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 March 23rd, 2016, 08:00 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,545 Thanks: 110 Cantor's Diagonal Argument Reconsidered Let T be the list of all natural numbers, (1,2,3,4,5,........) Perform an operation which takes account of every member of the list to come up with 4. 4 is not a member of the list. Proof: 4 is different from every member of the list except itself. Every rearrangement of the list produces another unique member of the list which is not on the list. Therefore the list is empty. The point being an analogy with Cantor's diagonal argument: The fact that you produce a number different from every member of the list does not prove it is not on the list. Last edited by skipjack; March 23rd, 2016 at 02:57 PM.
March 23rd, 2016, 09:14 AM   #2
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Quote:
 Originally Posted by zylo Let T be the list of all natural numbers, (1,2,3,4,5,........) Perform an operation which takes account of every member of the list to come up with 4.
Eh?

Quote:
 4 is not a member of the list.
um... it's in the list dude...

Quote:
 Proof: 4 is different from every member of the list except itself.
I'm not sure about "proof", but that statement makes sense because the list of natural numbers doesn't have duplicates.

Quote:
 The point being an analogy with Cantor's diagonal argument: The fact that you produce a number different from every member of the list does not prove it is not on the list.
What? If a number is different from every number on a list, it is not in the list.

Last edited by skipjack; March 23rd, 2016 at 02:58 PM.

March 23rd, 2016, 09:53 AM   #3
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Quote:
 Originally Posted by Benit13 What? If a number is different than every number on a list, it is not in the list.
4 is different than every member of the list (except itself, obviously)
4 is in the list.

EDIT
The point is, showing that a number (sequence) is different than every member of a list of numbers (sequences), doesn't prove it is not in the list, 4 above, for ex.

Last edited by zylo; March 23rd, 2016 at 10:05 AM.

 March 23rd, 2016, 01:18 PM #4 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,889 Thanks: 769 Math Focus: Wibbly wobbly timey-wimey stuff. Another thread? Really? -Dan
 March 23rd, 2016, 03:02 PM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,545 Thanks: 110 To bring things a little closer to home. Assume T is the set of all countably infinite binary digits, i.e., the set of all natural numbers, which is countable. But Cantor's argument says T is uncountable? What's wrong? See OP. REF: Cantor's diagonal Argument: The set of all infinite binary sequences is uncountable. Let T be the set of all infinite binary sequences. Assume T is countable. Then all its elements can be enumerated: 1 0 0 1 1 0............... 0 0 1 0 1 1............... 0 1 0 1 0 0.............. ............................ Let s be the binary sequence consisting of the complemented underlined digits: s = 0 1 1................... s is different from every member of the list and s belongs to T. Contradiction. Therefore T isn't countable. ____________________________________ https://en.wikipedia.org/wiki/Cantor...gonal_argument _____________________________________ Each thread is a different seminal argument which would otherwise get buried. Considering all that has been written on the subject in the last 100+ years, and the assertion that it is fundamental to the foundations of mathematics, a few weeks and threads are not unreasonable. Last edited by skipjack; March 23rd, 2016 at 03:15 PM.
March 23rd, 2016, 03:07 PM   #6
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Quote:
 Originally Posted by zylo . . . (except itself, obviously) . . .
Adding this means that what you're describing isn't analogous to what Cantor did, as Cantor didn't use "except itself". Hence it's your conjectured scenario that doesn't come up to scratch, whilst Cantor's proof remains untarnished.

March 23rd, 2016, 03:14 PM   #7
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Quote:
 Originally Posted by zylo Assume T is the set of all countably infinite binary digits, i.e., the set of all natural numbers
It doesn't make sense to add "i.e., the set of all natural numbers" because no countably infinite binary sequences are natural numbers. As Cantor didn't do that, you're countering your own approach, not what Cantor did.

 March 23rd, 2016, 03:43 PM #8 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,545 Thanks: 110 Guess you missed my previous post: "Assume T is the set of all countably infinite binary digits, i.e., the set of all natural numbers, which is countable. But Cantor's argument says T is uncountable? What's wrong? See OP." "except itself" is obvious, but in any event doesn't change above proposition. It's part of the explanation in OP of why Cantor's argument doesn't work. Thanks from Prakhar
 March 23rd, 2016, 05:01 PM #9 Global Moderator   Joined: Dec 2006 Posts: 19,713 Thanks: 1806 What is the OP that you are referring to? I didn't miss your post with "Assume T" in its second sentence as that's what I quoted. It's not what Cantor does, but instead something that you've devised that doesn't make sense, as no element of T is a natural number.
March 23rd, 2016, 06:59 PM   #10
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Zylo: why does your making a host of assumptions that Cantor didn't (and which happen to be nonsense) make Cantor's proof wrong?

If you wish to critique Cantor, you can start by learning what he did say, rather than making stuff up.

Quote:
 Cantor considered the set T of all infinite sequences of binary digits (i.e. consisting only of zeros and ones).
These are not natural numbers, they are infinite sequences. Because they are infinite, they cannot be identified in any obvious way with the natural numbers, which are all finite. Cantor makes no claims about the countability of T until he has used his diagonal argument to prove a different theorem.

Quote:
 If $s_1,\, s_2,\, \ldots ,\, s_n,\, \ldots$ is any enumeration of elements from T, there is always an element $s$ of T which corresponds to no $s_n$ in the enumeration. (My emphasis.)
"An enumeration of elements from T" is a countably infinite subset of T which is necessarily countable by its identification with the natural numbers as subscripts on the elements of the enumeration. We also identify the elements of the sequences in this way so that $s_k=\{a_{k,1},\,a_{k,2},\, \ldots,\,a_{k,n},\,\ldots\}$. We are thus guaranteed that the $m$th sequence has an $m$th member, $a_{m,m}$, for all natural numbers $m$ and that the sequence of these elements $s'=\{a_{1,1},\,a_{2,2},\, \ldots,\,a_{n,n},\,\ldots\}$ contains one element from each sequence of the enumeration. We form $s=\{b_{1},\,b_{2},\, \ldots,\,b_{n},\,\ldots\}$ by changing every element of $s'$ and thus guarantee that $s$ is different from each of the $s_k$. Specifically, $s$ differs from $s_k$ in the $k$th element because $b_k \ne a_{k,k}$ by construction. So $s$ is not in the enumeration, and the theorem is proved.

This means that any enumeration E of elements of T is, not equal to T. Thus any countably infinite subset of T is not equal to T. The corollary to this statement is that T is not countably infinite.

Why is this so difficult to grasp?

Last edited by skipjack; March 23rd, 2016 at 10:05 PM.

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