### (Un)countalby infinite many minibrots

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#### fractower

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#### Re: (Un)countalby infinite many minibrots

« Reply #15 on: March 15, 2019, 05:56:17 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."

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

I could do the same to all positive integers (move the decimal point to the left end) and get all possible decimals between 0 and 1. I think this is still a countably infinite set since it maps one to one with the integers.

I looked up Canter's decimal proof. It is pretty easy to understand how he gets to the contradiction, but I am not sure how/if it applies to this case.

• 3f
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#### Re: (Un)countalby infinite many minibrots

« Reply #16 on: March 15, 2019, 07:29:44 AM »
I could do the same to all positive integers (move the decimal point to the left end) and get all possible decimals between 0 and 1. I think this is still a countably infinite set since it maps one to one with the integers.
That will not get you all "decimals", for example you're missing $$\sqrt{2}$$.

An infinite sequence of digits does not represent an integer. It represents a real number.

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#### Re: (Un)countalby infinite many minibrots

« Reply #17 on: March 15, 2019, 01:11:55 PM »
I could do the same to all positive integers (move the decimal point to the left end) and get all possible decimals between 0 and 1.
This construction finds only a subset of all real numbers in the interval; namely all those with a finite number of digits (assuming the usual convention that trailing zeros are not written down).

#### fractower

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#### Re: (Un)countalby infinite many minibrots

« Reply #18 on: March 15, 2019, 07:51:03 PM »
This construction finds only a subset of all real numbers in the interval; namely all those with a finite number of digits (assuming the usual convention that trailing zeros are not written down).

Good point. Moving the decimal point to the left end will only give numbers between .1 and 1 (assuming decimal representation). Instead of moving the decimal point I could flip the digits around the decimal point (read it in a mirror). This will give numbers between 0 and 1 which are mapped one to one with the integers. Since the integers are infinite there are quite a few infinitely long decimal numbers. In theory there is an infinitely large integer whose digits match the fraction part of square root of 2, but that might be abusing the axiom of choice.

One difference with the integer argument and Garret's base 4 quad tree argument is that when I am counting dust particles I really want to increment the right most digit. Since this digit is at the end of an infinitely long string I am stuck with a contradiction. Since Garret starts with all infinitely long base 4 strings he skips the countable constructor step. This may be where I am messing up.

• 3f
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#### Re: (Un)countalby infinite many minibrots

« Reply #19 on: March 16, 2019, 06:15:15 AM »
I looked up Canter's decimal proof. It is pretty easy to understand how he gets to the contradiction, but I am not sure how/if it applies to this case.
For Cantor dust the diagonal proof goes something like this. Suppose you show me an infinite list of points p1, p2, p3, .... in Cantor dust, and claim this is complete, hence the dust is countable. No matter what your list is, I can refute that by constructing a point q in the dust that's not in your list. Simply choose the 1st sub-square  different from p1, the second different from p2, the third,... etc. Now q is in the dust but not on your list.

This shows merely that C-dust is uncountable, the mapping to the reals shows it's degree of infinity is the same as the real numbers, usually called "C".

#### Sharkigator

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#### Re: (Un)countalby infinite many minibrots

« Reply #20 on: March 16, 2019, 10:27:54 AM »
Since the integers are infinite there are quite a few infinitely long decimal numbers. In theory there is an infinitely large integer whose digits match the fraction part of square root of 2, but that might be abusing the axiom of choice.
No.
Every integer has a finite number of digits.
It can only be arbitrarily large, not infinitely large.

#### fractower

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#### Re: (Un)countalby infinite many minibrots

« Reply #21 on: March 16, 2019, 09:18:36 PM »
Ok you have convinced me. I hate being wrong but like learning new things.

Back to the original question. Are there uncountable many minibrots?

• 3e
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#### Re: (Un)countalby infinite many minibrots

« Reply #22 on: March 16, 2019, 10:29:13 PM »
At any given zoom level one could count all visible (apparent) mini brot locations.

Following that assumption, infinite zooming == infinite min brots ?

