Re: Cantor Diagonal Proof

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
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