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 had this strange notion some time ago, and I recently wrote a blog post about it, as a mere curiosity. I don't really consider it a "serious" mathematical question; but out of interest, I wondered if someone on this site could shed some light on what principle might be underlying the idea.

Basically, I envisioned a "pseudo-triangle" consisting of two straight edges and one jagged "edge" (not really an edge, since it's jagged, but I'm calling it that anyway):

Pseudo-triangle with a 4-step jagged edge

The above shape has 4 steps, its area is 10, and its perimeter is 16. Now let's increase the number of steps to 8:

Pseudo-triangle with an 8-step jagged edge

This shape has an area of 9 and a perimeter of 16. Now, without me having to write out a formal proof, I think it's pretty clear that as the number of steps increases, the area will approach 8 while the perimeter will remain constant at 16. And the resulting shape will look like this:

Pseudo-triangle with N steps (approaching infinity) along its jagged edge

Ultimately, there's nothing really "mysterious" about this; the shape above is not a triangle, and so it shouldn't be surprising that it doesn't have quite the same properties as a triangle. However, it does approach the same area as an analogous triangle; and, more to the point, it just seems odd.

Is there a concept in mathematics that describes this phenomenon (for lack of a better word)? That is, the effect of some kind of mathematical entity (e.g., a shape) converging to what resembles another entity but differs from it in a critically important and counter-intuitive way (in this case, having a completely different perimeter)?

If it seems that I'm having trouble articulating this question, that's because I am. But hopefully someone out there can see what I'm getting at and shed some light on the issue for me.

share|improve this question
1  
What do you mean for one shape to converge to another? I suggest you study the concept of limits more in-depth. –  Alex Becker Mar 25 '11 at 6:27
4  
Maybe this will help: math.stackexchange.com/questions/12906/is-value-of-pi-4 –  please delete me Mar 25 '11 at 6:30
    
@Alex: I realize that I used some terms in a very informal sense. I can remove my usage of that particular word ("converge"), but I don't think I have an insufficient understanding of limits. –  Dan Tao Mar 25 '11 at 6:37
1  
@Dan: The problem does lie in the "convergence" of the shape to the triangle, as the perimeter of the shape is actually not changing and so its perimeter does not converge to that of the triangle. –  Alex Becker Mar 25 '11 at 6:40
2  
@Dan: Well, it depends on the metric you apply to the space of "shapes" in question. This is what determines how limits behave in the space. The metric you have been using is essentially the area of the symmetric difference of the two shapes, which is actually a psuedometric. In order to convert it to a metric you have been considering shapes with a distance between them of zero equivalent. While this works fine for talking about area, this equivalence relation does not preserve information about the boundary of the shapes, and so you cannot talk about the perimeter of an equivalence class. –  Alex Becker Mar 25 '11 at 6:50

3 Answers 3

A Geometric Measure theoretic answer to your question could be the following.

Your example simply shows how things really go... There is nothing strange in it, nor in the "informal" notion of convergence you used: actually the trouble remains the same, even when all the underlying notions are well formalized.

The problem here is that the perimeter $\mathcal{P} (\cdot )$ is "only" a lower semicontinuous functional. This means: if a sequence of sets $E_n$ converges to another set $E$ in some sense (e.g. in the Hausdorff metric, which is a sort of uniform convergence for sets, or in some $L^p$ metric), then you have:

(*) $\displaystyle \mathcal{P} (E)\leq \liminf_{n\to \infty} \mathcal{P}(E_n)$;

moreover, in general the inequality is strict, even if the sequence at the RHside converges.

A less problematic example of this basic fact of GMT is the following. Let:

$E_n:=\{ (x,y)\in \mathbb{R}^2|\ x^2+y^2<1\text{ and } |y|\geq \frac{1}{n} |x|\}$,

so $E_n$ is the unit open circle with two symmetric slices (crossing the $x$ axis) removed and the slices become thinner and thinner as $n$ increases. Note that $E_n$ is a piecewise $C^1$ set with a finite number of corners, i.e. $5$, and this number does not increase with $n$ (on the contrary, in your example the number of corners becomes larger as $n\to \infty$).

Then $E_n$ converges to the unit open circle $D$ in the $L^1$ metric: in fact, the measure of the symmetric difference set $D\Delta E_n$ tends to $0$ as $n$ increases, i.e. $\lVert \chi_D -\chi_{E_n}\rVert_1\to 0$; on the other hand, one has:

$\displaystyle \mathcal{P} (E_n)=4+ 2\left( \pi -2\arctan \tfrac{1}{n}\right)$

(the summand $4$ appears because the boundary $\partial E_n$ contains four radii of lenght $=1$) therefore:

$\displaystyle \liminf_{n\to \infty} \mathcal{P}(E_n)=\lim_{n\to \infty} \mathcal{P} (E_n) =2\pi +4 >2\pi =\mathcal{P} (D)$.

For what is worth, $E_n$ converges to $D$ also in the strongest Hausdorff metric, because it is not hard to prove that the Haudorff distance:

$\displaystyle \text{dist}_H(E_n,D):=\inf \{ \epsilon >0| E_n\subseteq (1+\epsilon)D \text{ and } D\subseteq E_n+\epsilon D\}$

is given by:

$\displaystyle \text{dist}_H(E_n,D) =2-\frac{2}{\sqrt{1+\frac{1}{n^2}}}$,

and $\displaystyle \lim_{n\to \infty} \text{dist}_H(E_n,D) =0$.

If you choose $E_n$ as in your example, then you have the same situation: a sequence of sets which does converge to a triangle in some metrics (in particular, it converges in both $L^1$ and Hausdorff metric) and for which strict inequality holds in (*).

share|improve this answer

Not that I really know anything about it, but there's a concept called varifold, which I've seen loosely described as an "infinitesimally corrugated" manifold. I don't know how varifolds are defined rigorously, but apparently they arise in the calculus of variations when looking for surfaces of minimal area with a given boundary (like soap films on a wire loop). It might happen that a sequence of smooth surfaces converges to a would-be minimizing surface, except that the limit object is not smooth but infinitely wrinkled. This idea feels a bit like what you have here; the limit of your jagged lines should perhaps not really be considered as a normal straight line, but rather as some other kind of (infinitedly jagged) object?

share|improve this answer

Abstract geometry does not always match to intuitive geometry. If you are surprised from this triangle, you should be much more surprised from fractals with finite area and infinite perimeter, or fractals with a non-integer (and irrational) number as their dimension.

share|improve this answer

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.