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  • Supertask definition (Was: Winter Challenge 2023)

    From Julio Di Egidio@21:1/5 to Julio Di Egidio on Mon Nov 13 14:31:48 2023
    On 13/11/2023 13:23, Julio Di Egidio wrote:
    On Monday, 13 November 2023 at 13:08:43 UTC+1, WM wrote:
    On 13.11.2023 12:42, Julio Di Egidio wrote:
    On Monday, 13 November 2023 at 10:49:51 UTC+1, WM wrote:
    <snip>
    A supertask is a countably infinite sequence of operations that occur
    sequentially within a finite interval of time.

    Wrong, a "supertask" is the *limit* of any such sequence,
    and "time" is utterly irrelevant if not as an expository device.

    A supertask is a task, not a limit.

    Nope, you are wrong: but you are right that more nonsense has been
    written about what a supertask even is than there are stars in our galaxy.

    Thanks for bringing that up:

    <https://en.wikipedia.org/wiki/Talk:Supertask#Wrong_definition>

    Julio

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  • From WM@21:1/5 to Jim Burns on Mon Nov 27 12:52:57 2023
    Jim Burns schrieb am Montag, 27. November 2023 um 02:01:33 UTC+1:
    On 11/26/2023 3:20 PM, Transfinity wrote:
    Jim Burns schrieb am Samstag,
    25. November 2023 um 15:52:43 UTC+1:

    X contains only those natnumbers which,
    when added to X,
    leave |ℕ \ X| = ℵo.

    X contains only those natnumbers which,
    already being in X,
    cannot be added to X

    X is a set function. It can be increased. But never |ℕ \ X| < ℵo will be accomplished by adding individually definable numbers. Therefore
    contains dark numbers: |ℕ \ ℕ| = 0.

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Mon Nov 27 13:07:01 2023
    On 11/27/2023 6:52 AM, WM wrote:
    Jim Burns schrieb am Montag,
    27. November 2023 um 02:01:33 UTC+1:
    Message-ID: <[email protected]>
    Newsgroups: sci.math
    Subject: Re: The Principle of Mathematical Induction versus Infinity
    Date: Sun, 26 Nov 2023 20:01:24 -0500
    On 11/26/2023 3:20 PM, Transfinity wrote:

    X contains only those natnumbers which,
    when added to X,
    leave |ℕ \ X| = ℵo.

    X contains only those natnumbers which,
    already being in X,
    cannot be added to X

    X is a set function.

    X is a collection.

    You, upthread in sci.math
    Collect
    all natnumbers n with the above property
    into a collection X

    Collecting those is a supertask, not a task.
    Describing collecting those is a task.

    Describe the collection X of all n such that
    |ℕ \ {1, 2, 3, ..., n}| = ℵo

    ∀S: ∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ S ⟹ ∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ X ⊆ S

    Describing complete.

    It can be increased.

    A set increased from X
    is not X

    X described upthread in sci.math
    cannot be increased and also be X


    If you want to describe
    infinitely-many different sets
    which result from
    adding all the visible numbers one by one
    to the empty set,
    you can describe them.

    For each set so described,
    the visible successor to its last element
    is not-in that set.
    That set is not X, which all are in.

    You can describe them.
    Each set so described is not X

    But never |ℕ \ X| < ℵo will be accomplished
    by adding individually definable numbers.

    We who are not Chuck Norris can't do supertasks.
    That's a supertask.

    Therefore
    contains dark numbers: |ℕ \ ℕ| = 0.

    For each visibleᴶᴮ number n
    ordered ⟨0,…,n⟩ exists such that
    0‖n exists first‖last in ⟨0,…,n⟩ and,
    for each split Fᣔ<ᣔH of ⟨0,…,n⟩
    i‖i⁺¹ exists last‖first in F‖H

    i⁺¹ is non-0 non-doppelgänger non-final

    Ridiculously-large visibleᴶᴮ numbers exist.
    Do any darkᵂᴹ and visibleᴶᴮ numbers exist?

    Each ⟨0,…,n⟩ is not X because
    visibleᴶᴮ n⁺¹ is not in ⟨0,…,n⟩

    If darkᴶᴮ d ¬∃⟨0,…,d⟩ is in X
    X\{d} also contains all visibleᴶᴮ numbers
    Why isn't X\{d} the set of all visibleᴶᴮ ?

