Wikipedia:Reference desk/Archives/Mathematics/2016 May 7

= May 7 =

how can lines/curves of infinite length have positive area?
came up in editing an article...someone described that "There exist nonrectifiable curves of nonzero 2-dimensional Lebesgue measure" and therefore "curves with infinite length can have positive area." this makes no intuitive sense to me...can this be explained in any kind of intuitive manner? (I assume it's true as the editor seems genuinely knowledgeable)..68.48.241.158 (talk) 16:21, 7 May 2016 (UTC)


 * See Osgood curve.  Sławomir Biały  (talk)

Also, do there exist curves/lines of finite length that have area?68.48.241.158 (talk) 16:56, 7 May 2016 (UTC)


 * No, in general a set can only have one finite nonzero Hausdorff measure. See Hausdorff dimension.   Sławomir Biały  (talk) 17:46, 7 May 2016 (UTC)

The articles cited don't even explain technically why these curves have area, as far as I can tell...and they certainly don't offer any kind of intuitive explanation.....?????68.48.241.158 (talk) 18:01, 7 May 2016 (UTC)


 * You are asking for an intuitive explanation, but it might help if you said what point you are starting from. For example, are you already familiar with the Lesbegue measure?  Do you know calculus?  Or do you mean something really elementary like high school math?  Dragons flight (talk) 18:45, 7 May 2016 (UTC)


 * I took calculus in high school, studied humanities/philosophy in college..so not too advanced as far as math proper...and it's been a while since I've done/studied any math proper...I'm wondering if there's any even conceivable way to explain it intuitively...obviously, we're talking about an abstraction here..so such a thing could never exist in the real world/be measured for area in the real world...do all infinitely long curves have area? or only certain kinds? and how/at what point to they become imbued with area?? is this approaching it in a philosophical way that doesn't make much sense??68.48.241.158 (talk) 19:05, 7 May 2016 (UTC)


 * I encourage you to look at the concept of an improper integral first. --Jasper Deng (talk) 19:08, 7 May 2016 (UTC)


 * don't see how helpful as far what I'm asking for...granted, I may be asking the impossible..68.48.241.158 (talk) 19:16, 7 May 2016 (UTC)


 * But that's the most elementary way there is to calculate areas under infinite curves. I won't entertain any "philosophical" discussions of this subject because such discussions are necessarily hand-wavy (i.e. vague and not rigorous).--Jasper Deng (talk) 19:21, 7 May 2016 (UTC)


 * but we're not talking about calculating areas under infinite curves but the area of the curves themselves....?? does the area under the curve somehow become mixed up with the curve itself in particular infinite curves????68.48.241.158 (talk) 19:24, 7 May 2016 (UTC)


 * Then what you are talking about makes no sense. (Also, please stop using excessive question marks, I get it that you're confused, you don't need to have so many). I think the editor probably meant area enclosed, since the area of a one-dimensional curve itself is zero.--Jasper Deng (talk) 19:46, 7 May 2016 (UTC)


 * it now appears you don't understand it either and will be awaiting an explanation as I am! we're talking about one-dimensional infinite curves having area!68.48.241.158 (talk) 19:51, 7 May 2016 (UTC)


 * I agree with the OP - it seems you are unfamiliar with the concept of a Space-filling curve. That is, you can have a continuous function $$f(x):[0,\ 1]\to\mathbb{R}^2$$ such that its image $$f([0,\ 1])$$, a subset of $$\mathbb{R}^2$$, has positive area. -- Meni Rosenfeld (talk) 20:11, 7 May 2016 (UTC)


 * To say that a line or curve (whether finite or infinite) "has" an area seems wrong to me, and I suspect that's just a sloppy way of saying that the line or curve "bounds" an area. Of course a finite curve can bound an area, as in a circle, but also an infinite curve may, as in many fractals, such as the Koch snowflake. StuRat (talk) 20:05, 7 May 2016 (UTC)


