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Nothing contains everything

  1. Oct 10, 2008 #1
    how do we prove in set theory that ,nothing contains everything
     
  2. jcsd
  3. Oct 11, 2008 #2
    We prove "There is no set that contains every set.".
    For an arbitrary set U, we construct a set A not belonging to U.

    Let A = {x∈U | x ∉ x }.
    Then, x∈A <--> x∈U & x ∉ x (Axiom schema of specification).
    Let x be A.
    Then, A∈A <--> A∈U & A∉A.
    If A∈U, then this reduces to
    A∈A <--> A∉A, which is impossible.
    Thus, A ∉ U.
     
    Last edited: Oct 11, 2008
  4. Oct 11, 2008 #3

    HallsofIvy

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    that is, of course, a play on words. You can interpret "nothing contains everything" as "everything is contain in (something we call) nothing" or, with a slight change in meaning, there is nothing (no set as enigmahunter said) that contains everything.

    A non-mathematical variation: $1 is better than nothing, and nothing is better than a milliion dollars. Therefore $1 is better than a million dollars.
    (Actually, I've "cleaned" that up a bit.)
     
  5. Oct 11, 2008 #4
    In this forum when another guy asked for the ZFC axioms to be written down so any proof coming out of those axioms could be checked for its logical reasoning the reactions he got where:

    a) no person in the right mind would prove this straight from the axioms

    b) leave set theory to the set theorists

    So when you write , Axiom schema of specification,please write down this Axiom schema and show please how you formed set A out of this schema
     
  6. Oct 11, 2008 #5

    Hurkyl

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    Do you have a question, or are you just trolling?
     
  7. Oct 11, 2008 #6
    Please get rid of that expression,my question is how he gets A using the Axiom schema of specification
     
  8. Oct 11, 2008 #7
    Let B = [tex]\{x \in A | P(x)\}[/tex].

    Axiom schema of specification (comprehension) says,

    "Let P(x) be a property of x. For any set A, there is a set B such that [tex] x \in B[/tex] iff [tex]x \in A[/tex] and P(x)."

    The above proof basically used x [tex]\notin[/tex]x for P(x).

    If we simply define B = {x | P(x)} and x [tex]\notin[/tex] x for P(x), as you know, it ends up with a "http://en.wikipedia.org/wiki/Russell_paradox" [Broken]" (B is not a set).
     
    Last edited by a moderator: May 3, 2017
  9. Oct 12, 2008 #8
    The problem isn't that someone is asking for them; it's that they're asking the people on the forum when you can just look them up very easily. Mathematicians get worried when someone asks for help with looking up something that they believe the person already knows how to find.
     
  10. Oct 12, 2008 #9
    Trooooooooooooooooolllllllllllllllllllllllllllll.
     
  11. Oct 12, 2008 #10
    If what you wrote can be taken as one of the axioms of ZFC theory,why then when PoutsosA asked the forum to mention the axioms of ZFC theory as a point of reference,instead of that he/she got the answers that i mentioned in my post #4??
     
    Last edited by a moderator: May 3, 2017
  12. Oct 12, 2008 #11
    We can prove in my set theory, in which there is only 1 set, and it contains itself, that there is in fact something containing everything!
     
  13. Oct 12, 2008 #12
    I believe you can do everything, you can even call other people names and get away with it,but there is one thing you cannot do is to give a formal proof of the above
     
  14. Oct 12, 2008 #13

    CRGreathouse

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    Sure. Take predicate calculus with equality, add extensionality and the axiom
    [tex]\exists x\forall y, y\in x\wedge y=x.[/tex]

    The axioms of union and choice follow; the axioms of power set, infinity, and regularity (of course!) fail.
     
  15. Oct 12, 2008 #14
    Does infinity really fail? I gotta think about this a bit...
     
    Last edited: Oct 12, 2008
  16. Oct 12, 2008 #15

    OmCheeto

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    That reminds me of a paradox my brother told me about several decades ago.

    He said that the inside of the house was much larger than the outside of the house.

    This is because when you are inside, the entire house surrounds you. You cannot even touch the edges with your arms outstretched.

