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# Infinite Area and Perimeter

1. Can an object have infinite area but finite perimeter? Why, or why not?
2. Can an object have finite area but infinite perimeter? Why, or why not?
3. Can an object have infinite volume but finite surface area? Why, or why not?
4. Can an object have finite volume but infinite surface area? Why, or why not?

Thoughts?

I intentionally used the vague word 'object' (as opposed to polytope / shape), so there can be various answers depending on your interpretation. It boils down to considerations about the object being bounded, or unbounded. (I'm also avoiding a deep discussion of what perimeter and area actually refer to.)

Samuel provides good examples of 1 and 3, where the object is unbounded. Gabriel provides an argument for why 1 and 3 are false IF the object is bounded. (Where in his argument is the condition of bounded used?)

Mattias provides a example of 2 and 4 where the object is bounded. An example of a bounded object with infinite perimeter and finite area, is the object which is comprised of $$[0,1] \times [0, \frac{1}{2} ] \cup [\frac{1}{2^{2k+1}}, \frac {1}{ 2^k} ] \times [\frac{1}{2} , 1 ]$$, which is a fattened version of the topologist's comb.

Note by Calvin Lin
4 years, 10 months ago

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Not sure about all of them but I know that 2 and 4 are true. Some examples are the Koch snowflake (infinite perimeter, finite area) and Gabriels Horn (infinite surface area, finite volume). I'm definitely not sure, and I have no proof, but I think that 2 and 4 can be true, but 1 and 3 can't. This would be because an infinite area/volume can't actually be contained by a perimeter/area (in euclidean space), while an surface area/volume can be "chopped up", making the perimeter/surface area larger, but keeping surface area/volume constant.

- 4 years, 10 months ago

use Riemann approximation

- 4 years, 10 months ago

- 4 years, 10 months ago

If "object" is a real thing and we stop the resolution on molecule/atoms level(or any other level) none of them are true for simple reasons. Intuition tells me 2 and 4 are easy to build mathematically, while 1 and 3 need an extra dimension, easy to provide for 1, harder for 3.

- 4 years, 7 months ago

For 1:

Assume that an object has infinite area and (finite) perimeter P. Due to its infinite area, the object cannot be bound within a finite region. Assume the distance between any pair of points on the perimeter of the object is bounded above by P. We then have a contradiction as we can bound the object within a finite region (a circle of radius P centered on any point of the object).

Since there exist two points of distance at least P apart on the perimeter, it is clear that the object has perimeter larger than P, a contradiction.

(Alternatively, Isoperimetric inequality tells you for finite perimeter, the largest area possible is when its a circle (finite area). This inequality can be also used to prove 3. is impossible)

- 4 years, 10 months ago

1. Yes just let it be a sphere of infinite radius and bunch a disk on its surface. 2. Yes any space filing curve. 3. Yes the 3-sphere of infinite radius which is just 3-space and a point at infinity. To this just bunch out the unit 2-2-sphere at the origin. 4. Perhaps some sort of space filing surface.

- 4 years, 10 months ago

Hmm that's very interesting...although whether your answers for 1 and 3 would depend on how we define perimeter and surface area.

- 4 years, 10 months ago

I was confused by your description until I realized what you mean is to punch out a shape; i.e., you are taking a difference of sets.

Formally, we can describe such objects as follows:

[1] Let $$\overline{\mathbb{C}}$$ be the Riemann sphere, and let $$D = \{ z \in \overline{\mathbb C} : |z| < 1 \}$$ be the unit open disk in $$\overline{\mathbb C}$$. Then $$\overline{\mathbb C} \setminus D$$ has finite boundary but infinite area.

[3] This is defined analogously as in [1], just with a higher dimension; e.g., $$\{(x,y,z) \in \mathbb{R}^3 \cup \{ \infty \} : x^2 + y^2 + z^2 \ge 1 \}$$ suffices.

For [2] and [4], I believe that if we require the object to be bounded (i.e., be a subset of the interior of some disk or sphere of finite radius, respectively), then its boundary must have noninteger Hausdorff dimension; i.e., it must be a fractal. If boundedness is not required, then there are numerous examples as described elsewhere in this thread.

- 4 years, 10 months ago

For the second one, I have something in mind : $$\int\limits_0^1 |\ln x| \ \mathrm{d}x=1.$$ The length of the curve is infinite, but the area is finite. However, this may not be taken as a geometrical proof since the area is not closed.

- 4 years, 10 months ago

- 4 years, 10 months ago

Gabriel's Horn (also called Torricelli's trumpet) is a geometric figure which has infinite surface area, but finite volume. Gabriel's Horn.

- 4 years, 10 months ago

4: Torricelli's Trumpet

- 4 years, 10 months ago

For 2. This is one example: Given this sequence: \frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2}) ^3 + ... + (\frac{1}{2})^n = 1 This can be described in geometry as: a square, each line is 1m long; S = 1m2; split it into 2 equal rectangles; then continue splitting each smaller figures into 2 equal parts. The process would be endless but finally the area is always 1m2.

- 4 years, 10 months ago

1) Assume that we have a sheet of infinite area. If we cut a lamina of finite perimeter(for ex:a circle) from this sheet then, the perimeter of the remaining part is same as perimeter of the lamina taken out from it(i.e finite) and its area is still infinite. 2) Consider a straight line of infinite length. Its area is finite(i.e zero) and perimeter is infinite.(assumption: thickness of line is zero) 3) Consider an object with infinite volume and take out a solid(for ex: a cube) from it with finite surface area. Now the surface area of remaining part of original object is same as surface area of solid taken out from it(i.e finite) and its volume is still infinite. 4) Consider a sheet of infinite area. Its volume is finite(i.e zero) and surface area is infinite.(assumption: thickness of sheet is zero) So, i think all four are possible.