 * Stu, what you may be missing is that these "curves" are allowed to self-intersect. Think of taking a blue pen, and coloring in a square entirely blue, by moving it back and forth at random over the area of the square until you've scratched out a solid area.  It's not exactly like that, but it'll do for a start.  There are some nice pictures at space-filling curve that might help. --Trovatore (talk) 20:40, 7 May 2016 (UTC)


 * Yes, that's very much like my illustration down below, filling in a rectangle with an infinite number of infinitely thin V's (which doesn't require self-intersecting curves, BTW). However, the difficulty the OP is having in conceptualizing this is that, unlike the pen, the line segments have no width, so even a huge number of lines should not color in the rectangle even the slightest percentage.  It's difficult to make the jump from this to an infinite amount of lines completely filling it in. StuRat (talk) 21:09, 7 May 2016 (UTC)


 * seems wrong to me too...which is why I asked..but apparently it's somehow true...may need very expert math types to weigh-in (I believe the person who suggested it is a very expert math type)...68.48.241.158 (talk) 20:10, 7 May 2016 (UTC)


 * If we define "curve" as "the image of a continuous function from $$[0,1]$$ to a space (the Euclidean plane in our case)", then a curve can actually have a positive area. -- Meni Rosenfeld (talk) 20:11, 7 May 2016 (UTC)


 * John von Neumann famously said, "Young man, in mathematics you don't understand things. You just get used to them.". Many mathematical objects are pathological and defy our existing intuitions. If we study them hard enough, we develop new intuitions which can accommodate them.
 * I think the best way to gain intuition for space-filling curves is to study in detail the construction of a particular one, such as the Hilbert curve. -- Meni Rosenfeld (talk) 20:17, 7 May 2016 (UTC)


 * thanks, we're finally on to something with the "space-filling curves"...still don't intuitively understand it, but as you suggest may not be possible to....68.48.241.158 (talk) 20:20, 7 May 2016 (UTC)


 * Also, they might mean that an infinite curve approximates an area. Consider the following (poorly drawn) rectangle:

|   |


 * Now what happens if we start filling it in with V shapes ? Here's just 2:

|\/\/|


 * But, if you had an infinite number of infinitely narrow V's, couldn't you say that this approximates filling in the area of the rectangle, much as 0.9999 repeating approximates, and in a sense is, 1.0 ? StuRat (talk) 20:18, 7 May 2016 (UTC)


 * I have a feeling you're off here, Stu, but would have to defer to more expert math folk..68.48.241.158 (talk) 20:22, 7 May 2016 (UTC)


 * Well, if we consider the area as "the set of all points within a boundary", and my example above, with an infinite number of V's, does in fact occupy all points within the (rectangular) boundary, doesn't that meet the formal definition ? StuRat (talk) 20:50, 7 May 2016 (UTC)

what I don't understand is how the curve becomes imbued (so to speak) with area...as each distinct part of the curve would be said to have no area, wouldn't it??68.48.241.158 (talk) 20:30, 7 May 2016 (UTC)


 * I don't believe you can break it down into finite pieces like that, as is often the case in calculus. Also see Zeno's paradoxes and emergent property.  And to go to the next dimension, consider a solid of revolution, composed of an infinite number of cross sections, each of which have area, but no volume themselves.  StuRat (talk) 20:56, 7 May 2016 (UTC)


 * Such curves tend to have a fractal structure. So, while you certainly can zoom in, the magnified curve will appear qualitatively very similar to the original curve.  The irregularities never smooth out, as they do for differentiable curves.   S ławomir  Biały  21:30, 7 May 2016 (UTC)


 * If you think of the curve as having a width of 0, then the area of a curve of any finite length ℓ is 0 × ℓ = 0, but the area of a curve of infinite length is 0 × ∞, which is an indeterminate form. Getting an indeterminate form for an answer doesn't mean there is no answer; it means you can't tell by that technique what the answer is. A curve of any finite length has to have area zero, but a curve of infinite length doesn't have to, at least not by this line of argument. And it turns out that some actually don't. -- BenRG (talk) 21:41, 7 May 2016 (UTC)


