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Looking for information on fractals through google I have read several time that one characteristic of fractals is :

  • finite area
  • infinite perimeter

Although I can feel the area is finite (at least on the picture of fractal I used to see, but maybe it is not necessarly true ?), I am wondering if the perimeter of a fractal is always infinite ?

If you think about series with positive terms, one can find :

  • divergent series : harmonic series for example $\sum_0^\infty{\frac{1}{n}}$
  • convergent series : $\sum_0^\infty{\frac{1}{2^n}}$

So why couldn't we imagine a fractal built the same way we build the Koch Snowflake but ensuring that at each iteration the new perimeter has grown less than $\frac{1}{2^n}$ or any term that make the whole series convergent ?

What in the definition of fractals allows or prevent to have an infinite perimeter ?

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The "finite area" probably refers to the area enclosed by some closed fractal curve (where actually the curve is the fractal, not the enclosed area, and the area of the curve itself is $0$ — which admittedly is also finite).

Basically, an object of fractal dimension $f$ has the property that

  • if you calculate its measure in dimension $<f$, you always get $\infty$.
  • if you calculate its measure in dimension $>f$, you always get $0$.

For example, the Koch Snowflake curve has dimension $1 < \ln 4/\ln 3 < 2$, therefore its length (measure of dimension 1) is infinite, but its area (measure of dimension 2) is $0$.

Note however that the "filled snowflake" is an ordinary 2-dimensional object (although its border has fractal dimension — it's the Koch Snowflake curve, after all). Therefore the 2d measure (area) of that is not $0$. However it is not true that you always have to have a finite area; imagine that instead of putting 6 Koch curves to a snowflake forn, you'd just added them linearly one after the other. The border would still be a fractal (although it now is no longer astonishing that its length is infinite because it goes to infinity on both sides), but the area would be the area of one half of the plane, which is of courrse infinite as well.

On the other hand, the fractal dimension of the Cantor set is $\ln 2/\ln 3$, therefore even its 1d measure is $0$. Nevertheless it also is a fractal, although it doesn't fit the characterization you gave.

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I have a question... We know that the topological perimeter of a Koch Snowflake is infinite. But in geometric measure theory, we study the distributional perimeter. How to show that the distributional perimeter of the Koch Snowflake is infinite? Or finite? (This might be a dumb question, but it bothered me for a while.) – student Sep 17 '14 at 18:07

You can certainly define a fractal analogous to the Koch snowflake with finite perimeter.

The iterative rule for the standard Koch snowflake is to replace the middle third of each line segment with two line segments of equal length and meeting the original segment and each other at $60^\circ$ angles. This applies the rule $x \mapsto 4x/3$ to the overall length of the curve, so applying it repeatedly produces a divergent sequence of lengths.

A modified rule might be to start with a red line segment and iteratively replace the middle half of each red line segment with the top of a trapezoid made of one red and two blue line segments (all of equal length and again making $60^\circ$ angles). Then the overall red/blue lengths $(x,y)$ are subjected to the rule $(x,y)\mapsto(3x/4, y + x/2)$, which is convergent when iterated. It is a nice exercise to show that this process, starting from a red segment of length $1$, produces an entirely blue curve of length $2$; and starting from an equilateral red triangle of perimeter $3$ produces a closed blue curve of perimeter $6$. (Calculating the area inside the curve is another nice exercise. Moreover, since the trapezoid can be constructed either red-blue-blue or blue-red-blue at each step, there are any number of different fractals with the same perimeter and area.)

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Intrigueing, what is the Hausdorff dimension of this thing? – Hagen von Eitzen Sep 18 '12 at 17:27
by the next answer, if this has a Hausdorff dimension, it is necessarily 1. – Berci Sep 18 '12 at 18:11
does this have a name? – epeleg Jun 29 '14 at 5:49
You suppose "You could make an analogue to the mandelbrot equation with an finite circumference" it would not prove that fractal shapes can have finite perimiter. This answer doesnt create something that mandelbrot would call a fractal, because if it contains any kind of infinately repeating pattern, then it can't be a line, or something that isn't percieved to as a repeating pattern in the overall shape. If you use the iterations to delete the pattern from on top of itself, you won't get an infinite repeating pattern, the result would not have fractal properties. – comprehensible Sep 11 '14 at 14:41

The problem is: What is the perimeter of a point set after all? However, in the case of curve-like fractals like thesnowflake, we produce successive approximations by starting with a line segment and then repeatedly replace a line segment by a sequence of line segments which lengthens the path. In the case of the Koch snowflake, the lengtheing is by a constant factor of $\frac43>1$, hence the lengths of the approximating polylines grow without bound.

If on the other hand you have some set $S$ that is maybe quite zig-zag, but can be said to be of length $L$ in a suitable manner (that is: We can map $f:S\to \mathbb [0,L]$ injectively and with dens image, such that $|f(x)-f(y)|\ge |x-y|$), then its dimension is not fractal. Indeed, if $\epsilon>0$ is given, you can select the $\approx\frac L\epsilon$ points along the curve at length offsets $0, \epsilon, 2\epsilon, \ldots$ and observe that the curve is covered by $\frac L\epsilon$ balls of radius $\epsilon$, hence the Hausdroff dimension is $1$ and not fractal.

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Example. Start with a square. Then, on each iteration replace a square of side length $a$ with four squares of side length $a/4$. Clearly after any number of iterations, the perimeter is $4a$. Is this not a fractal? It would be Cantor Dust if I used $a/3$ instead. – Karolis Juodelė Sep 18 '12 at 17:05
But does the limit have a perimeter after all? – Hagen von Eitzen Sep 18 '12 at 17:20
@KarolisJuodelė Actually, I'd say the figure has no perimeter even after the first step. After the $n$-th step is consists of $4^n$ squares of which each one individually has the perimeter $a/4^{n-1}$. The sum of all the perimeters has the value $4a$. However, I agree that the set's Hausdorff dimension is $1$, and that it's 1d Hausdorff measure is $4a$. It's just that I wouldn't call it perimeter. Intuitively I'd also call it a fractal, despite its non-fractal dimension, however I'm not sure whether the standard definition of fractals demands a non-integer Hausdorff dimension. – celtschk Sep 19 '12 at 12:39

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