# Cantor set end points are and are not countable!

by Bbkowal
Tags: cantor set, end points
 P: 4 The number of end points of the cantor set double each time an iteration is performed, therefore the total number of end points after infinite iterations is ~ 2^N where N is cantor's aleph null. 2^N is, however, c (the number of the continuum) and is therefore uncountable but we know that the end points of the cantor set are countable. Hence the apparent contradiction. Any help?
 P: 181 There aren't 2^N endpoints in the Cantor set, but only N again. These endpoints correspond to all binary fractions with either finitely many zeroes or finitely many ones, for instance .0100110000000000000...(only zeroes)... and .1101001111111111...(only ones)... (Read each zero as "select the left third" and each one as "select the right third", and you'll be "led" to the endpoint in question.) But there are more real numbers between 0 and 1, e.g. .01010101010101010...(one-zero repeated infinitely)... and .01001000100001000001... (the number of zeroes increases by 1 each time)... So seen as a set of real numbers, the Cantor set only has countably many elements, and there are uncountably many others within the same interval, from 0 to 1.
 P: 4 I understand the argument for the cantor set end points being countable, but that still does not explain the fact that in order to generate the cantor set an infinite number of end point doublings must take place. If we double something repeatedly N times then we will end up with 2^N things. Since N in this case is aleph null we end up with c end points. The only explanation I can think of is that we somehow pass over to the full cantor set as N goes to aleph null and as is well known the full cantor set is uncountable. This, however, is not a very satisfying (or rigorous) argument.
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## Cantor set end points are and are not countable!

 Quote by Bbkowal If we double something repeatedly N times then we will end up with 2^N things.
This is simply not true. It may seem intuitive, but infinities don't work like this.
 P: 4 I disagree with Micromass. As an example take a proof that the reals between zero and one have a cardinal number of 2^N where N is aleph null: First write the reals in binary notation from 0.000000... to 0.111111... Then the first place after the decimal can take one of two possible values (0 or 1). The next place can take one of two values also so that brings the number of combinations to four. The next place brings it to eight etc. The Nth place brings the number of combinations to 2^N. If we proceed to N=infinity (aleph null) we get that the number of reals between zero and one is c.
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 Quote by Bbkowal I disagree with Micromass. As an example take a proof that the reals between zero and one have a cardinal number of 2^N where N is aleph null: First write the reals in binary notation from 0.000000... to 0.111111... Then the first place after the decimal can take one of two possible values (0 or 1). The next place can take one of two values also so that brings the number of combinations to four. The next place brings it to eight etc. The Nth place brings the number of combinations to 2^N. If we proceed to N=infinity (aleph null) we get that the number of reals between zero and one is c.
That proof is simply wrong. You can use the same "proof" to show that all numbers with terminating binary notation (such as 0.10100000...) are uncountable. That is simply false.

There is no reason to expect

$$\lim_{n\rightarrow \aleph_0} 2^n=2^{\aleph_0}$$
P: 704
 Quote by Bbkowal First write the reals in binary notation from 0.000000... to 0.111111... Then the first place after the decimal can take one of two possible values (0 or 1). The next place can take one of two values also so that brings the number of combinations to four. The next place brings it to eight etc. The Nth place brings the number of combinations to 2^N. If we proceed to N=infinity (aleph null) we get that the number of reals between zero and one is c.
The argument is wrong because the notation ##2^{\aleph_0}## does not mean the limit of ##2^N## as ##n \rightarrow \aleph_0##. It's the cardinality of all maps ##\mathbb{N} \rightarrow \{0,1\}##. See cardinal exponentiation.
 Sci Advisor P: 1,668 The cantor set includes all of the limit points of the end points of the middle thirds
 P: 4 Thank you micromass and pwshafu. I guess I had better study set theory more thoroughly.
 P: 1 I'm back from studying a little set theory and I have the answer to my apparent contradiction. First of all my proof of c=2^N where N is aleph null, was not wrong but merely incomplete. See http://en.wikipedia.org/wiki/Cardina..._the_continuum. The "Alternative Explaination" section uses a proof very similar to mine except the second part of the bijection is completed. Also my proof cannot be used to show that the rationals are uncountable because in that case there are serious constraints upon the values of the individual digits once past a finite number of them so that the total number of combinations of the digits is not 2^N. Secondly, I found that doubling something N times does indeed lead to 2^N things (the power set of N) even when N is aleph null or for that matter aleph n. Finally, the apparent contradiction goes away once it is realized that an end point is only created during an iteration n when n is finite. At n=aleph null the full Cantor set appears because the ternary expansions of the non-end points are a infinite (aleph null) pattern of zeros and twos which can be be shown, through a bijective mapping to the binary numbers between 0 and one, to have a cardinality of 2^N. I admit that all of the above arguments may be imprecise and non-rigorous by todays standards but I also believe that Cantor would have agreed with them.

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