves with non-zero area?

Yeah, your curves definitely converge to some function, which can be shown to be injective, but the curves don't converge uniformly, so there's no reason to believe your function is continuous.

I think you were on the right track when you said

Every Jordan curve is a continuous image of a compact set, hence compact, hence closed. The Jordan curve theorem states that every Jordan curve C in the plane separates its complement into two distinct connected components, which must both be open sets since they are connected components of the open set R

I think you were on the right track when you said

Jestingrabbit wrote:I forget how to prove it, but I'm pretty sure that if you take the symmetric difference between an open set and its closure then you have a set of measure zero. Every Jordan curve can be realised as such a set I believe.

Every Jordan curve is a continuous image of a compact set, hence compact, hence closed. The Jordan curve theorem states that every Jordan curve C in the plane separates its complement into two distinct connected components, which must both be open sets since they are connected components of the open set R

^{2}\C. So if we call the interior (the bounded component) U, and take the union of C and U to be K, then C is K\U where K is compact and U is the interior of K. So if it's possible to show that such things have measure 0, we're done.I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

- jestingrabbit
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### Re: Are there jordan curves with non-zero area?

I think I've got it now.

The places where the Hilbert curve (for instance) is self intersecting are on the lines x=a or y=a where a is a dyadic fraction. If you can arrange it so that the image is 'mostly' inside something like A

@skep: I don't think that theorem I half remembered is right (or I forgot some of the hypothesis), and if it is then that 1903 paper that someone linked to is wrong.

edit: if someone could get the rest of that paper I for one would appreciate it.

edit2: I think you're both right about my earlier attempt - its not a continuous curve I was describing (I don't think the image is closed).

edit3: Here's a picture of what I mean.

The places where the Hilbert curve (for instance) is self intersecting are on the lines x=a or y=a where a is a dyadic fraction. If you can arrange it so that the image is 'mostly' inside something like A

^{2}(ie apart from bits to join the sections together), where A is as in my post four posts up, then I think that will work. It will be continuous and it will contain A^{2}. Again, close the curve in an arbitrary way.@skep: I don't think that theorem I half remembered is right (or I forgot some of the hypothesis), and if it is then that 1903 paper that someone linked to is wrong.

edit: if someone could get the rest of that paper I for one would appreciate it.

edit2: I think you're both right about my earlier attempt - its not a continuous curve I was describing (I don't think the image is closed).

edit3: Here's a picture of what I mean.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

### Re: Are there jordan curves with non-zero area?

- jestingrabbit
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### Re: Are there jordan curves with non-zero area?

Looks like pretty much the same idea but using a Peano curve, instead of a Hilbert (and, you know, with proofs and stuff).

Thanks for posting.

Thanks for posting.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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### Re: Are there jordan curves with non-zero area?

jestingrabbit wrote:@skep: I don't think that theorem I half remembered is right (or I forgot some of the hypothesis), and if it is then that 1903 paper that someone linked to is wrong.

Ah, missed that link. You must have been misremembering that theorem, because the Jordan curve constructed in the 1903 paper is clearly the difference between it union its interior (which is closed) and its interior (which is open).

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

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