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I don't fully understand the concept behind...

(1) The Cantor Pairing Function and

(2) Cantor's Diagonalization Method.

I understand that (1) and (2) involve proving if a set is countable or not.

For example, let's consider the set of all real numbers, $\mathbb{R}$, which is known to be uncountable. We can prove this set is uncountable using Cantor's Diagonlization method.

To keep things simple, let's look at the set of real numbers between 0 and 1. We can prove that this subset is uncountable by using contradiction. Thus, we initially assume the subset is countable, like so:

$y_1$ = $0 \: . \: x_{11} \: x_{12} \: x_{13} \: . . .$

$y_2$ = $0 \: . \: x_{21} \: x_{22} \: x_{23} \: . . .$

$y_3$ = $0 \: . \: x_{31} \: x_{32} \: x_{33} \: . . .$

.

.

.

$y_k$ = $0 \: . \: x_{k1} \: x_{k2} \: x_{k3} \: . . .$

where each $x_{ij}$ is a decimal place.

Now we can consider $y' = 0. x_{11} \: x_{22} \: x_{33} . . .$, for $y' \in [0,1]$. This particular $y'$ is the diagonal... We can now immediately see that $y'$ is not in the subset, which proves that the set of real numbers between 0 and 1 is not countable.

So now I have 2 questions...

(1) How does Cantor's diagonalization method prove that this subset is uncountable? I understand that $y'$ is not in the subset, but how does this prove we can't count everything? Is this because we can create permutations of $y'$ that are also not in the subset? For example, let $y''$ and $y'''$ be permutations of $y'$:

$y'' = 0. \: x_{22} \: x_{33} \: x_{44} . . . \: x_{kk} . . . \: x_{11}$

$y''' = 0. \: x_{33} \: x_{44} \: x_{55} . . . \: x_{kk} . . . \: x_{11} \: x_{22}$

etc..

(2) Can I use Cantor's pairing function to prove the set is countable or not? If I can, how would I use it? If I can't, why can't I use it?

As a side note, I saw this question earlier about Cantor's pairing function: Prove that the union of countably many countable sets is countable.

It only confused me further. I don't understand how this method creates a one-to-one mapping to $\mathbb{N}$

Community
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maths_student15
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    The description you gave of $y'$ is not correct. It should differ from the $i$-th listed number in the $i$-th digit - not equal it. The point is that whenever we assume that the list is complete so that ALL numbers have been listed, we can construct another number not in the list. This is a contradiction. Just read it carefully ... – String Dec 13 '14 at 23:28
  • @String: There is absolutely no need to cast it as a proof by contradiction. I’s an algorithm that when given any enumeration ${x_n:n\in\Bbb N}$ of a set of real numbers in $(0,1)$ produces a real number in $(0,1)\setminus{x_n:n\in\Bbb N}$. Therefore there no such enumeration exhausts $(0,1)$. By definition this means that $(0,1)$ is uncountable. – Brian M. Scott Dec 14 '14 at 00:17
  • @BrianM.Scott: OK, fair point. Still the OP seemed to have trouble with the idea that producing a single number not in the list actually proves uncountability. I think the confusion is often that "why can't we just add that newly constructed number to the list?". – String Dec 14 '14 at 00:23
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    @String: That’s why I like to emphasize the universality of the algorithm: no matter how one tinkers with the list, the algorithm always produces a number not on it. Thus, no list can be complete. Overall I’ve had more success explaining it this way, though some people do seem to find the argument by contradiction more satisfying. – Brian M. Scott Dec 14 '14 at 00:27

1 Answers1

2

(1) It proves that the subset is uncountable since we assumed it was countable, so that all of its elements could be put in a list, and derived a contradiction (it turns out not all of the elements were in the list).

(2) Cantor's pairing function isn't useful here. This is only useful for showing $\mathbb N\times \mathbb N$ and $\mathbb N$ have the same size.

Mike Earnest
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