 * Or, if there is no other method to obtain a definitive solution, perhaps we just declare a solution and use that, by convention. StuRat (talk) 21:58, 7 May 2016 (UTC)

If these curves in fact have an area, then what is it? shouldn't areas be measurable? the area can't be infinite as these fractals can fit inside a small finite area...?68.48.241.158 (talk) 02:10, 8 May 2016 (UTC)
 * OK, so first of all, technically, it's not the curve that has nonzero area. The curve is a map.  What has nonzero area is the range of the curve; that is, the set of all points that the curve passes through at least once.
 * For a true space-filling curve that lives inside the unit square, [0,1]&times;[0,1], the range is just the entire square, and its area is 1. --Trovatore (talk) 04:08, 8 May 2016 (UTC)

In advanced mathematics, the concepts of length, area, volume, etc. are usually defined by the Lebesgue measure. That article is probably not very useful to you, but let's try and discuss some examples and perhaps that will help. First of all, one of the important things to understand about the Lebesgue measure is that the measure of an open set is defined to be equal to the measure of the closure of that set. For example, the length of the open interval (0,1) is equal to the length of the closed interval [0,1] is equal to 1. This is true, by definition, even though the open interval doesn't include the points 0 or 1. Similarly, the area of an open disk (e.g. all points less than R distance from the origin) is equal to the area of the comparable closed disk (e.g. all points less than or equal to R distance from the origin). Roughly, the closure of a set means all points in the set plus any points that are outside but "infinitely close" to the set. Specifically, a point is infinitely close if for any distance greater than 0 there is always a point in the original set less than that distance away. In other words, closing the set adds in the edges and fill in any simple holes. For example, if you take the interval [0,1] minus the point 0.5 = [0,0.5) + (0.5,1], then the closure is still [0,1]. Now consider a more complicated example, the interval [0,1] minus all rational numbers. That removes an infinite number of points but still leaves all the irrational numbers which is also an infinite number of points. For every rational number there is always an irrational number infinitely close to is. Hence the closure of all irrational number in the interval [0,1] is the whole interval [0,1]. Even though the irrational set has an infinite number holes, it's length is still defined to be the same as the whole interval, i.e. 1. Now, let's go a step further. What is the area of a fractal or space-filling curve? Well first, we have figure out what is the closure of that curve. So, we count not only the curve itself, but also any points that are infinitely close to the curve itself. Even though the curve itself has no "width", an curve that winds infinitely densely through a space can be infinitely close to every point in that area. When that occurs, the closure of the curve becomes whole area. At that point the area of curve, according to the Lebesgue measure, is the whole area of the space it fills. Hope that helps. Dragons flight (talk) 07:55, 8 May 2016 (UTC)
 * What now? No no no.  We're not taking the closure.  You can't do that.  The rational numbers in the unit interval have measure 0, but their closure has measure 1. Anyway, any continuous image of the closed unit interval is compact, and therefore itself closed.  So taking the closure, in this case, doesn't do anything.
 * No, the curve really traces out every single point in the unit square. It can't do that without self-intersection. --Trovatore (talk) 08:06, 8 May 2016 (UTC)


 * Yeah, I'm approximating a bit. For the purpose of an "intuitive explanation".  The measure of a set is usually the same as the measure of the closure of a set.  There are exceptions, e.g. your rationals example (which is why I gave the irrationals in the unit interval).  The Peano curve, our classic space filling curve, is of course the limit of an iterative process.  At any finite step in that process you have a non-self-intersecting curve of finite length.  However, we can imagine a curve in the infinite limit that fills the space (and also happens to be self-intersecting).  In particular, if I choose an irrational point in the plane (e.g. (pi/4, e/3) or something), there is no curve in the Peano generation process that reaches that point.  Such points are only reached in the infinite limit.  The points reached by the limit curve are "roughly" the closure of the set of points reached by the set of Peano generating curves.  Of course if you want to try a more precise explanation, please go ahead.  Dragons flight (talk) 08:34, 8 May 2016 (UTC)
 * I'm sorry, I think your explanation is worse than useless. You're saying things that just aren't true, and not even on point.  The space-filling curve quite literally traces out every point.  It doesn't just get arbitrarily close.  Getting arbitrarily close doesn't help at all. --Trovatore (talk) 08:45, 8 May 2016 (UTC)

