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Author Topic:  Length of the Mset boundary (orders of infinity)  (Read 279 times)

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fractower

• Fractal Phenom
• Posts: 54
Length of the Mset boundary (orders of infinity)
« on: November 18, 2019, 10:51:44 PM »
I have been thinking about the length of the Mset boundary in terms of order
of infinity and the implication of the dimension of the boundary.

I will start by trying to prove the Mset boundary is countably infinite in length.

Assumptions:
1. From the Canter dust though experiment we get X^N as N approaches
infinity is countably infinite. The canter set usually has an X of 2 or 4,
but the observation is valid for any finite X.

2. The boundary defined by some finite escape limit grows exponentially
every iteration. I have imperially found the growth to be X~=1.165 up to 30
or so iterations. 30 is far from infinity, but I am going to assume it is
bounded for this argument.

Proof that the Mset boundary is countably infinite in length:
At some iteration N normalize the length of the boundary to 1 unit, and
initialize a Canter duster to 1 segment. At each iteration the number of
units of boundary grows by 1.165 while the Canter dust doubles. Thus the
units of boundary length is bounded by the Canter dust count.

Tying this back to the dimension of the boundary. In order for a 1D boundary
to fill a 2D space it must be greater than countably infinite units in
length. For example consider a horizontal line segment 1 unit long. Replace
every point in the line with a vertical line segment 1 unit long to produce
a unit square. The total units of length of these lines is equal to the real
numbers. Contrast this with replacing every rational point with a vertical
line segment. The total length in this case is countably infinite but
occupies no area.

This is not sufficient to prove the boundary is not 2D for some
infinitesimal width.

« Last Edit: November 19, 2019, 01:07:50 AM by fractower »

claude

• 3f
• Posts: 1795
Re: Length of the Mset boundary (orders of infinity)
« Reply #1 on: November 19, 2019, 01:04:50 AM »
There is a conformal map between the exterior of the Mandelbrot set M and the exterior of the unit disk D. The Koebe 1/4 theorem provides a bound on the local stretching of a conformal map, which could maybe be enough to show something like the equipotential at iteration N+1 is at most 4x longer than the length of the equipotential at iteration N (not sure, I'm not expert enough at complex analysis).

But it has already been proved that the boundary of M has dimension 2.  Whether the boundary has non-zero area is uknown, afaik.

pauldelbrot

• 3f
• Posts: 2539
Re: Length of the Mset boundary (orders of infinity)
« Reply #2 on: November 19, 2019, 03:24:50 AM »
I don't think fractower's argument flies, because if it did, it could be used to "prove" that Peano space-filling curves aren't.

These curves have the properties that:

1. Each successive approximation curve has x times the length of the previous, for some finite constant x.

2. In the limit the curve has a non-zero area.

This suggests that the limit curve to the dwell-boundary lemniscates around M is not ruled out from having a non-zero area. And that limit curve is, of course, the boundary of M.

gerrit

• 3f
• Posts: 2337
Re: Length of the Mset boundary (orders of infinity)
« Reply #3 on: November 19, 2019, 03:49:44 AM »
How are "(un)countable infinite length" defined?
Uncountable infinite set is one you can't count (1to1 correspondence to integers).
How do you count a length?

fractower

• Fractal Phenom
• Posts: 54
Re: Length of the Mset boundary (orders of infinity)
« Reply #4 on: November 19, 2019, 04:22:33 AM »
I screwed up by trying to make a number into a set and apply set rules. Infinity is infinity no matter how you get there.
The Peano curve is the perfect counter example to what I was thinking.

Nothing to see here... Move along.

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