Re: Does the number of nines increase?

Le 26/06/2024 à 16:20, Jim Burns a écrit :

On 6/26/2024 3:15 AM, WM wrote:

Is the set of natural indices complete
such that
no natural number can be added?

 You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'

Why? There is nothing different. The set of numbers and of indices is ℕ.

 The well.ordered inductive natural.number.indices
are complete

Therefore the indices of the nines of 0.999... are the complete set ℕ. When they are shifted to 9.999..., none is added. One is missing at the right of the decimal point.

Nuance:
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.

If, by 'fixed', you mean that
each is not anything other than itself,
then yes.

Yes.

Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .

 No.

You believe in the magic of matheology. Try to think. "Countable" is an unsharp notion, too unsharp to measure the fact that {1, 2, 3, ...} has one number less than {0, 1, 2, 3, ...} although this is obvious: ℕ is a proper subsets of ℕ_0. 

"Move" the decimal point.
0.9⃒9999...  and  09.9⃒9999...
0.99⃒999...  and  09.99⃒999...
0.999⃒99...  and  09.999⃒99...
0.9999⃒9...  and  09.9999⃒9...
0.99999⃒...  and  09.99999⃒...
...
 

Not readable what you try here.

There isn't one 9 fewer.

As long as you are claiming that elements can be lost by exchanging, your arguments are not relevant in any respect.
Regards, WM