It seems that lines/curves are normally thought of as a kind of boundary/demarcation that don't actually exist in reality, but are abstractions..it makes intuitive sense that they have zero width and no area...to suggest there exist such things that do have an area flies in the face of intuition...I'm not even sure expert mathematicians can be said to have a genuine intuition about this (as an earlier poster referred to mathematical pathology)...it still doesn't quite make sense to me even as an abstraction...it would seem that for these curves to have an area then they should have a measurable width, which they don't seem to...otherwise one is simply measuring the area enclosed by a fractal...68.48.241.158 (talk) 13:17, 8 May 2016 (UTC)


 * No, it's not the area enclosed by a fractal that we mean here. There exist Jordan curves whose two-dimensional Lebesgue measure is not zero.  By the Jordan curve theorem, these are actually boundaries of open subsets of the plane.  So, yes, the boundary of an open domain can have positive area.  As to whether it "exists" in reality, all of mathematics represents an idealization which cannot be said to "exist".  However, I disagree with the (apparently widely-held) belief that pathologies are not present in nature.  Dynamical systems appearing in many applications exhibit pathological chaotic behavior.  Indeed, the "pathologies" studied by mathematicians in ergodic theory are fairly tame from the perspective of real-world dynamics.  Even geometrical pathologies exist in nature, such as corals, which exhibit isometric immersions of flat and hyperbolic surfaces, whose mathematical existence was only very recently proven.  Life imitates art; art imitates life.
 * The issue here is apparently how one defines concepts such as area and length in the first place. For planar regions, area is easier to define.  We know how to compute the area of a rectangle.  So we also know how to compute the area of an infinite collection of disjoint rectangles: we add their areas.  Now if a region is given, we may not know precisely how to calculate its area, but we can at least say that the area is not greater than a certain amount.  Indeed, all we need to do is cover the region with a collection of rectangles.  If a region is contained in a collection of rectangles, then the area of the region cannot be greater than the sum of the areas of the rectangles.  That seems like a very intuitive idea.  Now, the area of a region R can be defined as the smallest value of the area (actually the infimum) over all areas A such that the sentence "the area of R is not greater than A" holds.  This is the basic idea of the Lebesgue measure.
 * It is a standard theorem in analysis that there exist Jordan curves (that is, boundaries of bounded, connected, and simply connected open sets in the plane) whose area A, in this sense, is not zero. That is, any covering of the curve by rectangles has area is bounded below by the number A.
 * Length is actually much harder to define, because it is not possible to cover a curve with line segments. Instead, we cover a curve with rectangles, and (effectively) sum their perimeters rather than areas.  It is not at all obvious how to make this actually work in practice; it is the basis for the Hausdorff measure.   S ławomir  Biały  13:49, 8 May 2016 (UTC)
 * Really, you can get positive measure with a non-self-intersecting curve? I was wondering about that.  Do we have an article on that?  The examples in space-filling curve don't seem to help.
 * What does such a curve look like in a neighborhood of a point of density? --Trovatore (talk) 21:07, 8 May 2016 (UTC)


 * Wasn't my example with the infinite V's filling a rectangle a case of non-intersecting lines completely filling the rectangle ? StuRat (talk) 02:58, 9 May 2016 (UTC)


