Re: Cantor Diagonal Proof

On 4/7/25 5:42 PM, Lawrence D'Oliveiro wrote:

On Mon, 7 Apr 2025 02:10:23 -0600, Jeff Barnett wrote:
 

You are quite wrong about an elementary proof and various of your
misunderstandings have been pointed out to you ...

 I gave a proof by induction why the Cantor construction fails on that
flipped-integer list; no one has yet pointed out a flaw in that proof.
 Somebody kept insisting that, even if my proposition is true for every
element in the list, it somehow goes false at the end. The end of an
infinite list! Yeah, right.

And that is because while every element on the list has an algorithm to construct it, that list is infinite, so you can't just put them *ALL* in to one finite algorithm to compute any one you need at the moment.
Your algorithm needs to work for *ANY* n, no matter how large, and the algorithm doesn't get to change based on that n, so it needs to hold that full infinite set of algorithms, but still be, itself, finite.
The countable infinite set sits at an interesting point, all of its members are finite, but the set itself is infinite.
Some things have an induction property that lets us get to that infinity with finite step, because the induction property itself does that. But we can't just assume an inductive property exists, we need to find it.
Thus, your "inductive proof" showed steps to compute each digit, but failed to show that one finite algorithm could actually implement it, because it was based on a selection from an infinite list, which can't be done unless there does exist a induction that generates that list.