Trivial: is a straight line, a fractal?

[Vexer]

Is it?

[Daecon]

I think a fractal is definied by it's circumference being of infinite length, or something like that?

 

If the line was infinitely long, then maybe it could be considered a fractal...

[Vexer]

(I think I've had an original thought)

[D H]

No. A line is "self similar" but lacks all other characteristics of a fractal curve. It is easily described in geometric terms (what shape is easier to describe than a line?) and its Hausdorff dimension is the same as its topological dimension; i.e., one.

[Vexer]

No. A line is "self similar" but lacks all other characteristics of a fractal curve.

 

 

It is easily described in geometric terms (what shape is easier to describe than a line?) and its Hausdorff dimension is the same as its topological dimension; i.e., one.

 

"Hausdorff dimension is the same as its topological dimension; i.e., one." and that disqualifies it? What are you talking about?

 

 

(I should spend an hour (or two) Googling this, I know)

[D H]

"Hausdorff dimension is the same as its topological dimension; i.e., one." and that disqualifies it? What are you talking about?

 

I am talking about the key distinguishing characteristic of a fractal.

[Vexer]

Obviously? And because I am not agreeing with you in another thread, that's your answer.

 

Thanks.

[ajb]

DH is absolutely right on this one.

 

A line is not a fractal, but it is kind-of self-similar. (We need a metric I think to discuss this properly).

 

A fractal is defined to be a metric space that has Hausdorff dimension strictly greater than it's topological dimension.

 

Hausdorff dimension is a measure theoretic construction. I really suck at measure theory so you will have to read up on that yourself. But what I can say is that for Euclidean spaces [math]\mathbb{R}^{n}[/math] both the topological dimension and the Hausdorff dimension are equal to [math]n[/math] and so cannot be fractals.

[Vexer]

A fractal is defined to be a metric space that has Hausdorff dimension strictly greater than it's topological dimension.

 

It is? Oh. (Is that to do with infinite boundary within finite space?)

 

 

I accept that a 'line' ain't a fractal, because you (all) say so. But I'm yet to understand why. I guess I could Google this more effectively now. I guess there's an agreed 'definition', much further than I imagined.

 

Thanks all.

[alan2here]

http://www.scienceforums.net/forum/showthread.php?t=35696

 

Fractal suffers a little from being subjective.

 

I accept that a 'line' ain't happy

[Daecon]

Maybe a fractal requires 2 or more dimensions to be considered a fractal?

[John Cuthber]

I think there's a 1 dimensional fractal but I can't remember what it's called.

The way to make it is to start with a line segment.

Remove the middle third- this gives a couple of lines with a gap between them.

Remove the middle third of each of the 2 line segments left giving 4 shorter lines.

Continue to do this with each of the new lines until they are reduced to "dust".

[D H]

I think there's a 1 dimensional fractal but I can't remember what it's called.

What you described is Cantor's tenerary set, or the Cantor set for short. The result is an uncountably infinite set of points that has Hausdorff dimension of about 0.63, has set zero measure, is closed, and is nowhere dense.

[Mr Skeptic]

As I understood it, a fractal is something with a non-integer number of dimensions.

[John Cuthber]

What you described is Cantor's tenerary set, or the Cantor set for short. The result is an uncountably infinite set of points that has Hausdorff dimension of about 0.63, has set zero measure, is closed, and is nowhere dense.

 

Thanks for that, I understood almost every word, but not a single concept.

[D H]

Thanks for that, I understood almost every word, but not a single concept.

I'll explain it, bit by bit. Along the way I'll correct some bad grammar in my post.

 

What you described is Cantor's tenerary set, or the Cantor set for short.

That should be self-explanatory. It's such a neat set it has a name.

The result is an uncountably infinite set of points

I should have been a bit more explicit there. The process in question is repeatedly removing the middle third of the remaining parts of the set.

 

Start with the interval on the real number line the contains all points between 0 and 1, inclusive. This a closed set, which for a simple set such as this means the end points of the set are in the set. The set of all points exclusively between 0 and 1 is an open set. The end points of the set are still 0 and 1, but this time the end points are not themselves in the set. The closed interval is denoted [0,1] while the open interval is denoted, (0,1).