    Let Xᴶᴮ be the set of
    all and only visibleᴶᴮ numbers.
    For each n in Xᴶᴮ
    |Xᴶᴮ\⟨0,…,n⟩| = |Xᴶᴮ|

    For ℕ = Xᴶᴮ
    and ℵ₀ the "size" of any 1×1 1.ended
    Xᴶᴮ is the set of all and only
    numbers n such that |ℕ\⟨0,…,n⟩| = ℵ₀
    No darkᴶᴮ numbers are in ℕ

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  • From Jim Burns@21:1/5 to All on Mon Nov 27 13:05:02 2023
    On 11/27/2023 6:52 AM, WM wrote:
    Jim Burns schrieb am Montag,
    27. November 2023 um 02:01:33 UTC+1:
    Message-ID: <[email protected]>
    Newsgroups: sci.math
    Subject: Re: The Principle of Mathematical Induction versus Infinity
    Date: Sun, 26 Nov 2023 20:01:24 -0500
    On 11/26/2023 3:20 PM, Transfinity wrote:

    X contains only those natnumbers which,
    when added to X,
    leave |ℕ \ X| = ℵo.

    X contains only those natnumbers which,
    already being in X,
    cannot be added to X

    X is a set function.

    X is a collection.

    You, upthread in sci.math
    Collect
    all natnumbers n with the above property
    into a collection X

    Collecting those is a supertask, not a task.
    Describing collecting those is a task.

    Describe the collection X of all n such that
    |ℕ \ {1, 2, 3, ..., n}| = ℵo

    ∀S: ∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ S ⟹ ∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ X ⊆ S

    Describing complete.

    It can be increased.

    A set increased from X
    is not X

    X described upthread in sci.math
    cannot be increased and also be X


    If you want to describe
    infinitely-many different sets
    which result from
    adding all the visible numbers one by one
    to the empty set,
    you can describe them.

    For each set so described,
    the visible successor to its last element
    is not-in that set.
    That set is not X, which all are in.

    You can describe them.
    Each set so described is not X

    But never |ℕ \ X| < ℵo will be accomplished
    by adding individually definable numbers.

    We who are not Chuck Norris can't do supertasks.
    That's a supertask.

    Therefore
    contains dark numbers: |ℕ \ ℕ| = 0.

    For each visibleᴶᴮ number n
    ordered ⟨0,…,n⟩ exists such that
    0‖n exists first‖last in ⟨0,…,n⟩ and,
    for each split Fᣔ<ᣔH of ⟨0,…,n⟩
    i‖i⁺¹ exists last‖first in F‖H

    i⁺¹ is non-0 non-doppelgänger non-final

    Ridiculously-large visibleᴶᴮ numbers exist.
    Do any darkᵂᴹ and visibleᴶᴮ numbers exist?

    Each ⟨0,…,n⟩ is not X because
    visibleᴶᴮ n⁺¹ is not in ⟨0,…,n⟩

    If darkᴶᴮ d ¬∃⟨0,…,d⟩ is in X
    X\{d} also contains all visibleᴶᴮ numbers
    Why isn't X\{d} the set of all visibleᴶᴮ ?

    Let Xᴶᴮ be the set of
    all and only visibleᴶᴮ numbers.
    For each n in Xᴶᴮ
    |Xᴶᴮ\⟨0,…,n⟩| = |Xᴶᴮ|

    For ℕ = Xᴶᴮ
    and ℵ₀ the "size" of any 1×1 1.ended
    Xᴶᴮ is the set of all and only
    numbers n such that |ℕ\⟨0,…,n⟩| = ℵ₀
    No darkᴶᴮ numbers are in ℕ

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  • From Fritz Feldhase@21:1/5 to All on Mon Nov 27 10:31:34 2023
    On Monday, November 27, 2023 at 12:53:01 PM UTC+1, WM wrote:

    X is a set function.

    Nope. You defined X as a "collection" (set, class). [Hint: In ZF(C) everything is a _set_.]

    It can be increased.

    Nope. Sets ("collections") do neither grow nor shrink.

    Hint: If /a/ is !e X, then X u {a} =/= X. At least in the context of SET THEORY.

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  • From WM@21:1/5 to Jim Burns on Tue Nov 28 10:59:42 2023
    On 27.11.2023 19:05, Jim Burns wrote:
    On 11/27/2023 6:52 AM, WM wrote:

    It can be increased.

    A set increased from X
    is not X

    If you can identify a natnumber, then this number and all smaller
    numbers are automatically elements of X. (All last numbers of FISONs
    that you can reason about and all their predecessors belong to X.) You
    cannot reason about natnumbers as individuals which are in the
    difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it belongs to X. Nevertheless the difference remains actually infinite: ℵo natnumbers. Therefore they are dark.

    Regards, WM

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