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Math Help - Russell's paradox dilemma

  1. #1
    Newbie petike's Avatar
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    Question Russell's paradox dilemma

    Hi,
    I have a problem with "Russell's paradox" which I have just learned about at school. Simply said, it means that:
    Let's say we have sets R and A. And according to Russell's paradox (or theory) there cannot exist such set R, that:
    R = \{ A \ | \ A \notin A \}
    But I think that such set "exist"...

    What exactly means that A \in A? Then the set A could be one of these:
    A = \{ a, \ b, \ c, \ A \}
    A = \{ 4, \ 6, \ A, \ 9 \}
    A = \{ A, \ "string1", \ "string2" \}
    And what means that A \notin A? Then the set A could be one of these:
    A = \{ a, \ b, \ c, \}
    A = \{ 4, \ 6, \ 9 \}
    A = \{ "string1", \ "string2" \}

    So let's say that A = \{a, \ b, \ c\}.
    Then:
    R = \{\{a, \ b, \ c\}, \ \{a, \ b, \ c\}, \ \{a, \ b, \ c\}, \ ...\}
    ...so I have found such set R that:
    R = \{ A \ | \ A \notin A \}

    So where is the paradox ?


    Thanks.
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  2. #2
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    Russell Paradox

    There does not exist a set of all sets

    Let U = the universe and suppose U were such a set. Then if A is any set that is an element of U. The property P(A): A is not an element of A is a meaningful property for elements of U. Let M = {A; A is an element of U and A is not an element of A}. By the axiom of specification, M is a set; and therefore an element of U. By the law of the excluded middle M is an element of M or M is not an element of M.

    But if M is an element of M then, by definition, M is an element of U and M is not an element of M so that M is an element of M and M is not an element of M which is a contradiction.

    Alternately if M is not en element of M, then this is so only if M is not an element of U or M is an element of M. But M is an element of U so that we must have M is an element of M. Thus M is not en element of M and M is an element of M, again a contradiction.

    From Elements of Set Theory 2nd edition by Zehna and Johnson, 1972

    In either case it is a contradiction so that U does not exist
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  3. #3
    MHF Contributor

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    In c1904 Bertram Russell wrote to Frege to ask a simple question:
    “Is a set of tea spoons a tea spoon?”
    Upon receiving that note, Frege realized the falsity in his definition of extensionally.
    This is one of the most studied problems in the history of set theory.
    There is no point in beating this dead horse.
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  4. #4
    Newbie petike's Avatar
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    Thumbs up Solved...

    Hi again,
    I have understood it yet.
    I have read somewhere the story about the "barber who shaved himself and simultaneously didn't shave himself" and it gave me the answer to my question.
    But thanks also for your posts.
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  5. #5
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    Quote Originally Posted by petike View Post
    Hi,
    I have a problem with "Russell's paradox" which I have just learned about at school. Simply said, it means that:
    Let's say we have sets R and A. And according to Russell's paradox (or theory) there cannot exist such set R, that:
    R = \{ A \ | \ A \notin A \}
    But I think that such set "exist"...

    What exactly means that A \in A? Then the set A could be one of these:
    A = \{ a, \ b, \ c, \ A \}
    A = \{ 4, \ 6, \ A, \ 9 \}
    A = \{ A, \ "string1", \ "string2" \}
    And what means that A \notin A? Then the set A could be one of these:
    A = \{ a, \ b, \ c, \}
    A = \{ 4, \ 6, \ 9 \}
    A = \{ "string1", \ "string2" \}

    So let's say that A = \{a, \ b, \ c\}.
    Then:
    R = \{\{a, \ b, \ c\}, \ \{a, \ b, \ c\}, \ \{a, \ b, \ c\}, \ ...\}
    ...so I have found such set R that:
    R = \{ A \ | \ A \notin A \}

    So where is the paradox ?


    Thanks.

    A good example of a set belonging to itself is the set of all sets.

    Since the set of all sets is a set then it is a set belonging to itself.

    When Cantor formed his set definition one could write for a set S={ x: f(x)} where f(x) was a property for x,but then Russel came up with the monumental f(x)=~xεx (x does not belong to itself),and by writing S={ x: ~xεx}............................................. .............................1


    led to the well known contradiction: if SεS THEN ~SεS, If ~SεS THEN SεS BY using .......1..........

    Then mathematicians to get away from this and other paradoxes developed the axiomatic set theory.

    The axiom that bypasses the Russel's paradox is the axiom of specification ,where for any set we write S={ x: xεA & f(x) } instead S ={ x: f(x)},where A IS a set.

    If now again we put f(x)= ~xεx this time we have S={ x: xεA & ~xεx } and SεS or ~SεS DOES not lead to contradiction,but to ,as bclere showed to the rule "nothing contains everything" or there is no Universe
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