- 4 years, 10 months ago

w.r.t 1) Please explain your assumption that an infinite sheet has 0 perimeter. w.r.t 3) Please explain your assumption that an object with infinite volume has 0 surface area.

- 4 years, 10 months ago

Those are not assumptions. My assumptions are only for 2 and 4 without which for 2) area would also be infinite and for 4) volume would also be infinite. 1)Lets take a simple example. Take a rectangular sheet(of finite area) of dimension l * b and cut a circular region of radius r within its boundary. Here the remaining part(i.e rectangular sheet with a circular hole in it) has two boundaries: 1st is the outer rectangular boundary and 2nd is the inner circular boundary. So the perimeter of the remaining part (i.e rectangular sheet with a circular hole in it) is 2(l+b) + 2 * PI * r . If we consider sheet with infinite area, there is no outer boundary for it and if we make a hole in it as in previous case then there exists a inner circular boundary. So its perimeter is 2 * PI * r i.e finite. Well it need not be zero as u mentioned in your question obviously it can be zero when radius of hole is zero. Similar explanation can be given for (3). Even here surface area need not be zero but some finite value.

- 4 years, 10 months ago

If the n dimension measure of an object is infinite then the n-1 dimension measure of such object cannot be finite? Am I along the right tracks? (the n dimensional measure is like volume for 3d, area for 2d, perimeter or lenght for 1d).

- 4 years, 10 months ago

There are many functions like f(x) = e^x or f(x) = 1/x etc. which have area enclosed between them and the x axis as finite. But their length is infinite. So just revolve them about the x axis and you generate 3-D objects with finite volume but infinite area.

- 4 years, 10 months ago

Actually $$\int \limits_{0}^{\infty} \frac{1}{x}dx = \infty$$. But suppose you have a converging function f(x). $$\int \limits_{k}^{\infty} f(x)dx$$ is a finite value, but $$\lim \limits_{x \rightarrow \infty} f(x) = 0$$, which means the perimeter will be infinite, but area finite.

- 4 years, 10 months ago

Like for example $$\displaystyle \int_1^{\infty} \frac{1}{x^2} dx=1$$, which has an infinite perimeter.

- 4 years, 10 months ago

Oh thanks. Good explanation.

- 4 years, 10 months ago

If this dimension can be transmitted to some other dimension all this can be possible!!! Theory always need not 2 be true!!

- 4 years, 10 months ago

i think that 1 and 3 can't, if there is a finite perimeter/surface area, i think that the largest area/volume of it is in the shape of a circle/sphere

- 4 years, 5 months ago

Mattias, your picture is an object with finite area and infinite perimeter.

- 4 years, 5 months ago

Nobody has thought of 4) Menger Sponge? Infinite surface area, 0 volume. :)

- 4 years, 7 months ago

Updated

Staff - 4 years, 10 months ago

I hear that for number two, there's this idea that England has so many little inlets/breaks in the perimeter that because there are so many, one would have to measure in very small units to measure the perimeter. In fact, infinitely small. Because of the infinite smallness of the measurement, it would take an infinite amount of units to measure around the perimeter of England. Of course, England is an island, so there has to be a finite area.

- 4 years, 10 months ago

area , perimeter and surface area are related to each other according to there perspectives. simply they are directly proportional to each other.

- 4 years, 10 months ago

what is really meant by infinite?is 1/3 infinite?

- 4 years, 10 months ago

Is 1/3 infinite? Are you serious?

- 4 years, 10 months ago

infinite means no limits.1/3 has no limits.

- 4 years, 10 months ago

1. no thats impossible bcoz surface area is less to kept that object volume

- 4 years, 10 months ago

1. Infinite area means that there can be no bound on the perimeter, so the perimeter will also be infinite. So 1. is impossible.

2. If we take a rectangle of a fixed area, say $$A$$, then we may take the breadth to be infinitesimally small, say $$x$$, and length infinitely large, $$\frac {A}{x}$$. So the perimeter, $$2(x+\frac {A}{x})$$ is infinitely large, so 2 is possible.

3.An infinite volume cannot be bounded, so the surface area will also be infinity. So 3. is impossible.

4.If we take a cuboid of fixed volume V, and take the breadth say $$b$$, then the length and height can be made infintely large, say $$l$$ and $$h$$. So its surface area, $$2(lb+bh+lh)$$ will be infinitely large due to the term $$lh$$.So 4. is possible.

- 4 years, 10 months ago

- 4 years, 10 months ago

I don't think any of the statements can be true!!!!

- 4 years, 10 months ago

A Mobius Strip may have infinite area but has limited perimeter b/w points (Not sure)

- 4 years, 10 months ago

differentiation of area gives perimeter and that of volume gives surface area, so they are related. either both are finite, or both are infinite

- 4 years, 10 months ago

hmmm... why do we care? -_-

- 4 years, 10 months ago

Because we love maths.

- 4 years, 10 months ago

agreed with u Shaurya...:)

- 4 years, 10 months ago

YEAH!

- 4 years, 10 months ago

Because we live for Maths ... That's why we're here on Brilliant ...

- 4 years, 10 months ago

brilliant is not for guys like you pepe its for those who love maths and quoting dave we live for maths.

- 4 years, 10 months ago

hey shourya look at my profile :)

- 4 years, 10 months ago