Re: Cantor Diagonal Proof

On 4/10/25 8:33 PM, Lawrence D'Oliveiro wrote:

On Tue, 8 Apr 2025 18:59:02 -0400, Richard Damon wrote:
 

On 4/8/25 5:05 PM, Lawrence D'Oliveiro wrote:
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On Mon, 7 Apr 2025 06:47:43 -0400, Richard Damon wrote:
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And an infinite listing of values doesn't need to be computable, even
if every number in the list is computable.

>
Computability is a characteristic of particular numbers. It is a
characteristic of all the numbers in the list, and of the number that
the Cantor construction tries to construct from those numbers in the
list.
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The fact that you can’t apply that characteristic to the set as a whole
is irrelevant, since the set itself is not a number.

>
Right, so the DIAGONAL number, which you claim to be computable, needs a
finite algorithm to do so.
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The algorithm described is NOT FINITE, as it includes the infinite
number of algorithms to compute all the other numbers.

 Potentially, *any* computable number could need an infinite amount of
storage (as well as an infinite number of computation steps) to compute an
infinite number of digits. But the storage needed is always finite for a
finite number of digits. That is true of the Cantor construction as well.

No, by the definition, the MACHINE has finite state, but a potentially infinite tape.

 Remember, if you’re saying that the list cannot be constructed in the
first place, then you’re destroying the Cantor proof as well. I am
assuming it can be constructed, because the Cantor proof assumes it can be
constructed. You can’t knock out my proof and keep the Cantor one by
saying the list cannot be constructed.

Nope, because Cantor never said "computable", he said constructed, which doesn't imply the finite machine to create it.
You are just conflating the two ideas.
Remember, Cantor was showing that there was a number not in the countable list, and thus the reals were not-countable. He wasn't talking about computable, that is a later observation about his list.