Fractal Geometry

1. Dec 16, 2005

nate808

Lets take the cantor middle thirds set as an example (Iterate by removing the middle portion of a line segment) If this was done an ACTUAL INFINITE number of times, would you be left with line segments, points, or neither?

2. Dec 16, 2005

Tide

Neither! The fractal dimension of the "Cantor dust" is log(2)/log(3) which is greater than zero but less than one.

3. Dec 17, 2005

nate808

.99999

I get it, but there is something that is kind of bothering me. If instead of the middle thirds set, lets say we iterate a line segment of length one by removing a length of .9 every time. Why don't you get the same "dust"? And if you did, wouldn't that disprove that .99999...=1

4. Dec 18, 2005

Tide

You don't get the same set because there are numbers contained in each that are not contained in the other.

5. Dec 18, 2005

nate808

So what are you left with? nothing?

6. Dec 18, 2005

Hurkyl

Staff Emeritus
You get a different set that looks "the same" as the Cantor set. (in the same sense that a circle and a square are "the same": they're both simply closed loops)

7. Dec 18, 2005

nate808

I don't understand how you could be left with anything if you did it infinately many times. There shouldn't be anything left. What I understand, and I may be interpreting you wrong, but you would be left with a sort of "dust" similar to the cantor set. But the dust would be something. If you think about this iteration instead of taking away .9 every time but adding .9 every time, eventually, there should be and entire line with length of 1. Conversely, there should be nothing if you take away .9 every time. Where does my logic go wrong (By the way, the reason I think that you should either have a full line or nothing at all, is that the sum of the infinite sum .9 + .09 +.009 .... is actually 1)

8. Dec 18, 2005

Hurkyl

Staff Emeritus
The error in your logic is the assumption that a set with measure zero must be empty.

What you are essentially doing is computing how "long" the dust is. And just like the Cantor set, you find that it has zero "length".

(I use quotes because the technical term is measure. For nice subsets of the line, measure and length mean the same thing)

But it should not be a surprise that you can have a nonempty set with zero length. After all, how long is a single point?

9. Dec 18, 2005

nate808

If you take away a set with measure of zero which is not empty, from something, like a line segment, do you alter the characteristics of that line. For example, taking away a point from a line makes it unconnected. Would the length change? I also don't understand how the measure would be zero. I guess its just my perception, but it seems strange that fractals are supposed to be magnified versions of the original, so why wouldn't it still have a length no matter how many times you do it as long as the original senment had a length?

p.s. If this thread is becoming trivial, feel free to close it

10. Dec 18, 2005

Hurkyl

Staff Emeritus
Nope; measure is (countably) additive. If P and Q don't overlap, then $m(P) + m(Q) = m(P \cup Q)$. In particular, if Q is a set of measure zero, and P is what's left after removing Q from a line segment, then the measure of P is the same as of Q.

Because the sum of the lengths of the intervals you removed is 1!

Your usage of "it" is very ambiguous.

You can speak about different measures. For example, you already know of two useful measures on the line: length, and counting. (the counting measure of a finite set is simply the number of points it contains, and the counting measure of an infinite set is $+\infty$)

If we were in a plane, then you already know three measures: counting, length, and area.

There are "in-between" measures too. Counting is a "zero-dimensional" measure, length is a "one-dimensional" measure, and area is a "two-dimensional" measure. This dust we're discussing is too big to be zero-dimensional (i.e. point-like), and too sparse to be one-dimensional (i.e. line-like). I think it's somewhere around 0.23-dimensional. So, if you want to measure it "properly", you have to use a 0.23-dimensional measure.

11. Dec 18, 2005

Tide

Nate,

One thing you may have missed is that when you take out the "middle third" of the unit line segment to create the Cantor set you are in fact leaving behind all the end ponts. That means, e.g. the points {0, 1/3, 2/3, 1} remain after the first (and all subsequent) division(s).

This tells you that the Cantor set is at least countable but, in fact, it turns out to have the same cardinality as the set of real numbers, i.e. the same cardinality as the set of numbers with which you started.