(Un)countalby infinite many minibrots

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Offline marcm200

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« on: March 12, 2019, 09:31:36 AM »
I recently got interested in minibrots (thanks to pauldelbrot for the stating information) and was wondering, whether there are uncountably or countably infinite many minibrots.

On the one hand, I tend to say: countably, because the minibrots have surface area, hence a positive length in two prependicular directions and therefore any minibrot overlaps with at least one rational number in both the imaginary and real axis.
On the other hand, I feel like I am missing something - and, we're talking about the Mandelbrot set and weird stuff happens there.

Is the above argument valid and there are "only" countably many? And if so, can more be said about it? Countably like all even numbers - or countably many like the prime numbers or even more sparse?

Offline claude

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« Reply #1 on: March 12, 2019, 11:20:04 AM »
Yes the area + dense rationals thing is a valid argument, and probably the simplest.

There is also an association with rational angles that are periodic under angle doubling (they are strictly repeating when written in binary) - their "external rays" land in pairs on the roots of cardioids and discs.  These are sparser than rational angles in general, which are possibly pre-periodic with a non-zero preperiod.  I don't know of any simple test to check if an angle corresponds to a cardioid or a disc (I do have a method, but it's very complicated), but i think for any given period the cardioids usually outnumber the discs.

The sizes of minibrots shrink quickly.  I think this has been quantified for the main sequence along the antenna, but I don't know about other places.

Offline Dinkydau

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« Reply #2 on: March 12, 2019, 01:20:41 PM »
If you're interested in minibrots maybe you like this. Minibrots have the same shape (though slightly distorted) as the whole set. If you fool yourself and act like the minibrot is the whole set, you can go to the same minibrot again, or any other minibrot that exists in the whole set. I think of that as composing zoom paths (to minibrots). If you have two zoom paths \( s_1 \) and \( s_2 \) you can perform them one after the other. Then if the minibrots of the zoom paths have periods \( p_1 \) and \( p_2 \), the minibrot of the composition \( s_1 \circ s_2 \) has period \( p_1 \times p_2 \). That means there's a homomorphism between the set of minibrots and the positive integers: when \( f \) is the function that takes a zoom path and gives its period, then \( f(s_1 \circ s_2) = f(s_1) \times f(s_2) \). When I realized that I hoped I could use theories of algebra to gain insights but I've never gotten any further than this.

Offline claude

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« Reply #3 on: March 12, 2019, 01:35:29 PM »
This is called renormalization or tuning.  There is an algorithm to transform the binary expansion of an external angle t to be relative to a minibrot (or disc), given its periodic external angle pair with repeating blocks (lo, hi): replace each 0 in t with lo and each 1 in t with hi.  Unfortunately tracing high period rays is prohibitively expensive (O(p^2)) so I'm still searching for a better way to actually find coordinates for specific minibrots.

Offline lkmitch

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« Reply #4 on: March 12, 2019, 04:51:33 PM »
I made the mistake many years ago of thinking that there were uncountably many minibrots. There are two fairly simple ways to see that the number is countable:
  • Each minibrot has a cardioid whose center is the root of a (finite) polynomial. Thus, each center can be listed (counted), so there are only countably many.
  • Each minibrot has a finite area and they don't overlap, so a finite space (circle of radius 2) can only hold a countable number of them.

Offline fractower

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« Reply #5 on: March 13, 2019, 07:12:53 AM »
IKMITCH says
Quote
I made the mistake many years ago of thinking that there were uncountably many minibrots. There are two fairly simple ways to see that the number is countable:
Each minibrot has a cardioid whose center is the root of a (finite) polynomial. Thus, each center can be listed (counted), so there are only countably many.
Each minibrot has a finite area and they don't overlap, so a finite space (circle of radius 2) can only hold a countable number of them.

I can see how the polynomial argument can be extended to infinity. The order of the poly (and roots) grows as 2^N where N is the iteration count. As N->inf, 2^N becomes countably infinite.

I am not sure how to take the area argument to infinity since the area of the minibrots become infinitesimal as N>inf. Does anyone know of a finite area that can be subdivided in a way that is not countably infinite?

Offline marcm200

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« Reply #6 on: March 13, 2019, 09:42:04 AM »
First, thanks to all for your in-depth answers! That gives me a lot to think about as a beginner in looking for minibrots. Now I have a number of statements to start my list of "known or open facts about minibrots" with (I did that many years ago when I got interested in prime numbers).

