Talk:Cantor's diagonal argument/Archive 1

message: I think Cantor's argument as you describe it is fundamentally flawed. 
Let's look at statement (5) in the 'proof' which says:
"This set is therefore not an enumeration of all the reals in the interval (0,1)."
This statement seems to be exactly right.  The set Cantor has constructed is NOT an enumeration of all the reals. But there is no contradiction with (2)which says "We may then enumerate the numbers in this interval as a sequence, { r1, r2, r3, ... }"  Cantor has simply failed to find a way to construct them all.  In fact, this is the only thing his argument proves; that the diagonal method must of necessity fail.
Wherever Cantor stops on his way to infinity  (namely at n ), he will always find that he has failed to list the numbers he hasn't yet reached.  Why should this surprise anyone?
:You misunderstand utterly completely.  The diagonal method is not an attempt to enumerate the real numbers!  It is an attempt to show why any sequence whatsoever must always fail to enumerate them. There is no failure of any particular method to point to, since no particular method is proposed. Michael Hardy 19:22, 24 Oct 2003 (UTC)
thread_title:  Cantor's Error 

message: Mike:
I think you have missed the whole point here when you say "There is no failure of any particular method to point to, since no particular method is proposed".  It is Cantor himself who proposes the method, and it is he who finds that it fails!
When you say "The diagonal method is not an attempt to enumerate the real numbers" you again miss the point. That is the whole idea behind Cantor's argument! "Cantor's diagonal argument (was devised) to demonstrate that the real numbers are not countably infinite."   He does this by assuming the opposite (that they can be enumerated) and then looks for a contradiction. 
The numbers on the left (r) are intended to be from a countably infinite set like the natural numbers.  Cantor then tries to enumerate the reals in a chosen interval on the right, and place them in one-to-one correspondence with (r) on the left.  If he succeeds, the reals are countably infinite.
Now this is where Cantor goes wrong. Because his list fails to include some real number x on the right, he assumes that x can not be paired with any natural number (r) on the left.  He thinks he has found a contradiction where there is none. For ANY new number x Cantor creates on the right, there is also a new natural number (r) on the left with which it can be paired. He somehow thinks that his "diagonal" can be completed before he constructs his new number x. It can not be.  There is no such thing as a completed infinite set.
:You're wrong.  Cantor does not try to enumerate the reals; rather, he assumes it's already been done, and goes from there to a contradiction.  He was not proposing any particular enumeration of the reals.  Obviously, if he proposed some particular enumeration, and found that it fails to enumerate the reals, that would not prove that the reals are uncountable; it would only prove that one particular enumeration fails.  You have entirely misunderstood his argument. Michael Hardy 17:34, 31 Oct 2003 (UTC)
replies: text: Wrong: There is no last number in your list. Michael Hardy 21:26, 26 Aug 2004 (UTC)

text: (William M. Connolley 10:16, 24 Nov 2004 (UTC)

text: I would like to see more about this construction and issues arising on my shiny new talk page. --Flabdablet 05:17, 24 Nov 2004 (UTC)
thread_title:  Thanks Mike but... 

message: This is a question that I've had for some time, I've never seen anyone come across anything that really states that [0,1] is countable, but I kinda think that I have a counterproof. The proof is a proof by contradiction, I assume that the diagonal proof is true.
For simplicity, I'm doing this in binary (base 2), but the same result can be achieved in any base.
I have an ordering for the reals that maps it to the natural numbers. The ordering has a simple algorithm, take the binary number for the natural number (minus 1), and reverse the digits, and place a decimal point in front of it.
The numbers may look like this 
0.0
0.1
0.01
0.11
0.001
0.101
0.011
0.111
0.0001
...
Then via the diagnoal "proof" I choose an algorithm. If the xth digit is 0 I replace it with a 1 and vice versa. I will have to zero pad the numbers
       0.0000000000
       0.1000000000
       0.0100000000
       0.1100000000
       0.0010000000
       0.1010000000
       0.0110000000
       0.1110000000
       0.0001000000
...
A simple proof will show that all of the digits will be zero.
So using my algorithm, the number that the diagnoal proof says isn't in my list, is .111111111111111... (an infinite set of ones).
But isn't that number in this list? Yes, it will be the last number in the list, but it will be there right? If so, then this is a counter-proof for the diagnoal method
:Wrong: There is no last number in your list. Michael Hardy 21:26, 26 Aug 2004 (UTC)
replies: text: (William M. Connolley 10:16, 24 Nov 2004 (UTC)

text: I would like to see more about this construction and issues arising on my shiny new talk page. --Flabdablet 05:17, 24 Nov 2004 (UTC)
thread_title: Counterproof(?) for Diagonal method on [0,1]

