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How does Cantor's diagonal argument work?

It seems to me that if you have an infinite list of unending decimal numbers between $(0,1)$:

0.xxxxxxxxx...
0.xxxxxxxxx...
...

and you want to generate a number in $(0,1)$ that is not on the list, then Cantor's diagonalization does not work. For example, consider

0.xxy.....

where each x can be any value, and y, being the third, must be different from the the third digit in the third number, is any number except the third digit in the third number.

In this case, we know our number is different from the third number, and because of how we have been writing it (first two digits are appropriately distinct from the first two numbers), we know it is unique so far on the list. However, because this is an infinite list which claims to contain all real numbers between 0 and 1, the number we have written exists somewhere further down the list.

So we continue. We add a fourth digit, which differentiates it from the fourth number in the list. Facing the same problem, we continue to the fifth, then the sixth, and so on...

The point of this, as I understand it, is to show that we can write a number which is not already in this list. However, at any point, there are an infinite number of numbers left in the list, one of which is equal to what we have written. So we must continue on to infinite...

Since we will never reach a point where there are 0 numbers left in the list, we will never finish writing a number which is not on the list. Every digit we add makes it different from more and more numbers on the list, but never from EVERY number on the list. So, in practice, we CAN'T generate a number which is not on the list.

This seems to make perfect sense to me, but it also seems to be incorrect. Can someone please explain why?

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yoozer8
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    Have you read http://math.stackexchange.com/questions/39269/how-does-cantors-diagonal-argument-work ? – Willie Wong Sep 01 '11 at 17:06
  • Cantor's argument is not about "writing down" or a process. The diagonal argument defines a complete number, it does not construct it "step-by-step". – Arturo Magidin Sep 01 '11 at 17:07
  • @WillieWong I wish I had before I typed this whole thing up, but it didn't come up in my search results. – yoozer8 Sep 01 '11 at 17:08
  • @Arturo It seems to me that it doesn't really DEFINE it, probably because I'm more of a CS than math person, because it defines/generates it by a method that will never reach a conclusion or an actual number. Since it never actually PRODUCES a number, can we accept that the number it's defining actually exists and is on the list? – yoozer8 Sep 01 '11 at 17:10
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    @Jim: Read my answer in the other question. To give a "list of real numbers between $0$ and $1$" means to give a function $f\colon\mathbb{N}\to (0,1)$. Given such a function, we define a function $g\colon\mathbb{N}\to{0,1,2,3,4,5,6,7,8,9}$ as follows: $g(n)=5$ if the $n$th digit in the decimal expansion of $f(n)$ is not $5$; and $g(n)=6$ if the $n$th digit in the decimal expansion of $f(n)$ is $5$. This defines a function $g$ on its entire domain. Now let $c=\sum g(n)10^{-n}$. This defines a real number (in the usual sense of real numbers). That number is completely defined. – Arturo Magidin Sep 01 '11 at 17:13
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    @Jim: There is no "Process", there is no "generation", there is only definition. The number $c$ is defined in exactly the same sense that $\pi$, $\sqrt{2}$, or $\frac{1}{3}$ is defined, and it is provable that $f(n)\neq c$ for all $n\in\mathbb{N}$. Again, this is not a process. Don't confuse the way in which $c$ is explained with the way it is mathematically defined. – Arturo Magidin Sep 01 '11 at 17:15
  • @Arturo Thanks, this explanation actually makes a lot of sense. – yoozer8 Sep 01 '11 at 17:20
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    @Jim: Since you can never write down the full expansion of a real number, nor can you every write down a full list of real numbers, why aren't you just as concerned with "Given an infinite list of undending decimal numbers"? Since you can never write down such a list (even just write it down so that you only show the first $n$ decimals of the $n$th number on the list), shouldn't you be jumping all over that even before we get to the rest of the argument? That's a valid philosophical point, by the way ; but you can't let that go and then object to that same issue later. – Arturo Magidin Sep 01 '11 at 17:21
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    The classic An Editor Recalls Some Hopeless Papers by W. Hodges contains a remarkably (for its title) respectful analysis and discussion of this fallacy. – hmakholm left over Monica Sep 01 '11 at 17:45

3 Answers3

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Perhaps it is better to think of the Cantor diagonalization as defining a number rather than constructing one. Let me explain what I mean.

