Re: Does the number of nines increase?

On 6/29/2024 2:14 PM, WM wrote:

Le 29/06/2024 à 20:01, Jim Burns a écrit :

However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.

Each non.empty.set of 9s holds a first.in.set 9
Each 9 has a first.after 9 and a last.before 9,
except the first 9, which only has a first.after 9

We can reason about ℵo nines,
all missing the limit.

1 is near almost.all (all.but.finitely.many) of
  0.9  0.99  0.999  0.9999  0.99999  ...
for any sense > 0 of 'near'
For that reason,
we assign 1 to 0.999...
The values of infinite.length decimals
are assigned by a method different from how
the values of finite.length decimals are assigned.

We can reason about ℵo nines,
all missing the limit.

None of
  0.9  0.99  0.999  0.9999  0.99999  ...
is near almost.all of
  0.9  0.99  0.999  0.9999  0.99999  ...
for any sense > 0 of 'near'
but
the values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999...  =  1
0.999...  ≠  0.9
0.999...  ≠  0.99
0.999...  ≠  0.999
0.999...  ≠  0.9999
0.999...  ≠  0.99999
...

It's the same for infinitely.many 9s  in that
we can't go to them,  but
we can reason about them.

>
But most matheologians don't understand that
the sequence  0.9, 0.09, 0.009, ...  contains
ℵo nines without containing the limit 0.

1 is near almost.all (all.but.finitely.many) of
  0.9  0.99  0.999  0.9999  0.99999  ...
for any sense > 0 of 'near'
0.999...  =  1
0.999...  ≠  0.9
0.999...  ≠  0.99
0.999...  ≠  0.999
0.999...  ≠  0.9999
0.999...  ≠  0.99999
...

But
'infinite' is different from 'humongous' and
different conclusions get concluded.

>
If you think straight,
then only one conclusion follows:
0.999... < 1.

The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999...  ≠  0.9  <  1
0.999...  ≠  0.99  <  1
0.999...  ≠  0.999  <  1
0.999...  ≠  0.9999  <  1
0.999...  ≠  0.99999  <  1
...
0.999...  =  1
----

And Bob does not disappear.

The cardinal ℵ₀ of the set ℕ of
all cardinals.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
is a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
/ Assume otherwise
| Assume ℵ₀⁺¹ > ℵ₀
| There are at.least.ℵ₀⁺¹ cardinals.growable.by.1
| However, ℵ₀ is how.many cardinals.growable.by.1
\ Contradiction.
ℵ₀ = |ℕ| is not.growable.by.1
|ℕ| = |ℕ∪{Bob]|
exists f: ℕ∪{Bob} → ℕ : bijection
Bob ∉ f(ℕ∪{Bob})
And Bob disappears.