message: 1)You might insert a line after step 2. "Somewhere among the numbers is .0100.... If it is r1 leave it alone, otherwise exchange it with r1 so now r1=.0100... and the new  ri's count  (0,1) if the old ones did." Now make the necessary changes in r1 and the example. Now the x will be .1something and is not 0 and is in (0,1) as required and produces the contradiction.
2)Another way would be to use the given argument for [0,1) and give an explicit 1-1, onto function relating [0,1) and (0,1). For example: Let 
S={0, 1/2, 1/3, 1/4,...} and 
T={1/2, 1/3, 1/4,...}
so S is contained in [0,1) and T is contained in (0,1) and 
[0,1)-S is the same as (0,1)-T.
Define f:S to T by f(0)=1/2, f(1/2)=1/3, f(1/3)=1/4, etc. So f is 1-1, onto.
Define g:[0,1) to (0,1) by g(x)=x on [0,1)-S and g(x)=f(x) for x in S
Then g is the desired 1-1 correspondence.
Thus the main argument proved [0,1) not countable hence the above g shows (0,1) not countable.
----
One solution I have once seen to avoid the 0.09999... = 0.1 issue is to consider only the set S of numbers in [0, 1) that have only zeros and ones in their decimal representations. S is obviously a subset of [0, 1), and the diagonal argument works without exceptional cases.
:The entry itself should discuss the issue, as it's important to understanding variations of the proof. The goal is not to create the most asthetically pleasing presentation, but rather to instruct as best we can about the topic. MShonle 09:10, 30 Mar 2004 (UTC)
thread_title:  possible patches for (0,1) 

message: Wouldn't the more concrete proof of Cantor's theorem be more appropriate on Cantor's theorem? -- Jan Hidders 11:50, 23 Jun 2004 (UTC)
replies: text: Jan: you are right, I moved the concrete method to cantor's theorem and just provided a link to it in the cantor's diagonal argument article.--Dissipate 17:01, 24 Jun 2004 (UTC)
replies: text:  Great, thanks. -- Jan Hidders 17:46, 24 Jun 2004 (UTC)
thread_title:  Move proof of Cantor's theorem to Cantor's theorem? 

message: There's a slight problem with this digit replacement scheme: 
          * if the nth digit of rn is 0 then the nth digit of x is 1
          * if the nth digit of rn is not 0 then the nth digit of x is 0
Precisely, this can run afoul of the 0.09999999... = 0.1 problem.  Suppose our list of reals goes
       r1 = 0 . 0 9 9 9 9 9 9 ... 
       r2 = 0 . 4 9 3 2 0 4 3 ...
       r3 = 0 . 8 2 9 5 0 2 6 ... 
       r4 = 0 . 2 3 3 9 1 2 6 ...
       r5 = 0 . 4 1 0 7 9 4 6 ... 
       r6 = 0 . 9 9 3 7 8 9 8 ...
       r7 = 0 . 0 1 0 5 1 3 9 ... 
(with 9s the rest of the way on the diagonal).  Then x turns out to be 0.1000000..., which is equal to r1.  So I've changed the 0 and 1 to 4 and 5, which should cause no problems.  4pq1injbok 00:31, 2 Jul 2004 (UTC)
thread_title:  Replacement digit choices 

message: Tonight I had an argument with my girlfriend about infinites being equal and the countablity of real numbers.  She brought up Cantor's argument as proof of her position... After reading it, I am fairly confident that I found an obvious flaw.
Consider Cantor's proof but assume that due to the unspecified matching function the matched rn we're going after is n = ∞ by the proof we see that x and n would differ in the ∞th place.  If we accept that .99999.... = 1.000... then we must accept that x = n and thus this proof fails to show that you can't map reals onto the natural numbers.
If we deny that .9999... = 1.0000... then we break a lot of math including all of calculus.  
: (William M. Connolley 11:12, 6 Jan 2005 (UTC)
replies: text:  Right, no argument there.
I don't pretend to have a proof that you can map the reals onto the natural numbers, but rather I believe I've shown why the proof here fails to prove that it isn't possible.Gmaxwell 04:08, 6 Jan 2005 (UTC)
thread_title: A more simple argument

message: How could there be a infinitieth place? The sequence never terminates.
Grick 05:10, Jan 6, 2005 (UTC)
: Because if given a function f(n) (the mapping function) we can solve for f(n) as the limit approaching infinity.  A simple mapping function for x in the range [0,1] might be x = n * (1/∞). 
: (William M. Connolley 17:24, 6 Jan 2005 (UTC)
replies: text: (William M. Connolley 17:24, 6 Jan 2005 (UTC)

text: (William M. Connolley 17:24, 6 Jan 2005 (UTC)

text: (William M. Connolley 17:24, 6 Jan 2005 (UTC)

text: No, because those are not integers.  An integer is not represented by an infinite sequence of digits.  Even if the rows eventually terminate, your diagonal is still an infinite sequence of digits, and therefore does not represent any integer.  Michael Hardy 00:47, 7 Jan 2005 (UTC)
thread_title:  response Gmaxwell 

message: Substituted at 06:32, 7 May 2016 (UTC)
thread_title: Assessment comment