Re: Cantor



Thanks. I think I've got it and I appreciate your help and your patience.

In laymans terms:
- In the case of enumerable things (like natural numbers), I'll eventually
find it on the infinite list of naturals if only I look long enough. The
fallacy was in stopping and excalming that because I've not found it yet, it
must not be there.
- In the case real numbers, no matter how big I make the list or how long I
look there will always be at least one real that can't possibly be on the
list. And since there are an infinite number of ways to construct that
missing real, there are an infinite number of reals that aren't on even an
infinite list.

Thanks again.

"Randy Poe" <poespam-trap@xxxxxxxxx> wrote in message
news:1128519621.564953.271540@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
> Pedro wrote:
>> I think I'm getting sidetracked by giving poor examples. Can't see the
>> forest for the trees, and all that... Let me ask it one more way, if I
>> may.
>>
>> It's obviously impossible to actually present a "complete" list of reals
>> prior to initiating the diagonalization method as the list is infinitely
>> long. Rather, the diagonalization method is a thought experiment that
>> says,
>> in effect, "no matter how complete one makes the list, one can use this
>> typographical method to find with certainty a number not on the list".
>>
>> But I can make the same conceptual argument about the natural numbers (or
>> non-negative integers, if you prefer). In my example below, SSSS0 differs
>> in
>> the first position from 0, differs in the second position from S0,
>> etc...,
>> thereby giving a number that differs from all others on the list and
>> demonstrating an absurdity: that the natural numbers are not enumerable.
>> (The only new assumption / constraint I've placed on this example is that
>> the list be ordered. That doesn't seem to be an unreasonable thing to do
>> with the natural numbers.)
>>
>> So I'm obviously missing something fundamental about the whole
>> diagonalization thing. Any idea what it is?
>
> This point is made below by James Dolan. I'm just putting in
> my two cents worth.
>
> In the "diagonalization" proof that [0,1] is uncountable, the
> form of the proof is:
>
> 1. Let f:N->[0,1] be any mapping from N to [0,1].
> 2. There exists x in [0,1] such that x does not equal
> f(n) for any n in N.
>
> The construction of x in step 2 relies on a theorem that
> any string of digits 0.ssssss... represents an x in [0,1].
>
> Remember, the purpose of the construction is not to produce
> a sequence, but to generate a single value x which, by the
> construction, is not mapped by any f(n). This number x has
> the property that its n-th digit differs from the n-th digit
> of f(n).
>
> So the corresponding proof in the natural numbers should be
> trying to construct a corresponding m in N. But you do not
> have the corresponding theorem that any (possibly infinite)
> sequence of digits represents a natural number. Therefore
> the construction can not be guaranteed to produce such a
> number.
>
> - Randy
>


.



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