Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

WM <wolfgang.mueckenheim@tha.de> wrote:

On 26.11.2024 15:50, Richard Damon wrote:

On 11/26/24 8:58 AM, WM wrote:

 

An actual bijection is assumed: Consider the black hats at every 10 n
and white hats at all other numbers n. It is possible to shift the
black hats such that every interval (0, n] is completely covered by
black hats. There is no first n discernible that cannot be covered by 
black hat. But the origin of each used black hat larger than n is now
covered by a white hat. Without deleting all white hats it is not
possible to cover all n by black hats. But deleting white hats is
prohibited by logic. Exchanging can never delete one of the exchanged
elements.

 
But a bijection is NOT a set to itself, but between two sets.

 
The set of natural numbers divisible by 10 and the set of natural
numbers are two different sets.

Right, and a bijection is an operation matching the members of two sets.

NOT “moving” them between the set.

 
The white hats aren't on the "OTHER" numbers, (unless you bijection is
from numbers zero mod 10 to number non-zero mod 10), but ALL numbers
have a white hat, and every tenth has a black hat, and the bijection
shows we can pair every black hat to a white hat.

 
Let all numbers n have white hats and in addition the 10n have black hats.
Claim: We can cover every white hat by a black hat. But when we take a
black hat from 20 to give it to 2, then 20 has a white hat, and so on.
How can the white hats disappear?

Why do they need to? You started with ALL the numbers having white hats
PLUS the number divisible  by 10 having black hats too, so some numbers
have two hats.

The bijection shows that we can give EVERY number a black hats too, by
taking it from a higher number, of which there is an unlimited number.

 

 
Which element of which infinite set did not participate in the
bijection?

 
That is the crucial point! The white hats remain as long as logic is
valid. But their carriers cannot be found. They are dark.

 
And nothing in the logic says that white hats go away.

 
Nothing in logic allows that. But Cantor claims it erroneously.

So how do you make that something that doesn’t happen, that you say can’t
happen, be an excuse for claiming the thing that never had it happen to be
wrong

The only thing wrong is your argument saying that some of the hats need to
disappear, which you just admitted can’t happen.

Cantor NEVER had the white hats go away, he just showed you could match
every number with a black hat with a corresponding number with a white hat,
and thus the sets are the same size.

You err by trying to make your bijection not between to independent sets,
but with a set and a subset within it.

 
Regards, WM