• Re: More complex numbers than reals? (complex)

    From Jim Burns@21:1/5 to Ross Finlayson on Fri Jul 19 15:49:21 2024
    On 7/18/2024 10:37 PM, Ross Finlayson wrote:
    On 07/18/2024 02:51 PM, Jim Burns wrote:

    [...]

    Yet, there's a case for induction
    that there's no case for induction,
    which axiomless deduction usually
    arrives at as insufficient.

    Your (RF's) claim seems to be
    | Axiomlessly,
    | P(0) ∧ ∀j: P(j)⇒P(j+1)
    | is insufficient to conclude
    | ∀k: P(k)

    I don't know of anyone who'd object to
    that axiomless claim.

    Axioms describe what we are talking about.
    Axiomlessly,
    what our claim is about could be anything.
    For some things, induction is too strong a claim.
    Induction is unjustified
    without knowing what the claim is about.

    Maybe it helps to think of the numbers
    as ranging from zero to a large number,
    then that it's infinite in the middle.

    No, "large" is the opposite of helping,
    because infinite is after all the finites.
    Up.to.a.large.finite and after.all.finites
    have different properties.
    "Large" is misleading.

    A finite is accessible.in.principle.
    That includes numbers not accessible.in.fact,
    such as Avogadroᴬᵛᵒᵍᵃᵈʳᵒ.

    ⎛ A number one step beyond an accessible number
    ⎜ is also an accessible number.

    ⎜ A number on the way to an accessible number
    ⎝ is also an accessible number.

    That gives us some description of
    the set ℕ of accessible.in.principle numbers.
    (Axioms are description.)

    ⎛ ∀j ∈ ℕ: j+1 ≠ 0 ∧ j+1 ∈ ℕ

    ⎝ ∀k ∈ ℕ: k ≠ 0 ⇒ k-1 ∈ ℕ

    Add to that that N is well ordered,
    that is, each nonempty set holds a least element,
    and that is a thorough description of
    the familiar ℕ

    ( ∀A ⊆ ℕ: ∃j ∈ A: j ≤ᴬ A ⇐ A ≠ {}

    From those three claims about ℕ, induction follows for ℕ

    For whatever is outside ℕ, no claim is made.
    For whatever is inside ℕ, we know that
    it is accessible, and we can use its being accessible
    to derive further information like
    induction, unique prime factorization,
    N fitting some proper subsets, and so on.

    It is different, no matter what we call it,
    no matter what other thing we call 'infinite'.

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  • From Jim Burns@21:1/5 to Ross Finlayson on Fri Jul 19 16:36:15 2024
    On 7/19/2024 4:19 PM, Ross Finlayson wrote:
    On 07/19/2024 12:49 PM, Jim Burns wrote:

    [...]

    What's the set of
    all finite sets that don't contain themselves?

    What it isn't is Russell's paradoxical set,
    the set of all sets that don't contain themselves.

    An infinite set of finite sets
    is not a paradox.

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  • From Moebius@21:1/5 to All on Fri Jul 19 23:04:47 2024
    Am 19.07.2024 um 22:36 schrieb Jim Burns:
    On 7/19/2024 4:19 PM, Ross Finlayson wrote:
    On 07/19/2024 12:49 PM, Jim Burns wrote:

    [...]

    What's the set of
    all finite sets that don't contain themselves?

    What it isn't is Russell's paradoxical set,
    the set of all sets that don't contain themselves.

    Right.

    An infinite set of finite sets is not a paradox.

    Yes. But...

    In set theories like ZFC, NBG, MK, ... there are no sets "that contain themselves".

    Hence we can simplify Ross' question slightly:

    "What's the set of all finite sets."

    Well there simply is no such set (in ZFC, NBG, MK, ...)

    Let's just consider NBG for that.

    Consider V = {x : x is a set}, i.e. the universe of all sets.

    Then V* := {{x} : x e V} would be a class of finite sets (and hence a
    subclass of the class of all finite sets).

    But V* ~ V, and by "the principle of size", V* would be a proper class
    (just like V).

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  • From Jim Burns@21:1/5 to Moebius on Fri Jul 19 18:03:02 2024
    On 7/19/2024 5:04 PM, Moebius wrote:
    Am 19.07.2024 um 22:36 schrieb Jim Burns:
    On 7/19/2024 4:19 PM, Ross Finlayson wrote:

    What's the set of
    all finite sets that don't contain themselves?

    What it isn't is Russell's paradoxical set,
    the set of all sets that don't contain themselves.

    Right.

    An infinite set of finite sets is not a paradox.

    Yes. But...

    In set theories like ZFC, NBG, MK, ...
    there are no sets "that contain themselves".

    I think there is a pretty direct cause.and.effect
    between the use of ZFC, NBG, MK, ... and
    Russell's paradox.

    Hence we can simplify Ross' question slightly:

    "What's the set of all finite sets."

    Well there simply is no such set (in ZFC, NBG, MK, ...)

    Let's just consider NBG for that.

    We could also consider a theory fragment which
    avoids paradox by saying so little
    (but still says something about some questions here).

    -- Empty ∅ exists₁
    -- for each₁ X and Y: adjunct X∪{Y} exists₁
    -- if for each₁ Y: P(Y)∨¬P(Y) then {Y:P(Y)} exists₂

    I think we can write out 'Finite(Y)', and, thereby,
    the set {Y:Finite(Y)} of all finite sets exists₂

    I think I could show, from 0 and X∪{Y},
    that all hereditarily.finite sets exist₁
    I'm not sure that I need to.

    This is perhaps little messy, because
    none of that says what sets _don't_ exist or
    what sets _aren't_ elements,
    but,
    if the issue arose where it was contradictory
    for some set to be in {Y:Finite(Y)}
    then it wouldn't be in {Y:Finite(Y)}.

    But the issue does not arise here.

    ⎛ ∀X ∈ {Y:Finite(Y)}: X∪{X} ∈ {Y:Finite(Y)}\{∅}

    ⎜ ¬Finite({Y:Finite(Y)})

    ⎝ {Y:Finite(Y)} ∉ {Y:Finite(Y)}

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