Re: Replacement of Cardinality --- infinitesimal number system

On 7/30/2024 7:46 PM, olcott wrote:

On 7/30/2024 5:49 PM, Moebius wrote:

Am 30.07.2024 um 20:37 schrieb Jim Burns:

On 7/30/2024 1:30 PM, WM wrote:

[...] between [0, 1] and (0, 1].

>
Could you please explain to me
the meaning of the phrase "between [0, 1] and (0, 1]"?

The meaning of that phrase is {}.
In WM's mind, {} proves darkᵂᴹ numbers exist by
the application of (at least)
shifting quantifiers,
assuming what he intends to prove, and
thinking that 'finite' and 'infinite' mean something they don't.

One can construe them as line segments that
differ in length by a single geometric point
on the number line.
I came up with that a few years ago with my
infinitesimal number system that
seems to map integers to a contiguous set of
immediately adjacent geometric points
on the number line.

Consider a number line in which
each split is situated.
⎛ In a situated split F′ ᵉᵃᶜʰ<ᵉᵃᶜʰ H′
⎜ a number.line.point x′ exists
⎝ either last.in.F′ or first.in.H′
Consider the unit.fractions ⅟ℕ = {⅟j: finite j ∈ ℕ₁}
The set F′ of lower.bounds of ⅟ℕ  and
the set H′ of not.lower.bounds of ⅟ℕ
is a split of the number.line
x′ situates F′ ᵉᵃᶜʰ<ᵉᵃᶜʰ H′
0 is a lower.bound of ⅟ℕ
Therefore, x′ isn't < 0
x′ isn't > 0
⎛ Assume otherwise.
⎜ Assume x′ > 0
⎜ 2⋅x′ > x′ > ½⋅x′ > 0
⎜
⎜ ½⋅x′ < x′  is a lower.bound of ⅟ℕ
⎜
⎜ 2⋅x′ > x′  isn't a lower.bound of ⅟ℕ
⎜ finite unit.fraction ⅟k < 2⋅x′
⎜ finite unit.fraction ¼⋅⅟k < ¼⋅2⋅x′ = ½⋅x′
⎜ ½⋅x′ isn't a lower.bound of ⅟ℕ
⎜
⎝ Contradiction.
Therefore, x′ isn't > 0
x′ isn't < 0
x′ = 0
0 situates the split between
lower.bounds and not.lower.bounds of ⅟ℕ
No point δ > 0 is a lower.bound of ⅟ℕ
Each point δ > 0 has a finite unit.fraction
between δ and 0
No point δ > 0 is an infinitesimal.