# Re: Cantor's Diagonal

```Hi Dan,

Let me take your statements a few at a time.```
```
>> Let me see if I am clear about Cantor's  method.  Given a set S,
>> and some enumeration of that set (i.e., a no one-one onto map from
>> Z^+ to S) we can use the diagonalization  method to find an D
>> which is a valid element of S but is different from any particular
>> indexed element in the enumeration.

Not "given a set S and *some* enumeration..." but "given a set S and
*any candidate for an enumeration*..."  The idea is to show that no
candidate for an enumeration succeeds, hence the set S has no
enumeration, hence the set S is not enumerable.

>>  we can use the diagonalization  method to find an D which is a
>> valid element of S but is different from any particular indexed
>> element in the enumeration.

Yes.

>> Cantor's argument then goes on to say (and here is where I
>> disagree with it) that therefore D is not included in the
>> enumeration and the enumeration is incomplete.

Yes, except for your disagreement with it....

One step of Cantor's argument is to show, for a given candidate
enumeration of S, that some D in S is different from every single
enumerated member of S.  If you can't show this for some particular
candidate enumeration, you haven't done a Cantor diagonalization
argument that encompasses the candidate enumeration.  To do a
successful Cantor diagonalization argument, you must show that some D
in S is different from every single enumerated member of S, for each
candidate enumeration of S.  The value of D will typically depend on
the candidate.

>> I, on the other hand, would posit that the enumeration may include
>> elements whose index is not well defined. For example, 1 or 4 in
>> my second example.  They were in the enumeration prior to its
>> being shuffled, and always had a definite position during the
>> process.  They must still be in there, with definite positions,
>> despite the fact that their indices are now infinite and ill-defined.

I don't need to parse this shuffling enumeration to say that what
you're doing is not an instance of Cantor's argument.  Any instance
of Cantor's argument will debunk, not just one candidate shuffling
enumeration, to take your example, but each and every candidate
enumeration.  And you haven't even debunked the shuffling
enumeration.  In fact you're arguing that it's valid (it seems to me.)

>> If the diagonalization process does not produce the proffered
>> result in this case (i,e., it does not prove that the element D is
>> not included in the enumeration) then it does not prove it in any
>> case.  The number D found with this method may actually be in the
>> enumeration, but with an ill-defined index.

A defender of the validity of the form of Cantor's argument doesn't
*want* to use it to prove that Z^+ has no enumeration. To do so would
be to fall into the paradox you're trying to construct.  So it's
perfectly fine if the diagonalization process does not produce "the
proffered result" in this case.

Barry

On Dec 16, 2007, at 9:09 PM, Daniel Grubbs wrote:

> Hi Barry,
>
> Let me see if I am clear about Cantor's  method.  Given a set S,
> and some enumeration of that set (i.e., a no one-one onto map from
> Z^+ to S) we can use the diagonalization  method to find an D which
> is a valid element of S but is different from any particular
> indexed element in the enumeration.
>
> Cantor's argument then goes on to say (and here is where I disagree
> with it) that therefore D is not included in the enumeration and
> the enumeration is incomplete.
>
> I, on the other hand, would posit that the enumeration may include
> elements whose index is not well defined. For example, 1 or 4 in
> my second example.  They were in the enumeration prior to its being
> shuffled, and always had a definite position during the process.
> They must still be in there, with definite positions, despite the
> fact that their indices are now infinite and ill-defined.
>
> If the diagonalization process does not produce the proffered
> result in this case (i,e., it does not prove that the element D is
> not included in the enumeration) then it does not prove it in any
> case.  The number D found with this method may actually be in the
> enumeration, but with an ill-defined index.
>
> Dan

Dr. Barry Brent
[EMAIL PROTECTED]

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