• EnPassant
    336
    The problem is stated as "The set of all sets that do not contain themselves as subsets."
    Already we have a problem with the way the problem is stated because how do we know that "All sets that do not contain themselves as subsets" is a set? We cannot assume from the start that it is a set. And this assumption is the very cause of the problem that arises.
    What is "All sets that do not contain themselves as subsets" if it is not a set?

    In the following the symbol \x means 'without x' or 'excluding x'

    Let Set X = "All sets that do not contain themselves as subsets"\X

    I don't see anything wrong with this definition because it avoids the paradox by excluding X regardless of whether X can contain itself.

    Let Set X2 = (X U X)\X2

    Now X and X are united, but in X2\X2

    Let Set X3 = (X2 U X2)\X3

    Now X2 is included and X3 is included if we continue-

    Let Set X4 = (X3 U X3)\X4

    And we can continue in this way for an infinity of sets, thereby including every set that does not include itself as a subset.

    The conclusion is that "All sets that do not contain themselves as subsets" is a limit rather than a set. The limit of an infinity of sets.
  • Banno
    8.5k
    ...how do we know that "All sets that do not contain themselves as subsets" is a set?EnPassant
    Well, a set is an unordered collection of individuals. The unordered collections of individuals that do not contain themselves is an unordered collection of individuals; therefor it is a set.

    Unless of course you re-define set.
  • jgill
    651
    Set theory can drive a person nuts. :worry:
  • Snakes Alive
    638
    how do we know that "All sets that do not contain themselves as subsets" is a set?EnPassant

    It follows from the assumptions of a certain kind of naive set theory. The point is that it's not a set, but from the naive set theory it follows that it should be. Hence the problem.
  • ssu
    2.8k
    Yep. Noticing the link between Russell's paradox and limits means that you get it.
  • tim wood
    4.6k
    "The set of all sets that do not contain themselves as subsets."EnPassant

    That leads to a problem or at least a need for clarification. Sets contain elements, not subsets.

    Let S = {x l x is an integer and 0 < x < 4}

    S = {1,2,3}

    One subset of S is S' = {1,2,3}. But S does not contain {1,2,3}, and by the definition of S, cannot contain {1,2,3}.

    I am having some trouble thinking of any well-defined set that does contain itself. Help?
  • Banno
    8.5k
    The trouble is if one re-defines sets so as to rule out sets of sets, they cease to be useful for defining integers in terms of sets of sets.
  • jgill
    651
    I am having some trouble thinking of any well-defined set that does contain itself. Help?tim wood

    x={x}

    Nonsense, however.
  • Marchesk
    3.6k
    The infinite set? If infinity is the set of all numbers, then infinity is a member, unless it's not a number.
  • EnPassant
    336
    Well, a set is an unordered collection of individuals. The unordered collections of individuals that do not contain themselves is an unordered collection of individuals; therefor it is a set.Banno

    There is no need to redefine the set. All that is needed is to see that there are collections that are not single sets - as the paradox implies. I think Russel's Paradox is superficial and I never believed it "undermines mathematics" which strikes me as an unjustifiably dramatic statement.

    In fact it is a trick question because of the way it is stated: "The set of all sets that do not contain themselves as subsets." Why are they calling it a set? I don't think something can be called a set unless it can be demonstrated to be such. And the paradox shows that it cannot be a set. The entity should be defined as "All sets that do not contain themselves as subsets" There is no ambiguity in stating it this way and intuitively I feel that such an entity is possible. But what is it if it is not a set?

    If the logic of my first post is correct it is an infinite collection of sets, each nested within another.
    What seems to be the case is that there can be infinite sets that are not simply a set. I say this, not because of my reasoning but also because Russel's solution involves something similar: an infinity of "types", each nested one within the other. But it may be possible to resolve the issue with sets alone (Russel's solution seems very artificial and contrived)

    But the real question I am asking in my first post is: Is the logic I am using coherent? I don't see anything wrong with it, unless you can.
  • EnPassant
    336
    "I am having some trouble thinking of any well-defined set that does contain itself. Help?"jgill

    Set A = {a, w}
    Set B = {a, x}
    Set C = {a, y}

    Set X = the set of sets that have {a} as an subset.

