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From: Richard Heathfield <rjh@cpax.org.uk>
Newsgroups: comp.theory
Subject: Re: Cantor Diagonal Proof
Date: Fri, 4 Apr 2025 09:16:17 +0100
Organization: Fix this later
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On 04/04/2025 09:05, Lawrence D'Oliveiro wrote:
> On Fri, 4 Apr 2025 08:41:35 +0100, Richard Heathfield wrote:
> 
>> On 04/04/2025 08:21, Lawrence D'Oliveiro wrote:
>>>
>>> At every point N, we have the first N digits of our
>>> hypothetical number-that-is-not-in-the-list. But we have an infinitude
>>> of remaining numbers in the list we haven’t looked at, among which all
>>> possible combinations of those N digits will occur.
>>
>> Show me your first N digits, and I'll show you a counterexample.
> 
> Counterexample to what?

Your claim:
>>> At every point N, we have the first N digits of our 
>>> hypothetical number-that-is-not-in-the-list. But we have 
>>> an infinitude of remaining numbers in the list we haven’t
>>> looked at, among which all possible combinations of those
>>> N digits will occur.


> 
>>> Therefore there is guaranteed to be some number we haven’t looked at
>>> yet with all those first N digits the same.
>>
>> And yet you still won't post those first N digits.
> 
> Digit 1 is the first digit of entry 1 in the list.
> Digit 2 is the second digit of entry 2 in the list.
> .
> .
> .
> Digit N is the Nth digit of entry N in the list.

All right, from your data I deduce that your list is:

11
22

and that your Cantor construction to date is 12.

My counterexample for that rather unchallenging case is:

n=1: 11
n=2: 22
n>=3: 3*10^n

Since all elements (except your two openers) begin with a 3, none 
of them start 12, and so after just two iterations we have 
already constructed a number that's not in the infinite list.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
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