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

On 10/2/2024 7:10 AM, WM wrote:

On 01.10.2024 22:05, Jim Burns wrote:

On 10/1/2024 1:29 PM, WM wrote:

What is incorrect?

>
This is incorrect:
🛇⎛ ∀n ∈ ℕ: 1/n - 1/(n+1) > 0  shows that
🛇⎜ at no point x
🛇⎝ NUF can increase by more than one step 1.
>
∀n ∈ ℕ: 1/n - 1/(n+1) > 0  doesn't show that.

>
You [JB] believe that

if each unit.fraction is preceded
then no first unit.fraction exists.
You (WM) believe in quantifier shifts,
the opposite of that.

You believe that
more than one unit fractions
can occupy one and the same point
nevertheless?

No.

That would make the distance 0,
but it is > 0. Therefore you are wrong.

Therefore you are wrong about
what has and hasn't been said,
which greatly reduces the value of
anything you have to say.

∀n ∈ ℕ: 1/n - 1/(n+1) > 0
shows
∀n ∈ ℕ: 1/n  >  1/(n+1) > 0
which shows
each unit fraction 1/n is not first.

>
No.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
does not prove that n+1 is a natural number.
Note the infinite sequence
1, 2, 3, ..., ω-2, ω-1, ω.
It consists of infinitely many finite numbers.

A non.0 ordinal β with predecessor β-1: (β-1)+1=β
and each non.0 α < β with predecessor α-1: (α-1)+1=α
is a finite ordinal.
An infinite ordinal is an ordinal Not.Like.That.
⎛ You (WM) mean something else by 'infinite',
⎜ some vague 'reallyreallybig', which generates
⎝ plenty of gibberish for you to roll around in.

Note the infinite sequence
1, 2, 3, ..., ω-2, ω-1, ω

...is a _finite_ sequence.
As you have written it,
ω and each non.0 α < ω has a predecessor,
which makes ω finite.

∀n ∈ ℕ: 1/n - 1/(n+1) > 0
does not prove that n+1 is a natural number.

If non.0 β is finite
then
β and each non.0 α < β have predecessors
and
β+1 and each non.0 α < β+1 have predecessors
and
β+1 is finite.