    Set X = {A, B, C,...}

    {a} is in X (because {a} is in A, B, C,...)

    therefore X contains X
  • jgill
    651
    Set A = {a, w}
    Set B = {a, x}
    Set C = {a, y}

    Set X = the set of sets that have {a} as an element.
    EnPassant

    None of them do. They have "a" as an element. D={{a},z} does.

    This stuff is deleterious to mental health. :scream:
  • EnPassant
    336
    Ok, change it to 'subset'. Post edited.
  • jgill
    651
    {a} is in X (because {a} is in A, B, C,...)EnPassant

    No, {a} is not "in" A,B,C,...

    IMO this stuff is not worth the effort.
  • EnPassant
    336
    No, {a} is not "in" A,B,C,...jgill

    {a} is a subset of A and A is a subset of X therefore {a} is a subset of X

    Also, when it comes to set of sets, {a} can be an element.
  • jgill
    651
    {a} is a subset of A and A is a subset of X therefore {a} is a subset of XEnPassant

    {A} is a subset of X, not A. Ask fdrake or fishfry to explain this stuff to you. I'm done.
  • SophistiCat
    1.3k
    I think Russel's Paradox is superficial and I never believed it "undermines mathematics" which strikes me as an unjustifiably dramatic statement.EnPassant

    Who ever said that Russel's Paradox "undermines mathematics"? It undermines what is now known as "naive set theory" (an early attempt at an axiomatic set theory).

    In fact it is a trick question because of the way it is stated: "The set of all sets that do not contain themselves as subsets." Why are they calling it a set?EnPassant

    Because of the axiom of unrestricted comprehension, which is what had to be ditched in the wake of Russel's paradox:

    There exists a set B whose members are precisely those objects that satisfy the predicate φ.
  • SophistiCat
    1.3k
    Set A = {a, w}
    Set B = {a, x}
    Set C = {a, y}

    Set X = the set of sets that have {a} as an subset.

    Set X = {A, B, C,...}

    {a} is in X (because {a} is in A, B, C,...)

    therefore X contains X
    EnPassant

    You need to understand the difference between being a member of and being a subset of.

    Set X = {A, B, C} = { {a, w}, {a, x}, {a, y} }

    a is a member of A, B and C, but not a member of X. {a} is a subset of A, B and C, but not a subset of X.
  • EnPassant
    336
    {a} is a subset of A, B and C, but not a subset of X.SophistiCat

    Subset is transitive: If A is a subset of B and B is a subset of C, A is a subset of C.

    {a} is a subset of {A} and {A} is a subset of {X} ---> {a} is a subset of {X}
  • bongo fury
    471
    The problem is stated as "The set of all sets that do not contain themselves as subsets members."EnPassant

    Subset is transitiveEnPassant

    Exactly. Membership isn't.
  • Banno
    8.5k
    There is no need to redefine the set.EnPassant

    But that's what you did.
  • tim wood
    4.6k
    But the real question I am asking in my first post is: Is the logic I am using coherent? I don't see anything wrong with it, unless you can.EnPassant

    Actually, I do not understand it as presented.
    In the following the symbol \x means 'without x' or 'excluding x'EnPassant
    Let Set X = "All sets that do not contain themselves as subsets"\XEnPassant

    Is it correct to rewrite this as X = X\X ? Can you translate into English?
    I get, "the set of all sets that do not contain themselves as subsets" = "the set of all sets that do not contain themselves as subsets" and/but excluding "the set of all sets that do not contain themselves as subsets." And that looks like the empty set.

    But until the distinction between elements and subsets is laid out, I do not think you can get even this far.

    Let Set X2 = (X U X)\X2EnPassant
    And same problem
  • frank
    5.1k

    I think this is the correct answer from Snakes Alive:

    "It follows from the assumptions of a certain kind of naive set theory. The point is that it's not a set, but from the naive set theory it follows that it should be. Hence the problem"
  • bongo fury
    471
    I think this is the correct answer from Snakes Alive:frank

    That was one of several (prior to @tim wood's) where expertise of the reader had caused unconscious correction (and hence ignoring) of the mis-statement of the problem.



