(Un)countalby infinite many minibrots

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Offline fractower

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« 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.



Offline gerrit

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« 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.

Offline hobold

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« 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).

Offline fractower

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« 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.








Offline gerrit

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« 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".

Offline Sharkigator

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« 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.

Offline fractower

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« 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?





Offline 3DickUlus

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« 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.
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Offline gerrit

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« 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".

Offline gerrit

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« 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.


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