Is this really possible? Is there any other example of this other than the Koch Snowflake? If so can you prove that example to be true?
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24$\begingroup$ The Koch Snowflake has the "other problem": it is a finite area contained in an infinite perimeter. Is that what you meant? Is the Koch Triangle something else? $\endgroup$– Eric StuckyDec 1, 2014 at 6:10
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2$\begingroup$ Gabriel's horn is a 3-D equivalent. Infinite volume but finite surface area. $\endgroup$– CogitoErgoCogitoSumDec 1, 2014 at 6:34
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26$\begingroup$ @CogitoErgoCogitoSum Surely Gabriel's horn has finite volume and infinite surface area. $\endgroup$– Jeppe Stig NielsenDec 1, 2014 at 8:50
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5$\begingroup$ Your statement is wrong. As @EricStucky pointed out, the Koch Snowflake has finite area and infinite perimeter. $\endgroup$– JRNDec 1, 2014 at 13:10
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2$\begingroup$ Gabriel's horn would be an example in 3D in the sense that the perimeter (=boundary curve) at the opening has finite length, but the area is infinite. [/rimshot] $\endgroup$– Jyrki LahtonenDec 1, 2014 at 15:14
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10 Answers
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One can have a bounded region in the plane with finite area and infinite perimeter, and this (and not the reverse) is true for (the inside of) the Koch Snowflake.
On the other hand, the Isoperimetric Inequality says that if a bounded region has area $A$ and perimeter $L$, then $$4 \pi A \leq L^2,$$ and in particular, finite perimeter implies finite area. In fact, equality holds here if and only if the region is a disk (that is, if its boundary is a circle). See these notes (pdf) for much more about this inequality, including a few proofs.
(As Peter LeFanu Lumsdaine observes in the comments below, proving this inequality in its full generality is technically demanding, but to answer the question of whether there's a bounded region with infinite area but finite perimeter, it's enough to know that there is some positive constant $\lambda$ for which $$A \leq \lambda L^2,$$ and it's easy to see this intuitively: Any closed, simple curve of length $L$ must be contained in the disc of radius $\frac{L}{2}$ centered at any point on the curve, and so the area of the region the curve encloses is smaller than the area of the disk, that is, $$A \leq \frac{\pi}{4} L^2.)$$
NB that the Isoperimetric Inequality is not true, however, if one allows general surfaces (roughly, 2-dimensional shapes not contained in the plane. For example, if one starts with a disk and "pushes the inside out" without changing the circular boundary of the disk, then one can make a region with a given perimeter (the circumference of the boundary circle) but (finite) surface area as large as one likes.
answered Dec 1, 2014 at 6:10
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10$\begingroup$ The rigidity of $\Bbb R^2$ is really astounding to me sometimes. $\endgroup$ Dec 1, 2014 at 6:20
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2$\begingroup$ There are many versions, but the most general (in $\mathbb{R}^2$ anyway) applies to any simple, closed curve. I'm not an expert in the area, but I think the piecewise $C^1$ case is typically the one sees first, since some of the available proofs in that setting are satisfactorily elementary. $\endgroup$ Dec 1, 2014 at 10:16
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2$\begingroup$ @CameronWilliams $\mathbb{R}^2$ is surely quite rigid, but an isoperimetric inequality holds in any dimension and for both Lebesgue and Gaussian measure. For example in $\mathbb{R}^n$, for any compact set, $S/V^{(n-1)/n} \geq n\omega_n^{1/n}$ where $S$ is the measure of the boundary of the set (Minkowski content) and $V$ is the Lebesgue measure of the set. $\endgroup$ Dec 1, 2014 at 21:25
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19$\begingroup$ The isoperimetric equality is a bit of work to prove (especially for a maximally large class of curves), but that’s because it gives a sharp bound on the area. For this question, a much easier inequality suffices: $A \leq \pi L^2/4$. This follows just from the fact that a line is the shortest path (in the sense of your desired class of curves) between two points: from that fact, it is clear that a closed curve of length $L$ must lie entirely within the disc of radius $L/2$ around its starting point, and hence the region it encloses is a subset of that disc. $\endgroup$ Dec 2, 2014 at 9:23
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1$\begingroup$ Infinite perimeter? i.stack.imgur.com/VVXlW.png $\endgroup$– TimtechDec 2, 2014 at 20:25
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It's a bit of a matter of semantics.
What's a "shape" but a subset of the plane separated from the rest by a curve? But which subset?
A circumference (for example) is a finite closed curve (with finite perimeter) that separates and defines two subsets of the plane - we conventionally pick the one with finite area and we call it "circle". But the circumference also defines the subset with infinite area that lays "outside" (which is a conventional concept). That other "outside shape" would be an example of a finite-perimeter curve with an infinite area.
That sounds like cheating and playing with words. But think about it: what else could possibly an infinite area delimited by a finite curve look like? If you only allow yourself to look at the "inside" of any closed curve, it couldn't have an infinite area because you can always define a circumference "around it" whose circle would necessarily fully contain the first shape and also be of finite area. Any possible shape with infinite area and finite perimeter would have to be the "outside" delimited by a closed curve.
