• Michael
    8.7k
    Yet I am sure that a set that contains itself can be defined - I'm just not clever enough to think of it.tim wood

    Not given the axiom of regularity and the axiom of pairing.
  • frank
    5.1k
    set that contains itself seems like the Ouroboros making the last bite. How is that managed?tim wood

    A set that has only the moon as a member is a distinct entity from the moon itself. A set is an abstract object.

    A set is criteria, kind of like a club.

    So the set of all non-penguins, NP, intuitively has itself as a member because NP is not a penguin.
  • EnPassant
    218
    I'm not seeing how you can "without X" and still have any X left - in terms of the notation.tim wood

    It is not 'without X' it is 'without {X}' as a set. {X} is not the same as X, my bad notation in the beginning notwithstanding. X\{X} is every set in X but not the set {X} itself.

    X\{X} = {{a}, {b}, {c},...} but not {X}, regardless of whether {X} can be a member of X.

    Excluding {X} is not the same as excluding X.

    The paradox asks if {X} is a member of X but I am disposing of the paradox by defining X as X\{X} so there is no contradiction.
  • SophistiCat
    1.2k
    No, I am saying there are infinite collections of things that are not a set.
    See this link https://math.stackexchange.com/questions/24507/why-did-mathematicians-take-russells-paradox-seriously
    EnPassant

    That's an interesting discussion there. Most of us here are non-mathematicians, and among mathematicians only a small fraction are working in or at least interested in foundations.

    The paradox asks the question "Is X a member of itself?"

    Let's say Set X = {{a}, {b}, {c},....}

    If {X} is a member of X then

    Set X = {{a}, {b}, {c},....{X}}
    EnPassant

    Your notation is confusing. If you want to say that a is a member of X (a ∈ X), you would write that as

    X = {a, ...}

    which is not the same as

    X = {{a}, ...}

    {a} is a singleton set with a as the sole member.
  • jgill
    559
    Yet I am sure that a set that contains itself can be defined - I'm just not clever enough to think of ittim wood

    X={X}

    :roll:

    "...and among mathematicians only a small fraction are working in or at least interested in foundations." How true!
  • EnPassant
    218
    Your notation is confusing. If you want to say that a is a member of X (a ∈ X), you would write that as

    X = {a, ...}

    which is not the same as

    X = {{a}, ...}

    {a} is a singleton set with a as the sole member.
    SophistiCat

    Yes, but X is a set of sets so X = {{a}, {b}, {c},...} but {a, b, c, ...} might be correct too as long as the logic of what I'm saying holds up. Link: https://truebeautyofmath.com/lesson-4-sets-of-sets/
  • tim wood
    4.4k
    So the set of all non-penguins, NP, intuitively has itself as a member because NP is not a penguin.frank

    Bingo! - that's what I was trying to think of!

    Would you say this set is NP-complete?
  • frank
    5.1k
    What does that mean?
  • EnPassant
    218
    If we define things as follows it might make it clearer-

    a = {x}
    b = {y}
    c = {z}

    Set X is the set of sets a, b, c so

    Set X = {{x}, {y}, {z}}

    If X is included

    X = {{x}, {y}, {z}, {{x}, {y}, {z}}}

    If X is not included

    X = {{x}, {y}, {z}}

    So X\X is {{x}, {y}, {z}} which is what I originally meant by X\X or X\{X}
  • SophistiCat
    1.2k
    Yes, but X is a set of sets so X = {{a}, {b}, {c},...} but {a, b, c, ...} might be correct too as long as the logic of what I'm saying holds up.EnPassant

    I don't see what logic could imply that {{a}, {b}, {c},...} is the same as {a, b, c, ...}

    You keep making the same mistake over and over again:

    The paradox asks if {X} is a member of XEnPassant

    No!

    The paradox asks if X is a member of X.