 * It has to be a curve; that is, you have to find a continuous map from the interval [0,1] onto your set. I don't see that you specified a way to do that.  (To be fair, several of our articles don't, either, and I think that's a flaw &mdash; but the difference is that it can be done, whereas I don't think yours can.  But feel free to prove me wrong.) --Trovatore (talk) 09:48, 10 May 2016 (UTC)


 * I take it you mean [0,0] to [1,1]. My method would work for that or any other rectangular range you wanted in 2D.  Here's 3 line segments connecting [0,0] to [1,1]:

1+---+ |/\/| 0+---+  0   1


 * You can use any odd number of line segments at increasing steep angles to start from [0,0] and end at [1,1]. When you go to an infinite number of line segments, all points in the unit square are then covered, and none intersected, unless you consider touching to be intersecting. StuRat (talk) 22:34, 10 May 2016 (UTC)
 * You need to prove that the sequence of curves you have described converges uniformly. However, it clearly does not, because arbitrarily small subintervals of [0,1] must get stretched out to intervals of length greater than 1.  So, while in some sense the limit of this construction does cover the square, it is not a continuous image of the unit interval.  It's actually worse than this, because the construction you have given does not even converge pointwise, except at the endpoints.  So it doesn't even define a discontinuous function from the interval to the square.   Sławomir Biały  (talk) 22:51, 10 May 2016 (UTC)
 * I'm not sure we're at the right stage of the discussion to bring in pointwise v uniform convergence. Let's just focus on getting across what needs to be accomplished, at a high level.
 * Stu, no, I really did mean the interval [0,1]; that is, the set of all x with 0&le;x&le;1. You have to find (or at least, show that there exists) a continuous function f that takes in values between 0 and 1 inclusive, and spits out points in your rectangle.  You haven't described such a function, not even in vague terms.  That's what needs to be accomplished, first. --Trovatore (talk) 23:17, 10 May 2016 (UTC)
 * I anticipate the retort as "pass to the limit", which was what seems to be intended here. The functions are the graphs of a sequence of triangle waves whose union has a closure that fills out the square.  I think the observation that the limit does not exist as a function is very relevant to this discussion.   Sławomir Biały  (talk) 23:34, 10 May 2016 (UTC)


 * Yes, for example the boundary of the Mandelbrot set is believed to have this property. But, more generally, according to the Denjoy–Riesz theorem, every compact totally disconnected planar set is a subset of a Jordan arc.  It's easy to construct such "fat" Cantor sets in $$\mathbb R^2$$.   Sławomir Biały  (talk) 21:26, 8 May 2016 (UTC)
 * Added after the fact, but lost in an edit conflict: Up to sets of measure zero, these latter examples probably just resemble fat 2-dimensional Cantor sets near each point of density. (Measure cannot "see" the connectedness of the set.)   Sławomir Biały  (talk) 21:32, 8 May 2016 (UTC)
 * Oh, so I saw your link up above to Osgood curve. The picture is nice; it's clear that you can get positive measure.  What's less obvious to me is that what's left, after you take out the wedges, is actually (the range of a) curve.  Same thing for the boundary of the Mandelbrot set.  Can you say something about that?  (It would be nice to add it to the article, as well.  Also, the Hilbert curve and Peano curve articles could say something about what the actual curve is, as opposed to the stages.) --Trovatore (talk) 21:29, 8 May 2016 (UTC)
 * It seems like an interesting question in general: is the boundary of a bounded connected and simply connected domain homeomorphic to the circle? I believe the answer is yes, but regardless of the answer to this general question, for the curve depicted at Osgood curve (due to our own David Eppstein), the boundary is already a uniform limit of curves.  For each n, there is a compact connected simply-connected set $$K_n$$ that is a union of $$4^n$$ mutually congruent triangles, $$T_n$$.  Eppstein's Osgood curve is the intersection of the $$K_n$$.  It is hopefully obvious what the $$T_n$$ are in the picture.  For instance, $$T_0$$ is the largest triangle that contains the picture.  $$T_1$$ is any of the four smaller triangles, and so forth.  Naturally, the triangles $$T_n$$ and $$T_m$$ are not similar unless $$n=m$$.  However, we still have
 * $$\operatorname{diam}(T_n) = 2^{-n/2}\operatorname{diam}(T_0). \qquad (1)$$
 * Now, let $$\phi_n$$ be a sequence of continuous homeomorphisms so that $$\phi_n$$ maps each of the $$4^n$$ equal subintervals forming a partition $$[0,1]$$ into each of the $$T_n$$ that comprise $$K_n$$. Then, owing to (1), any such sequence $$\phi_n$$ is uniformly Cauchy.  It is easy to show that the limit function is one-to-one (hence a homeomorphism).  Sławomir Biały  (talk) 23:03, 8 May 2016 (UTC)