    But when you go outside, walk a bit away, and face the house with your arm held out straight, you can squash the house like a little bug with just your thumb and forefinger.

    I've yet to run across the proof though.
     
  17. Oct 12, 2008 #16
    Can I borrow that dollar to buy a ham sandwich?
     
  18. Oct 12, 2008 #17

    CRGreathouse

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    I guess it depends on how you define the axiom. The usual version asserts the existence of sets we don't have, like the empty set and the ordinals up to omega.

    I think of the axiom of infinity as "there exists an inductive set containing the empty set" which requires an empty set.
     
    Last edited: Oct 13, 2008
  19. Oct 13, 2008 #18
    I think so too, since Aczel's non-well-founded set theory still requires the infinity axiom.

    I was thinking that if you have, say, [itex]x\in x[/itex], then you can "expand" it into x->x->x->x->... (an accessible pointed graph, as Aczel calls it), and then use "replacement" to change each x into a different set, then define a set that contains all of them, which is infinite.

    But I'm guessing if we modify the axiom of replacement to be able to do what I suggest, it would probably be equivalent to infinity anyway.
     
  20. Oct 13, 2008 #19

    Hurkyl

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    For an amusing aside, there are characterizations of the empty set that would allow it to exist in Dragonfall's one-set universe. e.g

    . Z is the empty set iff, for any set Y, there is exactly one function Z --> Y
    . Z is the empty set iff Z is a subset of every set

    And even the usual characterization:

    . Z is the empty set iff every x is not an element of Z

    holds in one-valued logic. :smile:
     
  21. Oct 13, 2008 #20
    From A<--->AεU & Aε'Α, and AεU how do you get AεΑ <----> Aε'A ( A does not belong to A)

    I am sorry i cannot follow.

    Aε'Α means A does not belong to A
     
  22. Oct 13, 2008 #21

    CRGreathouse

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    Yes, that is amusing. With either of those definitions the axiom of infinity holds in the model, of course, along with the null set axiom.

    I tried to be cautious in my phrasing, since it wasn't obvious what the null set would represent in this one-set universe.
     
  23. Oct 13, 2008 #22
     
  24. Oct 13, 2008 #23
    Interesting conversations are going on here ( I could not follow some of them).
    It is known that "http://en.wikipedia.org/wiki/ZFC" [Broken]". Anyhow, ZFC still has some strengh over naive set theory as shown above (ex.Russell's paradox).

    I am interested in constructive set theory (rather than ZFC) embed their set axioms in intuitionistic logic. I still don't know what advantages we can expect from it over traditional set theory (ex. ZFC) and first order logic.

    Any opinion or example?
     
    Last edited by a moderator: May 3, 2017
  25. Oct 13, 2008 #24

    Hurkyl

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    Gödel's second incompleteness theorem.


    I don't know if this is related to your interests, but the 'natural' internal logic of Cartesian categories is intuitionistic. (Most interesting mathematical theories take place in Cartesian categories) By internal, I mean that there is a way to interpret logical statements in terms of objects and arrows in the category. (As opposed to interpreting them in terms of sets and functions) Topos theory subsumes the idea of intuitionistic set theory (the ordinary category of sets, for example, is a topos. Don't forget that Boolean logic is intuitionistic!) -- which is very interesting because toposes describe other things too, such as the category of sheaves on a topological space.

    While I don't know the field of constructive set theory, I would have imagined its logic would actually be weaker than intuitionistic logic....
     
  26. Mar 24, 2011 #25
    We prove "There is no set that contains every set.".
    For an arbitrary set U, we construct a set A not belonging to U.

    Let A = {x∈U | x ∉ x }.
    Then, x∈A <--> x∈U & x ∉ x (Axiom schema of specification).
    Let x be A.
    Then, A∈A <--> A∈U & A∉A.
    If A∈U, then this reduces to
    A∈A <--> A∉A, which is impossible.
    Thus, A ∉ U.

    I understand how this answer was derived, but can anyone tell me how this holds up it seems that the definition of A doesn't holds up from the start violating the initial restraints of A not belonging to U. any help is appreciated Thanks
     
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