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May 13th, 2016, 12:11 PM   #1
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Cantor's Diagonal Argument and Infinity

Cantor's argument fails because there is no natural number greater than every natural number.

Last edited by skipjack; May 20th, 2016 at 02:22 PM.
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May 13th, 2016, 12:38 PM   #2
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It doesn't fail, as it effectively uses and relies on that fact.
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May 14th, 2016, 01:27 PM   #3
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Quote:
Originally Posted by skipjack View Post
It doesn't fail, as it effectively uses and relies on that fact.
Assume T consists of all countably infinite binary sequences, ie, the natural numbers.

Assume T is countable and enumerate ALL of T

Create Cantor's diagonal number. It is not in the list. Contradiction.
Therefore the natural numbers are not countable.

_______________________________________________

So what's the problem?

Is Cantor's diagonal number in T, which would be a contradiction?
NO, because it doesn't exist.

By Cantor's construction, Cantor's diagonal number is greater than every number in the list. But there is no natural number greater than every natural number.

Last edited by skipjack; May 14th, 2016 at 02:38 PM.
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May 14th, 2016, 02:56 PM   #4
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The "problem" you ask for is that what you've presented isn't Cantor's argument. Also, obtaining a contradiction shows that one of your assumptions must be wrong, but it needn't be (and isn't) the one that you picked.

The assumption in your post that all countably infinite binary sequences are natural numbers is wrong.

Last edited by skipjack; May 20th, 2016 at 02:27 PM.
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May 16th, 2016, 07:28 AM   #5
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From:
https://en.wikipedia.org/wiki/Cantor...gonal_argument

"He assumes for contradiction that T was countable. Then all its elements could be written as an enumeration s1, s2, … , sn, … . Applying the previous theorem to this enumeration would produce a sequence s not belonging to the enumeration. However, s was an element of T and should therefore be in the enumeration. This contradicts the original assumption, so T must be uncountable."

s is not an element of T because there is no natural number greater than every natural number. The proof fails.
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May 16th, 2016, 07:44 AM   #6
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None of the elements of $T$ are natural numbers, they are infinite binary sequences in that formulation. Neither does Cantor claim his $s$ to be any "bigger" than any of the enumerated elements of $T$.

Your argument (as previously stated ad nauseam) is that "the" enumeration is:
0000000... = 0
1000000... = 1
0100000... = 2
1100000... = 3
...
1111111... =?
This suffers from numerous problems:
  1. there is no single enumeration, the theorem uses any enumeration;
  2. every element in the enumeration suggested by the first four lines terminates in an infinite string of zeros, so the last element given is not actually in the enumeration;
  3. if it were in the enumeration, it would not represent a natural number as you have assumed the elements of the enumeration to do;
  4. if it did represent a natural number, it would be one that is greater than all the others: which you yourself have stated does not exist.
There are probably more.

Last edited by v8archie; May 16th, 2016 at 07:58 AM.
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May 16th, 2016, 08:11 AM   #7
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Quote:
Originally Posted by v8archie View Post
None of the elements of T are natural numbers, they are infinite binary sequences in that formulation.
Infinite is non-finite. Countably infinite satisfies that criteria.
Any other interpretation makes Cantor's diagonal argument meaningless.

If $s_i$ is any sequence in the enumeration, it has a maximum length n followed by all zeros. No matter what n is, sooner or later Cantor's construction produces a non-zero element of s greater than n. There is no infinite sequence greater than any infinite sequence.

If infinite means more than countably infinite, the countably infinite sequences will be a subset of the infinite sequences and those will be the only ones you can enumerate.

Last edited by skipjack; May 23rd, 2016 at 03:59 PM.
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May 16th, 2016, 08:36 AM   #8
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Quote:
Originally Posted by zylo View Post
If $s_i$ is any sequence in the enumeration, it has a maximum length n followed by all zeros.
This is trivially not all of T as you claim. Futhermore, the diagonal sequence is guaranteed to end in all zeros and thus Cantor's s ends in an infinite string of 1s which by your definition is not in the enumeration, as Cantor claims.

Last edited by skipjack; May 23rd, 2016 at 04:00 PM.
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May 16th, 2016, 08:42 AM   #9
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Quote:
Originally Posted by zylo View Post
Infinite is non-finite. Countably infinite satisfies that criteria.
Any other interpretation makes Cantor's diagonal argument meaningless.
If this is your justification as to why all the elements of T are natural numbers, it is clearly a circular argument. You are also, yet again, confusing infinite sequences with an infinite number of sequences.

Last edited by skipjack; May 20th, 2016 at 02:39 PM.
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May 16th, 2016, 09:36 AM   #10
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Quote:
Originally Posted by zylo View Post
If infinite means . . .
If you're unsure what Cantor meant by it, why not take the trouble to find out? If you think he wrote something that's incorrect, simply state the precise point at which his mistake occurs instead of repeating "if this means this... if this means that..." arguments that don't even mention the sequences that Cantor uses.

If you can't specify the precise point at which Cantor (in your opinion) makes a mistake, you aren't justified in asserting that it's Cantor rather than you who has slipped up, especially if you insist on misquoting available accepted translations of Cantor's work. As Cantor initially explains his diagonal construction without any use of T, why not start by telling us whether you agree with that part of his work? If you disagree with that part, say where exactly he goes wrong and do so without mentioning T, as T is not used by Cantor in the first part of his work (the part where he isn't using proof by contradiction).
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