• Michael
    9.1k
    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
    430
    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
    1.4k
    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
    788
    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
    430
    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
    5.3k
    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
    430
    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
    1.4k
    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
    430
    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
    430
    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
    788
    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
    1.4k
    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
    430
    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
    430
    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
    1.4k
    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
    430
    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.
  • Tommy
    13


    Would your definition of imply



    Indeed, it seems to imply



    where containment is strict. You should consider checking out NBG class-set theory which is an alternative formulation of set theory. The theory uses the notion of classes to avoid Russell's paradox. I'm not sure I can say I'd prefer it over the standard ZFC but it's interesting to having an alternative point-of-view. There is a good dover book by Smullyan and Fitting on it as well. It can be used for self-study.

    EDIT: fixed the second equation.
  • EnPassant
    430
    You should consider checking out NBG class-set theory which is an alternative formulation of set theory.Tommy

    My idea is that it can be framed in terms of set theory alone without the invention of classes. My definition is that if X as a subset of itself is excluded this leaves X\X' where X' is the subset.
  • Tommy
    13
    My idea is that it can be framed in terms of set theory alone without the invention of classes.EnPassant

    Hmm, I'm not sure I follow. ZFC does not deploy the use of classes, it is set-based. One can avoid the difficulties of Russel's Paradox without the invention of the class object. Indeed, I believe the use of classes was for alternative purposes, namely, to simplify some aspects of ZFC. From what I understand, NBG is generally regarded as more elegant than ZFC--again for more info see the book I linked in my last post. If you don't like classes, then just avoid the use of a class-set theoretic system.

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

    I don't see anything wrong with this definition...
    EnPassant

    The problem with this definition is that the set of all sets that do not contain themselves as subsets is shown, by Russell's Paradox, to be logically contradictory. Your definition requests that we posit an object which is logically contradictory, and then remove from it. This is akin to requesting the reader to take the smallest prime number with exactly three divisors, subtract it from itself, and then insist that the answer is 0.
  • Gregory
    1.6k
    I think the paradox would have been more powerful if it said "the set of all sets that DO contain themselves".

    The barber shaves those and only those who do not shave themselves. Actually this is obvious if he is a barber. So the barber's hair grows long, or another barber does shaves him. Where is the paradox with regard to the barber?
  • Olivier5
    243
    Russel's parafox is simply a proof by the absurd that sets cannot contain themselves.
  • EnPassant
    430
    The problem with this definition is that the set of all sets that do not contain themselves as subsets is shown, by Russell's Paradox, to be logically contradictory. Your definition requests that we posit an object which is logically contradictory, and then remove
    X
    X from it. This is akin to requesting the reader to take the smallest prime number with exactly three divisors, subtract it from itself, and then insist that the answer is 0.
    Tommy

    My argument is to define X without X as a subset of itself regardless of whether it can or can't be such. In this way the paradox is avoided by defining a set that contains 'All sets...' but not X. X is then included in X2 and the paradox is avoided. The same argument is then applied to X2, X3... The result is that the 'Set of all...' is really an infinity of sets each containing the other.
  • ssu
    3.2k
    My argument is to define X without X as a subset of itself regardless of whether it can or can't be such. In this way the paradox is avoided by defining a set that contains 'All sets...' but not X. X is then included in X2 and the paradox is avoided. The same argument is then applied to X2, X3... The result is that the 'Set of all...' is really an infinity of sets each containing the other.EnPassant
    What's the difference to Russell's Type theory?
  • EnPassant
    430
    What's the difference to Russell's Type theory?ssu

    I don't know enough to say but as far as I can see this question can be resolved with simple set theory alone - if my ideas are coherent that is. The problem is with the way the paradox is stated: "The set of all sets that are not members of themselves." Why assume it is a set? Because it turns out that 'The set' is not a set at all. But we can intuitively grasp the concept of 'All sets...etc'. It must be something so what is it if it is not a set? Seemingly it is an infinite collection of sets.
    This is similar to Russell's types because there is also a hierarchy of types, each containing the ones below it. So why not just frame the whole thing in terms of sets alone?
  • EnPassant
    430
    Your definition requests that we posit an object which is logically contradictory, and then remove X from it.Tommy

    I've had great difficulty in this thread with notation but the concept I'm trying to define is simple, namely X\X.

    Suppose set A = {a, b} the subsets of this set are:

    {{0}, {a}, {b}, {a, b}} Now redefine set A as:

    A\A. That is, A\{a, b}

    Now the subsets of A are:

    {{0}, {a}, {b}}

    This is what I mean by X\X.
  • Tommy
    13
    In this way the paradox is avoided by defining a set that contains 'All sets...' but not X.EnPassant

    I see what you mean. I'm still sticking to my previous statement on why this set is ill-defined. In any case, I don't think one can avoid Russell's Paradox proceeding in the way you have--regardless of the logical consistency of the argument.

    We have to remember that Russell's Paradox established the inconsistency of Frege's set theory, in particular, the abstraction principle which loosely stated that, given any property P, there exists a (unique) set A consisting of those and only those things that have property P.

    Russell suggested we consider the collection "the set of all sets which are not subsets of themselves". Note that, by Frege's abstraction principle, this is necessarily a set. Asking that we, in effect, look the other way and consider instead another set, as you've proposed, doesn't prevent us from considering Russell's set.

    The paradox establishes that the abstraction principle is unsound. Historically, this lead Zermelo to suggest the limited abstraction principle to prevent the issue--one would need to clarify which set from which the elements satisfying P are being taken. Thus we cannot speak of the set of all x having property P, but we can speak of the set of all x in A that have property P. For an explanation of how this avoids the paradox, I recommend looking at pg. 12 of Smullyan's book, linked in my first post.
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.