As for the area: My argument relies on the fact that the rational numbers lie dense in the real numbers, hence in any surrounding of a real number of positive length, there is a rational number. So, no matter how infinitesimal small the minibrots become, they still have a length in both the real and the imaginary axis, so they overlap with a rational number in both directions, hence they carpet the plane and can only be countably many (if the minibrots do not overlap that is - has that been proven?)


Offline marcm200

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« Reply #7 on: March 13, 2019, 10:00:10 AM »
@fracttower: I just had a (raw) thought: The Cantor dust is uncountable and disconnected. Is it somehow possible to derive a partitioning of the unit interval from that set, something like a number of sets so that the union is the Cantor set and the sets themselves are in some sense "the smallest possible"? Wouldn't then an uncountable infinite number of sets be needed? (And one big set that contains everything that has been thrown out during the construction process of the Cantor set).

Offline fractower

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« Reply #8 on: March 13, 2019, 07:37:46 PM »
I thought Cantor dust was countable. The number of dust particles grows as a power law. The standard square alg. grows as 4^N where N is the iteration step.

It is possible I am wrong about the limit N->inf of 4^N being countable. It has been a few years since I took that class.

Offline claude

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« Reply #9 on: March 13, 2019, 07:45:23 PM »
I thought Cantor dust was countable.

Nope.  Countable sets have dimension 0, but dimension of Cantor dust (formed by iteratively keeping 4 subsquares out of 9 in a 3x3 grid) is log 4 / log 9 ~= 0.631 > 0.

Offline gerrit

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« Reply #10 on: March 14, 2019, 07:49:31 AM »
I thought Cantor dust was countable. The number of dust particles grows as a power law. The standard square alg. grows as 4^N where N is the iteration step.

It is possible I am wrong about the limit N->inf of 4^N being countable. It has been a few years since I took that class.
That's not how it works. To specify a point in a Cantor dust you have to specify an infinite sequence of which of the 4 small sub squares you are going to next. You can encode this in an infinite row of digits of base 4, i.e., a real number. So C dust is uncountable like the real numbers.


Offline marcm200

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« Reply #11 on: March 14, 2019, 09:41:56 AM »
@gerrit: I like the digit argument. It gives one a nice visualization.

I was trying to un-raw my thought from yesterday to partition an uncountable set into uncountably many countable/finite ones.

Let's start with the Cantor set C. Construct the set {C} that contains the Cantor dust as its sole element. Now the axiom of choice in ZFC set theory states, that there is a function f that selects an element from C (if there are more functions, we choose one randomly). Let's say that function chooses a number a. We put aside the one-element set {a}, remove the number a from our starting set C and repeat that process over and over until the Cantor dust has been emptied out. Now we have an uncountably infinite number of sets we have put aside (all have 1 element) whose union is the Cantor set.

It's only a proof (? Any corrections?) of existence. I couldn't think of a more direct way that allows one to actually do the partitioning. The above reasoning works with every uncountable set, however I found it most appealing using the Cantor dust, because there I have the impression that the numbers already lie somewhat lonely on the real axis.

Offline fractower

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« Reply #12 on: March 14, 2019, 05:06:58 PM »
I think a countable set of points has zero dimension while a uncountable set of points will have a dimension of > 0. For example the rational numbers are zero dimension while the irrational numbers are dimension one.

[The elements of the Cantor set are squares that become infinitesimal in the limit but still 2 dimensional. If we are just counting elements then I think we have to replace each element with a point and ask "what is the dimension of these points?"
« Last Edit: March 14, 2019, 08:06:19 PM by fractower »

Offline fractower

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« Reply #13 on: March 14, 2019, 10:41:24 PM »
That's not how it works. To specify a point in a Cantor dust you have to specify an infinite sequence of which of the 4 small sub squares you are going to next. You can encode this in an infinite row of digits of base 4, i.e., a real number. So C dust is uncountable like the real numbers.

Your description is a way to label each dust partial with a base 4 integer which I think makes them countable.

Offline gerrit

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« Reply #14 on: March 15, 2019, 12:19:32 AM »
Gerrit: "That's not how it works. To specify a point in a Cantor dust you have to specify an infinite sequence of which of the 4 small sub squares you are going to next. You can encode this in an infinite row of digits of base 4, i.e., a real number. So C dust is uncountable like the real numbers."

Your description is a way to label each dust partial with a base 4 integer which I think makes them countable.

Put a "." in front and you will hopefully see it's a real number (with \( \infty \)ly many digits), so not countable.


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