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Math Help - Problems with proofs in set theory.

  1. #1
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    Problems with proofs in set theory.

    Hello I am attempting to teach myself abstract algebra using FM Hall's "Abstract Algebra". The first section is an introduction to set theory which I get concepotually but when goigng through the excercises at the end of the chapter I got completely thrown by the following,

    Proove:

    A is contained completely in B <=> for all elements x which are not elements of A are elemets of B

    How do I go about prooving this. Thanks for any help with this.

    (This place could do with a maths notation font or summit )
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  2. #2
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    A is contained completely in B <=> for all elements x which are not elements of A are elemets of B
    These are not the same. The left-hand side says that A is a subset of B; the right-hand side says that the complement of A ("all elements x which are not elements of A") is a subset of B.

    (This place could do with a maths notation font or summit )
    You can use LaTeX between math and /math tags (tags use square brackets here unlike angular brackets in HTML). There is also a \Sigma button on the toolbar above the text area where one types posts; it inserts the math tags.
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  3. #3
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    Quote Originally Posted by Bwts View Post
    Prove:
    A is contained completely in B <=> for all elements x which are not elements of A are elemets of B
    That statement is false.
    This is true A \subseteq B \Leftrightarrow B^c  \subseteq A^c
    Quote Originally Posted by Bwts View Post
    (This place could do with a maths notation font or summit)
    As you can see we do have symbols. You just have to know LaTeX.
    [tex]A \subseteq B \Leftrightarrow B^c \subseteq A^c [/tex]
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  4. #4
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    OK thanks I will retype the question..

    Prove A \subseteq B \iff \forall x.x \notin A, x \in B

    Forgive me if my notation isnt standard as I am learing from a book. The two why arrow is ment to signify a two way implication.
    Last edited by Bwts; January 18th 2010 at 08:48 AM. Reason: clarity (I hope)
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  5. #5
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    Quote Originally Posted by Bwts View Post
    Prove A \subseteq B \iff \forall x.x \notin A, x \in B
    That is also clearly false.This is true: A \subseteq B \iff \forall x.x \notin B, x \notin A
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  6. #6
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    Hi Plato at the risk of sounding dumb could you please explain?
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  7. #7
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    Quote Originally Posted by Bwts View Post
    Hi Plato at the risk of sounding dumb could you please explain?
    Example: A = \left\{ {2,3,5,7} \right\}\,\& \,B = \left\{ {0,1,2,3,4,5,6,7,8,9} \right\}
    Now A \subseteq B but 0\notin A but 0\in B.

    It must be true that x \notin {\rm B} \Rightarrow \quad x \notin {\rm A}.
    Every element in A must also be in B.
    Every element not in B must also be not in A.

    This may be a translation problem or maybe a notation problem.
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  8. #8
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    I am guessing here that the question is implying that B is the universal set, so in this case would it be true that :

    \forall x.x \notin A, x \in B ?
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  9. #9
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    Quote Originally Posted by Bwts View Post
    I am guessing here that the question is implying that B is the universal set, so in this case would it be true that :

    \forall x.x \notin A, x \in B ?
    There is absolutely nothing in your first post that would imply that.
    Once again you must be complete in the question.
    Not doing so has wasted a good deal of time.
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  10. #10
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    My apologies Plato but the question was exactly how I put it and it seems to be unproovable if it is as you say. I am merely trying to ascertain whether the book is in error or that I am not understanding the procedure.

    Most sorry for wasting your time.
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  11. #11
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    Quote Originally Posted by Bwts View Post
    My apologies Plato but the question was exactly how I put it and it seems to be unproovable if it is as you say. I am merely trying to ascertain whether the book is in error or that I am not understanding the procedure.

    Most sorry for wasting your time.
    Write EXACTLY what the books says.
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  12. #12
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    Quote Originally Posted by drexel28 View Post
    write exactly what the books says.
    i have
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  13. #13
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    Well, if:

    \forall x: x \in B your whole statement of \forall x: x \notin A: x \in B is completely trivial.
    Go with Plato's correction, that actually DOES say something.
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  14. #14
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    OK thankyou all for your time.

    Iff (se what I did there ) it is not an inconvieniance I will probably be asking more questions as the book, although well written, does now appear to have issues with the excercises.
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