• Re: Replacement of Cardinality (defining numbers as half finite, half i

    From Jim Burns@21:1/5 to Ross Finlayson on Tue Aug 6 07:57:20 2024
    XPost: sci.logic

    On 8/5/2024 11:05 PM, Ross Finlayson wrote:
    On 08/05/2024 03:19 PM, Jim Burns wrote:
    On 8/5/2024 3:04 PM, WM wrote:

    For every number of unit fractions
    NUF(x) gives the smallest interval (0, x).

    For  each real x > 0
    for each finite cardinal k > 0
    there is a finite.unit.fraction uₖ such that
    |⅟ℕᵈᵉᶠ∩[uₖ,x)| = k
    uₖ = ⅟⌊k+⅟x⌋

    For  each x > 0
    for each finite cardinal k > 0
    there is a finite.unit.fraction uₖ such that
    |⅟ℕᵈᵉᶠ∩[uₖ,x)| = k
    ⅟(1+⅟uₖ) ∈ ⅟ℕᵈᵉᶠ∩(0,uₖ)
    k < |⅟ℕᵈᵉᶠ∩(0,x)|

    For  each x > 0
    for each finite number k
    k ≠ |⅟ℕᵈᵉᶠ∩(0,x)|

    For  each x > 0
    ⅟ℕᵈᵉᶠ∩(0,x) is not finite.

    Not.finite what?

    Not finite set, not finite number?

    Not.finite set ⅟ℕᵈᵉᶠ∩(0,x) of
    not.infinite.and.not.infinitesimal unit.fractions.

    ----
    Definition.

    A finite order is a total order in which
    each non.{}.subset is 2.ended ==
    each non.{}.subset holds its max and its min.

    An infinite order is a total order which
    is not finite.

    ⎛ Lemma:
    ⎝ No set has both a finite and an infinite order.

    A set with a finite order is finite.

    A set with an infinite order is infinite.

    ----
    Definition.

    Each non.{} S ⊆ ℕᵈᵉᶠ has min.S ∈ S.

    Each j ∈ ℕᵈᵉᶠ has successor j+1 ∈ ℕᵈᵉᶠ

    0 = min.ℕᵈᵉᶠ
    Each non.0 k ∈ ℕᵈᵉᶠ has predecessor k-1 ∈ ℕᵈᵉᶠ

    ----
    Consequences.

    Each bounded subset of ℕᵈᵉᶠ is finite.
    ℕᵈᵉᶠ and each of its unbounded subsets is infinite.

    Each element of infinite ℕᵈᵉᶠ is
    finitely.preceded and infinitely.succeeded.

    Each is finite, all are infinite.

    Similarly,
    for each u ∈ ⅟ℕᵈᵉᶠ
    ⅟ℕᵈᵉᶠ∩[u,1] is finite
    ⅟ℕᵈᵉᶠ∩(0,u] is infinite

    Sometimes it's hard to convey intended inflection

    Very true.

    I depend upon others to point out those times to me.
    Of course 𝐼 know what I mean.
    That won't tell me that others know. Or don't.

    And, perhaps I merely have room for improvement in
    how I express my ideas.

    But perhaps I didn't carry the 2. https://www.gocomics.com/bloomcounty/1988/07/10

    It's a significant help to me.

    What I would ask,
    is that you surpass,
    the inductive impasse,
    with the infinite super-task.

    Induction is a finite task of
    reasoning about infinitely.many.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)