So countable minibrot locations is a factor of zoom resolution.
Resistance is fertile... you will be illuminated!

https://en.wikibooks.org/wiki/Fractals/fragmentarium

• 3f
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#### Re: (Un)countalby infinite many minibrots

« Reply #23 on: March 16, 2019, 11:46:46 PM »
Back to the original question. Are there uncountable many minibrots?
This question has been answered here many times already with the correct "no".

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#### Re: (Un)countalby infinite many minibrots

« Reply #24 on: March 16, 2019, 11:48:30 PM »
At any given zoom level one could count all visible (apparent) mini brot locations.

Following that assumption, infinite zooming == infinite min brots ?

So countable minibrot locations is a factor of zoom resolution.
That's not even wrong.

#### Anthony Lahmann

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#### Re: (Un)countalby infinite many minibrots

« Reply #25 on: March 24, 2019, 05:25:26 AM »
IKMITCH says
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?

The cardinality of the power set is 2 to the power of the cardinality of the original set. The cardinality of the integers is countably infinite. The power set of the integers is also infinite, but because it cannot be paired up with every integer, it must be uncountably infinite. This means that 2^(countably infinite)=uncountably infinite.

#### tavis

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• Bill Tavis

#### Re: (Un)countalby infinite many minibrots

« Reply #26 on: April 01, 2019, 10:04:42 PM »
Consider the whole Mandelbrot set. It's got a cardioid and main bulb which are adorned with bulbs that vary in period number. Each of those bulbs has tendrils where we find the minibrots in question. So the first place to start counting is with the bulbs.
Ok, well in Elephant Valley the periods of the bulbs increase by one, so you can count up to infinity by positive numbers.
In Seahorse Valley the periods are odd and increase by 2 so you can count up to infinity with the odd numbers.
So it seems pretty obvious that there is a countably infinite number of bulbs going around the set. But...
Every single one of those bulbs has an infinite number of more bulbs attached around it, and each of those bulbs have infinite bulbs attached, and those bulbs have bulbs, etc...
Every single one of these bulbs and the bulbs on bulbs has an infinite number of minibrots attached to each one.
And then...
Every single one of those minibrots has just as many new minibrots around them as could be counted for the overall set! Which of course adds exponentially to the total for the overall set which then adds the same amount to each minibrot counted which then adds to the total which then adds to each one... so the number of them just completely explodes. I don't understand how it could be countable?
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• 3e
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#### Re: (Un)countalby infinite many minibrots

« Reply #27 on: April 01, 2019, 10:24:30 PM »
each hyperbolic component (cardioid or disc) has a root point that is the landing point of two rational external rays between 0 and 1.  each rational ray lands at exactly one point.  rational numbers are countable.

each hyperbolic component has a unique nucleus, a root of a monic polynomial, which is an algebraic integer.  algebraic numbers are countable.

each hyperbolic component has positive area and they overlap only in isolated points.  thus each contains a distinct rational point, and rational numbers are countable.

#### tavis

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#### Re: (Un)countalby infinite many minibrots

« Reply #28 on: April 02, 2019, 12:26:10 AM »
well ok, the external angle thing makes the most sense to me in terms of establishing that the number is countable. But, why then is everyone comparing this to the Cantor set and saying that one is uncountable? It makes no sense and I think that is what got me the most confused. According to a quick google search, it seems that the endpoints of the segments in a Cantor set are countable after all. Forget the boxes for now, consider the simple lower dimensional case. There is a line that gets divided in thirds, right? So then you could identify an endpoint as being at 1/3 and another at 2/3 and another at 1/9, etc. There is a rational fraction for every endpoint of every segment. How is that any different than specifying a rational angle for a minibrot?

The Cantor set as a whole is considered uncountable specifically because it contains points that are not endpoints. But for this conversation it seems like the endpoints are what make a useful comparison to external rays landing at a root point. In this way, counting the number of minibrots in the Mandelbrot set is akin to counting the number of segments in the Cantor set (countable), not to counting the total number of numbers in the Cantor set (uncountable).

• 3f
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#### Re: (Un)countalby infinite many minibrots

« Reply #29 on: April 02, 2019, 04:25:13 AM »
completely explodes. I don't understand how it could be countable?
It has nothing to do with "exploding". They're countable because you can count them. For example start with all period 1 minis, then period 2, then period 3, etc. There are only finitely many minis of a given period so this list (number each item) will be complete.

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