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July 3rd, 2016, 02:06 PM   #1
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Reals are uncountable - without Cantor diagonal argument

Gist of proof: length of unit interval = 1, if the reals are countable, the length = 0, contradiction.

Proof: Arrange reals into countable list. Let x be an arbitrarily number > 0. Cover first point on list by interval of length x/2, second point by interval of length x/4, third point by interval of length x/8 , etc. $\displaystyle n^{th}$ point by interval of length $\displaystyle x/2^{n}$. All points in the unit interval are covered - sum of the lengths of cover = x, so unit interval has length < x. Because x can be arbitrarily small, unit interval has length = 0. Contradiction.
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July 4th, 2016, 05:19 AM   #2
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I'm afraid that I don't understand this argument. Couldn't you replace "reals" with "rationals" and conclude that that the set of rational numbers in the unit interval were uncountable?
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July 4th, 2016, 06:39 AM   #3
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Isn't this related to the Banach-Tarski paradox? Certainly in this case, the concept of length on the number line has to be very well understood. I'm not convinced that I have the knowledge to explain the problem with the proof.
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July 4th, 2016, 07:51 AM   #4
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I wonder whether the problem is the mixture or actual and potential infinities. In particular, when you sum the lengths, you are trying to sum an actually infinite number of elements. This is not the same as taking the limit of finite sums as the number of elements grows without bound.
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July 4th, 2016, 09:42 AM   #5
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I believe we are being trolled by Zylo, who is very probably trolling himself.

Cantor provides a perfectly adequate and very elegant proof of the non-denumerability of the real numbers. Zylo does not point out a flaw in that proof. He merely presents alternative arguments that he asserts prove a contrary result.

His latest, if you separate out all the binary notation rigmarole, is that once you posit a denumerably infinite list of numbers between 0 and 1 in infinitely expanded form and further posit putting the members of that list into 1-1 correspondence with the list of natural numbers, those natural numbers can also be expressed in an infinitely expanded form. OK, I buy that. He then asserts without any proof that his list of numbers between 0 and 1 inclusive includes all reals between 0 and 1.

It is quite clear that his assertion is absurd with respect to finite lists.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a list with 10 elements, can be put into 1-1 correspondence with

0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, another list with 10 elements. But the second list does not include all the real numbers between 0 and 1. It does not even include all the rational numbers.

Cantor showed that what is false for finite sets may be true for infinite ones, but that does not entail that what is false for finite sets must be true for infinite ones. Cantor showed constructively how to put the integers into 1-1 correspondence with the natural numbers and how to put the rationals into such a correspondence. Zylo thinks he has done the same with real numbers, but he has not.

Let's go back to the finite example I gave above. 0.81 is a real number in the interval from 0 through 1, but there is nothing corresponding to it in the list of natural numbers of natural numbers from 0 through 9. Zylo says that there is a corresponding natural number in the list from 0 through 99, namely 81. I say what about 0.811. I can always add one more significant digit. Ultimately, I think Zylo believes that the process must terminate at infinity, but infinity is non-terminating by definition.

I'll put this a different way. Cantor shows a way to put a finite set of natural numbers into 1-1 correspondence with a finite set of integers, which set entirely spans a range of integers. He further shows that the process of expanding to wider spans is non-terminating and is thus extensible from the finite into the infinite realm. This is a demonstration in two steps. Zylo skips the first, crucial step. He has never shown how to put even a finite set of natural numbers into 1-1 correspondence with a set that includes all the real numbers within a non-zero span.

Last edited by JeffM1; July 4th, 2016 at 09:47 AM.
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July 4th, 2016, 01:41 PM   #6
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Quote:
Originally Posted by Petek View Post
I'm afraid that I don't understand this argument. Couldn't you replace "reals" with "rationals" and conclude that that the set of rational numbers in the unit interval were uncountable?
No, since the rational numbers do not fill up the unit interval. The argument is based on the fact that the reals do fill up the entire unit interval.

The proof is essentially using measure theory in a simplified way. The rationals have measure 0, while the reals have measure 1.

Last edited by mathman; July 4th, 2016 at 01:43 PM. Reason: Expand on discussion.
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July 4th, 2016, 01:47 PM   #7
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Quote:
Originally Posted by v8archie View Post
I wonder whether the problem is the mixture or actual and potential infinities. In particular, when you sum the lengths, you are trying to sum an actually infinite number of elements. This is not the same as taking the limit of finite sums as the number of elements grows without bound.
I don't understand the distinction you are trying to make. A sum of a countable number of terms is the limit of finite sums.
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July 4th, 2016, 03:29 PM   #8
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That's a definition. It's a definition that is based on some sensible intuition but the fact is that nobody knows how to sum an infinite number of terms.

The rational case highlighted by Petek may not be correct. The conclusion could be that there are numbers between the rationals that determine the "lengths". On the other hand, every pair of irrationals have infinitely many rationals between them. And even if we do accept this argument, where is the proof that there are no non-real numbers between the reals? If we work in the hyperreal numbers, for instance, there are.

This has similarities to Zeno's paradox. And I think it's for the same reason. We are trying tomix the ideas of continuous and discrete and it just doesn't work.

Also, the first post says that because $x$ can be arbitrarily small, the length is zero. But $x$ can also be arbitrarily large. I think that the dichotomy between continuous and discrete is messing everything up.

Either way, I don't see the first post as containing a valid proof.
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July 5th, 2016, 09:24 AM   #9
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Quote:
Originally Posted by mathman View Post
Gist of proof: length of unit interval = 1, if the reals are countable, the length = 0, contradiction.

Proof: Arrange reals into countable list. Let x be an arbitrarily number > 0. Cover first point on list by interval of length x/2, second point by interval of length x/4, third point by interval of length x/8 , etc. $\displaystyle n^{th}$ point by interval of length $\displaystyle x/2^{n}$. All points in the unit interval are covered - sum of the lengths of cover = x, so unit interval has length < x. Because x can be arbitrarily small, unit interval has length = 0. Contradiction.
There are countably infinite many points between 0 and $\displaystyle \delta$ no matter how small $\displaystyle \delta$ is. You'd never get past $\displaystyle \delta$.
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July 5th, 2016, 09:46 AM   #10
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Tjat obviously an incorrect amalysis. If you were covering the rationals in ascending order, you'd never get reach any given $\delta$ either. But nobody said that the enumeration has to be in ascending order. It is only necessary that an enumeration exists, and the existence of such an enumeration is guaranteed by the fact that the reals are assumed to be countable.
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