If one defines a function $f(x) = x^2$, you presumably have no objection to the fact that a graph of this function exists (a parabola). There are, however, a vast infinity of possible values for $x$. If you bring your complaint about diagonalization to bear on the function $f$, then you would have to conclude something along the lines of "There are an infinite number of possible inputs, so I can't ever graph them all. Thus, no such graph exists."

There is a way to describe the function without listing each of its ordered pairs, however. Given any value of $x$, one can quickly compute the output of the function by squaring it. Thus, one never needs to graph the entirety of the parabola to demonstrate its existence. It is simply defined as "that function whose output is the square of its input".

In the same way, the number defined by the Cantor diagonalization process never need be written out in its entirety to exist. It is simply defined as "that number whose $n^\text{th}$ digit is obtained by adding 1 (mod 10) to the $n^\text{th}$ digit of the diagonal".

Austin Mohr
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What you show is that if you stop diagonalization after a finite number of steps it doesn't show that the list is incomplete. This is correct; you need an infinite number of steps to obtain the result.

Charles
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  • That's my problem with this - you will never reach an infinite number of steps, and therefore never obtain a result (never find a number between 0 and 1 that isn't already on the list) – yoozer8 Sep 01 '11 at 17:11
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    @Jim: Is it your belief as well that proofs by mathematical induction don't actually prove that something holds for all natural numbers? – Arturo Magidin Sep 01 '11 at 17:17
  • I had never considered the parallel, since with induction one would say that being true for x necessitates being true for x + 1, whereas in this situation it seemed more of a "it will be true for the next one on the list because we change this" without a clear definition of what the "next" element is, other than what was listed next. – yoozer8 Sep 01 '11 at 17:22
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    @Jim: It's exactly as I thought, then. What you call Cantor's diagonalization is not, in fact, Cantor's diagonalization. You're right that the method you refer to ("Jim's diagonalization") fails. In particular: using that method you can neither conclude that [0, 1] is uncountable nor that it is countable. When you use Cantor's actual diagonalization procedure you'll find that they are indeed uncountable. – Charles Sep 01 '11 at 17:25
  • @Jim: The procedure doesn't say what the next thing on the list is because it's not defining a list but rather showing a property of any list: that is does not contain all numbers in [0, 1]. – Charles Sep 01 '11 at 17:27
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Diagonalization defines a real number, which is an infinite object. You seem to be doubting the existence of numbers with nonterminating decimal expansions (such as 1/3).

Brendan Cordy
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  • What does an "infinite object" mean? – Asaf Karagila Sep 01 '11 at 17:23
  • Admittedly, this is a difficult point. To rigorously define the reals you need to talk about infinite sequences. However, you're probably used to dealing with these "infinite objects" (such as 1/3, or $\pi$) all the time. – Brendan Cordy Sep 01 '11 at 17:29
  • @Brendan: I would question the claim 1/3 is an 'infinite object' - or even, arguably, that $\pi$ is. In fact, there are countably many reals that we can talk about, each of which has a finite description(!) - it's a question of what language the description is in; if 'the' description of a real is its decimal expansion, then 1/3 has an infinite description, but obviously it has a finite description via other means. But these are subtle distinctions... – Steven Stadnicki Sep 01 '11 at 18:29
  • Yes, yes, I understand these issues, and they are beside my point. I'm merely pointing out that regardless of how much joe blow might know about the real numbers, he is probably comfortable with the idea that an infinite decimal, periodic or not, can represent a real number. – Brendan Cordy Sep 01 '11 at 18:41