A question on Cantor's second digitalization argument.

  • Thread starter Organic
  • Start date
  • Tags
    Argument
In summary, Cantor's second diagonalization argument is used to show that the set of real numbers between 0 and 1 is uncountable. This is done by constructing a real number that cannot be put in a one-to-one correspondence with any natural number. The proof proceeds by assuming the set is countably infinite and then using a proof by contradiction to show that this is not possible. The argument relies on the fact that there are an infinite number of decimal expansions for any real number, and by constructing a number that differs from every number in a countable list, the set must be uncountable. The construction of this number is achieved by using Cantor's switching function. While there may be different ways to construct a list
  • #1
Organic
1,224
0
A question on Cantor's second diagonalization argument.

Hi,

Cantor used 2 diagonalization arguments.

On the first argument he showed that |N|=|Q|.

On the second argument he showed that |Q|<|R|.

I have some question on the second argument.

From Wikipedia, the free encyclopedia:
http://en.wikipedia.org/wiki/Cantor's_diagonal_argument

The proof by contradiction proceeds as follows:
• (1) Assume that the interval (0,1) is countably infinite.
• (2) We may then enumerate the numbers in this interval as a sequence, { r1, r2, r3, ... }
• (3) We shall now construct a real number x between 0 and 1 by considering the nth digit after the decimal point of the decimal expansion of rn. Assume, for example, that the decimal expansions of the beginning of the sequence are as follows.
r1 = 0 . 0 1 0 5 1 1 0 ...
r2 = 0 . 4 1 3 2 0 4 3 ...
r3 = 0 . 8 2 4 5 0 2 6 ...
r4 = 0 . 2 3 3 0 1 2 6 ...
r5 = 0 . 4 1 0 7 2 4 6 ...
r6 = 0 . 9 9 3 7 8 3 8 ...
r7 = 0 . 0 1 0 5 1 3 0 ...
...
The digits we will consider are indicated in bold. From these digits we define the digits of x as follows.

o if the nth digit of rn is 0 then the nth digit of x is 1
o if the nth digit of rn is not 0 then the nth digit of x is 0
For the example above this will result in the following decimal expansion.
x = 0 . 1 0 0 1 0 0 1 ...
The number x is clearly a real number between 0 and 1.
• (4) However, it differs in the nth decimal place from rn, so x is not in the set { r1, r2, r3, ... }.
• (5) This set is therefore not an enumeration of all the reals in the interval (0,1).
• (6) This contradicts with (2), so the assumption (1) that the interval (0,1) is countably infinite must be false.

Note: Strictly speaking, this argument only shows that the number of decimal expansions of real numbers between 0 and 1 is not countably infinite. But since there are expansions such as 0.01999... and 0.02000... that represent the same real number, this does not immediately imply that the corresponding set of real numbers is also not countably infinite. This can be remedied by disallowing the decimal expansions that end with an infinite series of 9's. In that case it can be shown that for every real number there is a unique corresponding decimal expansion. It is easy to see that the proof then still works because the number x contains only 1's and 0's in its decimal expansion.


My question is:

The first assumption was that we have a bijection between |N| and |R| iff R list is complete.

Then Cantor showed that there is a way to construct a real number x between 0 and 1, that cannot be put in 1-1 correspondence with any natural number.

Therefore, there are more real numbers between 0 and 1 then all natural numbers.

Now I construct the list in another way.

...
r7' = 0 . 0 1 0 5 1 3 8 ...
r6' = 0 . 9 9 3 7 8 0 8 ...
r5' = 0 . 4 1 0 7 0 4 6 ...
r4' = 0 . 2 3 3 5 1 2 6 ... New list
r3' = 0 . 8 2 0 5 0 2 6 ...
r2' = 0 . 4 0 3 2 0 4 3 ...
r1' = 0 . 1 0 0 1 0 0 1 ...(= Cantor's switching function resultes)
----------------------
r1 = 0 . 0 1 0 5 1 1 0 ...
r2 = 0 . 4 1 3 2 0 4 3 ...
r3 = 0 . 8 2 4 5 0 2 6 ...
r4 = 0 . 2 3 3 0 1 2 6 ... Original list
r5 = 0 . 4 1 0 7 2 4 6 ...
r6 = 0 . 9 9 3 7 8 3 8 ...
r7 = 0 . 0 1 0 5 1 3 0 ...
...

