Re: Cantor Diagonal Proof

On 05/04/2025 08:38, Lawrence D'Oliveiro wrote:

On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:

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.

Remember that the hypothesis of the Cantor “proof” is that the list is
already supposed to contain every computable number.

In the original Cantor proof, it was every /real/ number.  Cantor
had no concept of computable numbers.  But the case of computable numbers
is an easy edit.

The fact that the
contruction succeeds for your list examples does not mean it will succeed
with mine. Remember, the “proof” depends on it succeeding in the general
case, with every possible list.

It does succeed with every possible list.  The constructed number
differs from the n-th member of the list in the n-th digit.  This does not
imply that the first k digits of the constructed number differ from those
of all members of the list, for arbitrary k.  Indeed, if the list is of a
suitably dense set, then such a starting set of digits will indeed recur
arbitrarily often;  but the continuations will eventually differ.  [As is
well-known, there is a problem if (eg) 0.123999... is constructed and
0.124000... is in the list, but that is easily circumvented.]
Note that if the list is of all rationals, this provides a simple
proof of the existence of irrational numbers.  Details left as an exercise.
[The boundary between confusion, trollery and crankiness looms ever
nearer.]
--
Andy Walker, Nottingham.
    Andy's music pages: www.cuboid.me.uk/andy/Music
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