Re: How many different unit fractions are lessorequal than all unit fractions?

On 9/5/2024 9:53 AM, WM wrote:

Le 04/09/2024 à 22:52, Jim Burns a écrit :

On 9/4/2024 3:10 PM, WM wrote:

Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.

>
...or
natural numbers aren't what you think they are.

>
That is possible.

My arguments hold only under
the premise of actual infinity

Apparently, it's because
(in)finiteness isn't what you think it is
that
natural numbers aren't what you think they are.
⎛ In a set with an infinite linear order,
⎜ at least one nonempty subset has fewer than two ends.
⎜
⎜ In a set with a finite linear order,
⎜ no nonempty subset has fewer than two ends.
⎜
⎜ Sets have different orders, but
⎝ no set has both an infinite and a finite order.
⎛ < is a finite linear order of B
⎜ < has,
⎜ for each nonempty S ⊆ B, two ends.
⎜
⎜ x ∉ B
⎜ Extend < to <ₓ for B∪{x}
⎜
⎜ nonempty Sₓ ⊆ B∪{x}
⎜ Sₓ\{x} ⊆ B
⎜ Sₓ\{x} has two ends (except Sₓ\{x}={})
⎜  (min.Sₓ\{x} and max.Sₓ\{x})
⎜
⎜ x in Sₓ
⎜ either
⎜ replaces an end of Sₓ\{x}
⎜  (Sₓ has two ends, one is x)
⎜ or
⎜ doesn't replace an end of Sₓ\{x}
⎜  (Sₓ has two ends, neither is x)
⎜
⎜ <ₓ has,
⎜ for each nonempty Sₓ ⊆ B∪{x}, two ends.
⎝ <ₓ is a finite linear order of B∪{x}
Therefore,
if B has a finite order,
then B∪{x} has a finite order.
A set with a finite order is a finite set.
A set with an infinite order is an infinite set.
A set with neither a finite nor an infinite order
is unusual, and
a counter.example to the Axiom of Choice, and
not any set or collection which we are discussing.
A set with both a finite and an infinite order
requires impossibilities, and
is not.

My arguments hold only under
the premise of actual infinity
showing that Hilbert's hotel is nonsense
because the set of natural numbers cannot be extended.
If all rooms are occupied
than no guest can leave his room for a not occupied room.
(When I was in USA or the first time,
I asked in a Hilton whether they had free rooms.
They laughed.)

Yes, imagine
someone "teaching" about mathematics
who thinks that
an infinite ordered set has
two ends in each of its nonempty subsets.
Very funny.

Peano has been generalized from
the small natural numbers.

>
Peano describes the finite natural numbers.
'Finite' doesn't need to be 'small'.

>
Finite is much larger than
Peano or you could/can imagine.

Consider an ordinal β as {α:α<β}
β = {α:α<β}
β+1 = {α:α<β}∪{β}
Define ω as the first transfinite ordinal.
β < ω  ⇔  {α:α<β} is finite
 From up.post,
finite {α:α<β}  ⇒  finite {α:α<β}∪{β}
finite β  ⇒  finite β+1
β < ω  ⇒  β+1 < ω
⎛ Define (ω-1)+1 = ω
⎜
⎜ ω-1 < ω  ⇒  (ω-1)+1 < ω
⎜ ω-1 ≥ ω  ⇒  (ω-1)+1 > ω
⎝ (ω-1)+1 ≠ ω
Therefore, ω-1 doesn't exist.

Finite is much larger than
Peano or you could/can imagine.

Insisting that ω-1 exists and that,
for b ≠ 0 and β < ω, β-1 exists
is
insisting that ω is finite.
The most frugal explanation of your claim is that
you simply do not know what 'finite' means.
Billions of people don't know what 'finite' means,
and also live successful, fulfilling lives.
However,
nearly all of them aren't doing their best
to spread their ignorance as far as they can.

"All different unit fractions are different"
however is a basic truth.
Therefore I accept the latter.

>
You also accept quantifier shifts,
which breaks your logicᵂᴹ.
Quantifier shifts are unreliable.

 Do you believe that it needs a shift to state:
All different unit fractions are different.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
I can see no shift.

It needs a shift to conclude from
( for each ⅟j: there is ⅟k≠⅟j: ⅟k < ⅟j
that
( there is ⅟k: for each ⅟j≠⅟k: ⅟k < ⅟j
Have you evolved on that topic?