So the answer to your question depends on whether you're interested in considering the "outside" of a closed curve (in which case all closed curves delimit such shapes), or whether you're not (in which case there cannot be any such shape).
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10$\begingroup$ Excellent observation! I suppose, then, that the "real" question would be whether there exists a closed curve of finite length such that both the interior and the exterior of the curve have infinite area--which would appear to be prohibited by the isoperimetric inequality, as stated in Travis's answer. $\endgroup$ Dec 1, 2014 at 19:42
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Consider a 1-by-1 square, let your shape be everything on 2-dimensional Euclidean space except this square. Perimeter is 4, area is $\infty$.
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67$\begingroup$ Now that's literally thinking outside the box! $\endgroup$ Dec 1, 2014 at 20:57
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$\begingroup$ Do you mean Riemann's sphere of complex plane without preimage of that square ? $\endgroup$ Jan 1, 2015 at 6:40
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The geometric figure which occupies the largest area given a fixed perimeter is the disk of a circle. But all circles with finite perimeters have finite areas. $($In three dimensions, the geometric shape which occupies the largest volume given a fixed surface is the sphere. So, if your question would have been about three-dimensional objects with infinite volume and finite perimeter, the answer would still have been a no$)$. Both these results $($as well as countless other optimization problems, about finding the extrema $[$i.e., minima and maxima$]$ of a function$)$ are obtained using calculus.
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Any bounded subset of $\mathbb{R}^2$ clearly has finite area (assuming $A \subseteq B$ implies $Area(A) \leq Area(B)$, which is true for instance with Lebesgue measure or any other reasonable sense of "area"), so for such a set to exist, it would have to be unbounded.
Perimeter is harder to define, but I think it's believable that for any meaningful sense of perimeter $p$, $p(A) \geq \sup \{ |x_1 - x_2| : x_1,x_2 \in A \}$, i.e. the perimeter of a set is at least as great as the distance between any two points in the set. Thus, if the set is unbounded, its perimeter cannot be finite.
answered Dec 1, 2014 at 21:02
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If instead of the plane, you considered the Riemann sphere, the equator has perimiter = 1, but it also bounds an infinite region on the top.
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1$\begingroup$ This is essentially the same answer as James Worthington's, except using a unit circle instead of square. But good answer either way! $\endgroup$– CharlesDec 4, 2014 at 17:14
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Consider a sphere of infinite radius, with a hole of radius R cut in it.
The interior or exterior surface has infinite area, but the bounding edge has a finite length of the circumference of the hole...
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29$\begingroup$ This reminds me of the old joke about a shepherd who asked three people to build a pen in the most efficient way possible. On day 1, the physicist builds a perfectly circular fence to give the sheep plenty of room to move. On day 2, the engineer builds a square pen of equal area. Since the fence only has four posts, it can be built much more cheaply despite the longer perimeter. On day 3, the farmer returns to find sheep roaming everywhere and the mathematician standing in the middle of the field, wrapped in wire. "Ah," says the mathematician, "but you see, I am on the outside!" $\endgroup$– r3mainerDec 1, 2014 at 22:02
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There isn't—on a plane. However we may take a 3D cylindrical coordinate system and define a surface $z=\sqrt r \sin\tfrac 1r$. The $\tfrac 1r$ term makes infinite sequence of waves towards the center, $\sqrt r$ term reduces their height (amplitude) so that the surface has no singularity there.
Then we cut a part of the surface, containing the center point – it will possibly have an infinite area.
The function above is just a sketch, I didn't check if the area is actually infinite. In case I'm wrong you might either increase the waves' frequency growth, say by replacing $\frac 1r$ with $\frac 1{r^3}$ or with $\exp(\tfrac 1r)$, or take the steeper amplitude term, say $\sqrt[4] r$ instead of $\sqrt r$.
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Take an Euclidean plane. Draw a circle. Now the region outside the circle has infinite area but finite boundary.
Interestingly enough we even can find the area of that infinite shape in closed form. But it would depend on the filtering (for some shapes the placement of the coordinate origin also would make difference but not in this case).
Let the circle have the radius of 1.
If we take the filtering along the polar radius, the area of the whole plane is
$$S_p=\int_0^\infty 2\pi r dr= 2\pi \left(\frac {\tau^2}2+\frac1{24}\right)=\pi\tau^2+\frac\pi{12}$$
where $$\tau=\int_0^\infty 1 dx,$$
a divergent integral.
Now, we take the area of the circle, $\pi$ from the area of the plane. We get
$$S=\pi\tau^2-\frac{11\pi}{12}$$
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$\begingroup$ This is identical to James Worthington's answer from six years ago. $\endgroup$ Apr 19, 2020 at 0:21
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$\begingroup$ @Oh thanks. In that case I think I can expand it. $\endgroup$– AnixxApr 19, 2020 at 0:22
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It depends on the geometry. In Euclidean: definetely no. On a fractal surface: sure.
