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

On 10/21/2024 4:12 AM, WM wrote:

On 20.10.2024 23:41, Jim Burns wrote:

On 10/20/2024 4:42 PM, WM wrote:

On 20.10.2024 22:23, Jim Burns wrote:

If n is in the set ℕ≠{1,...,n} of finites,
then 2⋅n is in the set ℕ≠{1,...,n} of finites.

>
When the elements of any set of naturals are multiplied by 2,

>
∃{1,2,...,n-1,n}  ⇔
∃{1,2,...,n-1,n,n+1,...,n+n-1,n+n}
>

their density is halved,
their reality remains the same,
their extension is doubled.
Therefore the image contains
numbers which are not in the range.

>
n ∈ ℕ  ⇔  ∃{1,2,...,n-1,n}

>
n is not among the dark numbers.

For each n ∈ ℕ,  exists n+1 ∈ ℕ
For each n ∈ ℕ,  exists n-1 ∈ ℕ or n = 0
For each S ⊆ ℕ,  exists first.S ∈ S or S = {}
There is a general rule not open to further discussion:
Things which aren't natural numbers
shouldn't be called natural numbers.

∃{1,2,...,n-1,n}  ⇔  ∃{1,2,...,n-1,n,n+1,...,n+n-1,n+1}
>
∃{1,2,...,n-1,n,n+1,...,n+n-1,n+1}  ⇔  n+n ∈ ℕ
>
n ∈ ℕ  ⇔  n+n ∈ ℕ

>
That and more holds for all definable numbers.

Then your objection isn't to
what we mean by 'natural numbers'.
You propose dark numbers in order to object.
Compare your activity to
someone claiming to be an auto technician, who,
seeing what they consider bad auto.behavior,
disconnects the battery, in order that
they can say "Your problem is the battery".
⎛ More importantly,
⎜ how you apparently think of axioms is reversed.
⎜ Axioms do not pop things into existence.
⎜ Axioms narrow what.it.is which
⎜  we are currently discussing.
⎜
⎜ Are they ordinals which we're discussing?
⎜⎛ Sets of these things are minimummed or empty.
⎜⎝ Each thing has a successor.
⎜
⎜ Are they finite ordinals which we're discussing?
⎜⎛ Sets of these things are minimummed or empty.
⎜⎜ Each thing has a successor.
⎜⎜ Each thing and each thing before it
⎝⎝  has a predecessor or is 0

But:
When doubling numbers,
their distance increases in positive direction,
hence larger numbers are in the image
than in the range.

n ∈ ℕ  ⇔  n+n ∈ ℕ  ∧  n+n > n

If the range was complete, the image shows that the range was not complete.