Re: Cantor Diagonal Proof

On 4/11/25 6:15 PM, Lawrence D'Oliveiro wrote:

On Fri, 11 Apr 2025 09:15:38 -0400, Richard Damon wrote:
 

On 4/11/25 3:27 AM, Lawrence D'Oliveiro wrote:

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On Thu, 10 Apr 2025 21:29:51 -0400, Richard Damon wrote:
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On 4/10/25 8:30 PM, Lawrence D'Oliveiro wrote:
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On Wed, 9 Apr 2025 22:00:15 -0400, Richard Damon wrote:
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The problem with the "Cantor Construction" is that one of the steps,
"Compute to the Nth digit of the Nth number" isn't computable with a
finite algorithm.

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But N is always finite, therefore the construction is always finite.

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N my be finite, but is unbounded, and thus the construction is
unbounded in size, and thus not finite.

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Why is N allowed to be finite but unbounded, but not the construction?

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Because N is a variable, but the construction is supposed to be a
constant.

 Do you know how loops work in algorithms? A finite amount of code can be
executed an unbounded number of times. Without this capability, we would
have no “halting problem”. The construction can be conveyed with a finite
amount of information, which makes it finite.
 (See also “primitive recursive”.)

Yes, but since you need the algorithms to compute ALL the numbers in your code, you can't put them all in.
The problem is that we CAN'T write ONE algorithm to compute all of the numbers based on an input value.
This is hard to prove, but to make your claim, it is on YOU to prove that you can do it, which you will not be able to do.
Loops don't help you here.