    Ironically, one (Lesniewski's) response to the paradox was to try to base arithmetic on a transitive part-whole relation. One that wasn't, like subset, derived from a non-transitive one. Part of the motivation, if I understood and now recall correctly, was to separate out a distributive sense of "set" (by which to attribute properties to each member) from a collective one (for attributing to the whole collection as a thing). A bit like the possible difference between talking (distributively) about,

    All sets that do not...EnPassant

    and talking (collectively) about

    The entityEnPassant

    so defined.
  • frank
    5.1k
    mis-statement of the problem.bongo fury

    Misstatement?
  • bongo fury
    471
    Misstatement?frank

    The problem is stated as "The set of all sets that do not contain themselves as subsets members."EnPassant

    Hence @tim wood's and @SophistiCat's clarifications.
  • frank
    5.1k
    Yea, I didn't notice that.
  • jgill
    651
    In the article in Wikipedia:

    In 1923, Ludwig Wittgenstein proposed to "dispose" of Russell's paradox as follows: The reason why a function cannot be its own argument is that the sign for a function already contains the prototype of its argument, and it cannot contain itself.

    In practice this is nonsense as well, IMO, although in some abstract sense it may have weight. In iteration the first application is f(f(z)). So the outer f acts upon a function value and not on the function itself. Fractals arise from these processes. Perhaps f(f) makes less sense. If F is a functional, then neither F(F(z(t))) nor F(F) is normally well-defined.
  • EnPassant
    336
    There is no need to redefine the set. — EnPassant
    But that's what you did.
    Banno

    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

    Is it correct to rewrite this as X = X\X ? Can you translate into English?tim wood

    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}}

    If {X} is a not member of X then

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

    But I am excluding {X} regardless of whether it can be a member of itself.

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

    {X} can only be a member of itself according to the definition of X. I am explicitly excluding {X} from being a member of itself by definition. X\{X} excludes {X} as a member, not as the entire set.

    Then I unite Set X with {X} in

    X2 = (X U {X})\{X2} (See my next post)

    Since the paradox shows that X is a kind of 'pathological' set we don't know where to put it.
    I am creating X2 and putting it in there. Then the process is repeated infinitely so that all relevant sets can be contained. The result is an infinite progression of sets that contain "All sets that are not members of themselves" And this entity turns out to be an infinity of sets, each nested within the other.

    (It may also be that every Xi contains every Xj but not {Xi} but I have not got this far with it yet.)
  • EnPassant
    336
    I get, "the set of all sets that do not contain themselves as subsets" = "the set of all sets that do not contain themselves as subsets" and/but excluding "the set of all sets that do not contain themselves as subsets." And that looks like the empty set.tim wood

    Yes, you are correct. Since we are talking about sets of sets, a better notation would be-

    Set X = X\{X}

    Set X2 = (X U {X})\{X2}

    Set X3 = (X2 U {X2})\{X3} and so on.

    Apologies for the sloppy notation.
  • tim wood
    4.6k
    Set X = {{a}, {b}, {c},....{X}}EnPassant

    Or X = {{a}, {b}, {c},..., {{a}, {b}, {c}...}} \ ???

    I'm not seeing how you can "without X" and still have any X left - in terms of the notation.

    Let's try this:

    Let L be a listing of all the combinations K of integers such that each is ordered on <, with duplicates eliminated so that in each combination no number appears more than once, and that in the listing the combinations be ordered, perhaps shortest first and lexicographically with combinations of the same length. L, then, is a listing K1, K2, K3,..., KN.

    We can ask, then, if an integer i is in some Ki. If i is in Ki then put i in T, unless i is already in T, in which case do nothing. Similarly, If i is not in Ki then put i in F unless i is already in F. Both T and F, then, appear in L. That is, for each of T and F there is a unique Ki.

    Let F be Ke.

    Is e in Ke? Oops. This construction a sketch due to Emil Post, (The Undecidable, Ed. Martin Davis, 1965, pp. 305 - 315).

    The moral of the story seems to be that sets of everything can lead to problems. A set that contains itself seems like the Ouroboros making the last bite. How is that managed?

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

    edit: small but important correction.
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