 

Removing the middle third is a bit ambiguous: Do we take 1/3 and 2/3 out, or leave them in? If you take them out the result is a pair of sets that is neither open nor closed. Leaving those points in makes things cleaner, so leave them in. The result after the first iteration is the union of the closed intervals [0,1/3] and [2/3, 1]. The second iteration involves removing the open intervals (1/9,2/9) and (7/9, 8/9), leaving four intervals [0,1/9], [2/9,1/3], [2/3, 7/9] and [8/9, 1]. The third iteration removes four middle thirds, leaving eight intervals, and so on.

 

What's left will be a lot clearer by working in base 3. The base 3 representations of 3, 9, 27, 81 are 10, 100, 1000, and 10000. The first iteration removes (1/10₃, 2/10₃). The second iteration removes (1/100₃, 2/100₃) and (21/100₃, 22/100₃), and so on.

 

Some background: we write the fraction 1/2 as 0.5 in base 10 (or 0,5 if you are from Europe), and 1/3 as 0.333... (the ... means the 3s continue forever). In base 2, 1/2 is 0.1 and 1/3 is 0.0101.... In base 3, 1/2 is 0.111... and 1/3 is 0.1.

 

Back to the problem at hand: Except for 1/3 = 0.1 in base 3, the first iteration throws out all numbers whose first ternary digit is 1. Except for 1/9=0.01₃ and 8/9=0.21₃, the second iteration throws out all numbers whose second ternary digit is 1. The n^th iteration throws out all numbers whose n^th ternary digit is 1 (except for those nasty endpoints).

 

Those endpoints look problematic. You may have heard the argument that 0.999... = 1. This is also true for lots of other numbers: 0.1999... = 0.2, for example. Terminating decimals such as 1.0, 0.2, and 0.125 have an equivalent non-terminating form: Just subtract 1 from the terminal digit and tack on an infinite sequence of 9s. I'll call 0.999... the non-terminating decimal expansion of 1.

 

This same concept applies to other bases. For example, 1/3 in base 3 is written compactly as 0.1[/sub]3[/sub], but can also be written as 0.0222...₃. If we do that to the endpoints in the construction of the Cantor set, the n^th iteration of the construction can be viewed as throwing out all numbers whose n^th ternary digit is 1. The end result is the set of all numbers between zero and one inclusive whose non-terminating ternary expansion does not contain a 1.

 

So, how "big" is this set? Change all of the twos in the non-terminating ternary expansions of the members of the set to ones. The result looks like a set of numbers between 0 and 1 written in binary. So, view this set as a set of binary numbers. The result is the set of all numbers between 0 and 1 (inclusive) written in binary. Now take the non-terminating binary expansion of the real numbers between 0 and 1 (inclusive) and change all of the ones into twos. The result, when viewed in base 3, is the Cantor set. This little trick maps the Cantor set to all of the the real numbers between 0 and 1 (inclusive) one-to-one and onto. Since the set of all real numbers between zero and one is uncountably infinite, so is the Cantor set.

that has Hausdorff dimension of about 0.63,

While the Cantor set maps one-to-one and onto with the set of all points between zero and one, the Cantor set obviously is not the same as the set of all points between zero and one. The Cantor set is all broken up while the set of all points between zero and one is nice and continuous. The Hausdorff dimension is a measure of the dimensionality of some set. For example, the set [0,1] has a Hausdorff dimension of 1 (as does the entire real number line), and the unit square (or even the 2D plane) has Hausdorff dimension of 1. The Cantor set truly is "smaller" than the set [0,1] when viewed in terms of this dimensionality metric.

has set zero measure,

That's embarassing. The Cantor set is a set of measure zero, not has set measure zero. Think of measure as an extension of the concept of length. The set [0,1] has a length of 1. The first iteration of the construction whacks a 1/3 of the total length, leaving 2/3. On the second iteration, 4/9. On the third, 8/27. On the n^th, all that is left is (2/3)ⁿ of the original length of 1. This remaining length vanishes as the number of iterations increases. In the limit, there is no "length" left: All that is left is a bunch of dust.

is closed

Each stage of the construction of the Cantor set entails removing an open set from a closed set. The result of doing this is a closed set. (The endpoints are in the set.) When the process is complete, the result is still a closed set. A very funky closed set, but a closed set nonetheless.

and is nowhere dense.

The rationals are everywhere dense in the reals. What this means is that given any real number, a neighborhood of that number, no matter how small, will always contain at least one rational number. The contrary is true for the Cantor set. Members of the Cantor set are isolated points ("dust"), and any real number in (0,1) that is not in the Cantor set is part of a non-zero open interval that has no points in common with the Cantor set.