By constructing the above list I show that there are infinitely countably many r' numbers that are not covered by Cantor's switching function.

Therefore Cantor's digitalization's second argument doesn't hold.


Please someone show my mistake.


Thank you.


Organic
 
Last edited:
Mathematics news on Phys.org
  • #2
Let's write down an enumeration of your list:

Code:
r1  = 0.0105110...
r1' = 0.1001001...
r2  = 0.4132043...
r2' = 0.4032043...
r3  = 0.8245026...
r3' = 0.8205026...
...

Then, just like any other enumeration, you apply Cantor's diagonal argument and you can generate a number that cannot appear in the list. In this case, 0.110010...
 
  • #3
It doesn't help to post the same question twice.
 
  • #4
Hi HallsofIvy,


By complete I mean that there are no numbers left out not in R and not in N (means bijection between N and R).

The R list is arbitrary and it is one and only one list.
"New list" and "Original list" are the same one list.

All what I showed is that we can construct it in such a way that Cantor's function can never reach all of it.

Here it is again:
...
r7' = 0 . 0 1 0 5 1 3 8 ...
r6' = 0 . 9 9 3 7 8 0 8 ...
r5' = 0 . 4 1 0 7 0 4 6 ...
r4' = 0 . 2 3 3 5 1 2 6 ... "New list"
r3' = 0 . 8 2 0 5 0 2 6 ...
r2' = 0 . 4 0 3 2 0 4 3 ...
r1' = 0 . 1 0 0 1 0 0 1 ...(= Cantor's function resultes)
----------------------
r1 = 0 . 0 1 0 5 1 1 0 ...
r2 = 0 . 4 1 3 2 0 4 3 ...
r3 = 0 . 8 2 4 5 0 2 6 ...
r4 = 0 . 2 3 3 0 1 2 6 ... "Original list"
r5 = 0 . 4 1 0 7 2 4 6 ...
r6 = 0 . 9 9 3 7 8 3 8 ...
r7 = 0 . 0 1 0 5 1 3 0 ...
...

You Wrote:
Let's write down an enumeration of your list:

r1 = 0.0105110...
r1' = 0.1001001...
r2 = 0.4132043...
r2' = 0.4032043...
r3 = 0.8245026...
r3' = 0.8205026...
...

Then I use r'', r''' ,r'''' ,... on top of your list.


Organic
 
  • #5
Last edited:

1. What is Cantor's second digitalization argument?

Cantor's second digitalization argument is a mathematical proof that shows there is an infinite number of real numbers between any two real numbers. This argument is based on the concept of Cantor's diagonalization, which states that there are more real numbers than natural numbers.

2. How does Cantor's second digitalization argument work?

Cantor's second digitalization argument works by assuming that there is a finite number of real numbers between any two real numbers. It then uses a proof by contradiction to show that this assumption leads to a contradiction, thus proving that there must be an infinite number of real numbers between any two real numbers.

3. What is the significance of Cantor's second digitalization argument?

Cantor's second digitalization argument is significant because it demonstrates the concept of infinity in a concrete and mathematical way. It also has implications for other areas of mathematics, such as set theory and analysis.

4. Are there any criticisms of Cantor's second digitalization argument?

Yes, there are some criticisms of Cantor's second digitalization argument. Some mathematicians have argued that it relies on the assumption that real numbers are discrete and countable, which is not necessarily true. Others have questioned the validity of using a proof by contradiction in this argument.

5. Can Cantor's second digitalization argument be applied to other areas of mathematics?

Yes, Cantor's second digitalization argument has been applied to various other mathematical concepts, such as the Cantor set, fractals, and the cardinality of different infinite sets. It has also been used in computer science and philosophy to explore the concept of infinity.

Similar threads

Replies
11
Views
366
  • General Math
Replies
4
Views
698
Replies
3
Views
377
  • General Math
Replies
24
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
Replies
5
Views
803
  • General Math
Replies
2
Views
871
Replies
4
Views
281
Replies
5
Views
1K
Replies
4
Views
825
Back
Top