 * you state some interesting things; some of which make some sense to me...I do think at least some of the concepts raised by the question are of a philosophical nature as opposed to of a strict mathematics nature (of course it's difficult to ever entirely separate out the two)..but I don't doubt the mathematical truth of what's being discussed...68.48.241.158 (talk) 14:19, 8 May 2016 (UTC)
 * To clarify my earlier comment: Expert mathematicians do have an intuition for things like this. It's just a different intuition than the one they started out with. Sometimes you need to adapt mathematical concepts to your intuition, sometimes you need to adapt your intuition to the mathematical concepts. -- Meni Rosenfeld (talk) 14:30, 8 May 2016 (UTC)

Some comments in Alexander Grothendieck's Esquisse d'un Programme may be of interest on "intuitive" versus "modern":

"I would like to say a few words now about some topological considerations which have made me understand the necessity of new foundations for “geometric” topology ... (256)

After some ten years, I would now say, with hindsight, that “general topology” was developed (during the thirties and forties) by analysts and in order to meet the needs of analysis, not for topology per se, i.e. the study of the topological properties of the various geometrical shapes. That the foundations of topology are inadequate is manifest from the very beginning, in the form of “false problems” (at least from the point of view of the topological intuition of shapes) such as the “invariance of domains”, even if the solution to this problem by Brouwer led him to introduce new geometrical ideas. Even now, just as in the heroic times when one anxiously witnessed for the first time curves cheerfully filling squares and cubes, when one tries to do topological geometry in the technical context of topological spaces, one is confronted at each step with spurious difficulties related to wild phenomena.

This [stopgap way to avoid difficulties] is a way of beating about the bush! This situation, like so often already in the history of our science, simply reveals the almost insurmountable inertia of the mind, burdened by a heavy weight of conditioning, which makes it difficult to take a real look at a foundational question, thus at the context in which we live, breathe, work – accepting it, rather, as immutable data. It is certainly this inertia which explains why it took millennia before such childish ideas as that of zero, of a group, of a topological shape found their place in mathematics. It is this again which explains why the rigid framework of general topology is patiently dragged along by generation after generation of topologists for whom “wildness” is a fatal necessity, rooted in the nature of things.... The only thing in all this which I have no doubt about, is the very necessity of such a foundational work, in other words, the artificiality of the present foundations of topology, and the difficulties which they cause at each step. ...

... It need (I hope) not be said that the necessity of developing new foundations for “geometric” topology does not at all exclude the fact that the phenomena in question, like everything else under the sun, have their own reason for being and their own beauty. More adequate foundations would not suppress these phenomena, but would allow us to situate them in a suitable place, like “limiting cases” of phenomena of “true” topology. 

On Grothendieck’s tame topology comments on recent developments & has more quotations. I would say that Robert Ghrist's Elementary Applied Topology is an accessible text in a similar spirit and neighborhood.John Z (talk) 02:25, 9 May 2016 (UTC)
 * That is an interesting quote. I would associate this philosophy also with Michael Gromov, particularly the h-principle.   S ławomir  Biały  16:45, 9 May 2016 (UTC)