    X ≠ {X}

    {X} is a set with one member: X

    Set X = {{x}, {y}, {z}}

    If X is included

    X = {{x}, {y}, {z}, {{x}, {y}, {z}}}
    EnPassant

    No, that's not how it works.

    X = {{x}, {y}, {z}}

    X' = {{x}, {y}, {z}, {{x}, {y}, {z}}}

    X ≠ X'

    X ∈ X'

    X' ∉ X'
  • EnPassant
    218
    No!

    The paradox asks if X is a member of X.
    SophistiCat

    Let 'All sets that do not contain themselves as members' be

    a = {x}
    b = {y}
    c = {z}
    d = ... and these sets go on for as long as is necessary, e, f, g, h,...

    Set X = {{x}, {y}, {z},...}

    Suppose for some set h, h = {X}

    I am saying X = {{x}, {y}, {z},...}\h

    That is, X = {{x}, {y}, {z},...}\{X}

    There may be h such that h = {X} or there may not.

    I am saying X\h regardless and this is the definition of X.

    In simple language X = "All sets that do not include themselves as members, except {X}"

    You seem to be assuming that {X} is included in X but by definition it is not.

    Or suppose Set V = {{x}, {y}, {z},...{X}}

    Set X = V\{X} and there you have it.
  • frank
    5.1k
    Ah, tim wood does a math joke.
  • EnPassant
    218
    No, that's not how it works.

    X = {{x}, {y}, {z}}

    X' = {{x}, {y}, {z}, {{x}, {y}, {z}}}

    X ≠ X'

    X ∈ X'

    X' ∉ X'
    SophistiCat

    No, I am saying IF X is included in X then

    X = {{x}, {y}, {z},... {{x}, {y}, {z}}}

    But IF X is not included

    X = {{x}, {y}, {z},...}

    I am only saying this to clarify things. But by definition X is NOT included so

    X = {{x}, {y}, {z},...}

    Precisely X = {{x}, {y}, {z},...}\{{x}, {y}, {z}}

    {x}, {y}, {z} and {{x}, {y}, {z}} are different sets so excluding the set {{x}, {y}, {z}} does not exclude {x}, {y}, or {z}
  • jgill
    559
    Would you say this set is NP-complete?tim wood

    Create an algorithm to list all the penguins. Then take the complement in the Universal Set.
  • SophistiCat
    1.2k
    Let 'All sets that do not contain themselves as members' be

    a = {x}
    b = {y}
    c = {z}
    d = ... and these sets go on for as long as is necessary, e, f, g, h,...
    EnPassant

    Why are they all singletons?
  • EnPassant
    218
    Ok, I'll put it this way. List the sets that are not members of themselves as

    a1, a2, a3, ...

    X is going to be {a1, a2, a3, ...}

    But for some i,

    ai = {a1, a2, a3, ...} = X

    So the sets in question are-

    a1, a2, a3, ...ai...

    = a1, a2, a3, ...{a1, a2, a3, ...}...

    So X is to be defined as {a1, a2, a3, ...}\ai

    = X\{a1, a2, a3, ...} = X\X

    That is, X is defined as not being a member of itself.

    Don't worry about the notation. X is defined as not being a member of itself, that is all.
  • EnPassant
    218
    X ∈ X'

    X' ∉ X'
    SophistiCat

    By the way, is how do I type these set symbols? Latex? Is there a guide?

    Edit: Found it. Logic and Philosophy of Mathematics sub forum
  • SophistiCat
    1.2k
    Those symbols are just Unicode characters that you can copy/paste from anywhere (e.g. the first Google hit for "set symbols"). But this site also supports Latex.
  • EnPassant
    218
    Those symbols are just Unicode characters that you can copy/paste from anywhereSophistiCat

    Thanks. The first thread in the Logic and Philosophy of Mathematics forum (this forum) explains how to use the math tag and how to create symbols.
12Next
bold
italic
underline
strike
code
quote
ulist
image
url
mention
reveal
youtube
tweet
Add a Comment

Welcome to The Philosophy Forum!

Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.