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Proofs in set theory

  1. Oct 14, 2008 #1
    Let A,B,C,X,Y be subsets of E,and A' MEAN the compliment of A in E i.e A'=E-A,and

    A^B = A [tex]\cap[/tex] B

    Then prove the following:

    a) (A^B^X)U(A^B^C^X^Y)U(A^X^A') = A^B^X

    b) (A^B^C)U(A' ^ B^C)U B' U C' = E

  2. jcsd
  3. Oct 14, 2008 #2


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    Yep, that looks like a homework problem. It's curious that you would simply copy it here, rather than explain what you've done, and where you're having trouble, so that we could help you work through the problem....
  4. Oct 15, 2008 #3
    I tried to solve the problem by using the definition of equality of sets and it gets very lengthy ,is there another way??


    A=B <====> (xεΑ <------>xεB)
  5. Oct 17, 2008 #4

    a) (A^B^X)U(A^B^C^X^Y)U(A^X^A') = A^B^X

    b) (A^B^C)U(A' ^ B^C)U B' U C' = E

    Let xε[(A^B^X)U(A^B^C^X^Y)U(A^X^A')] <======> (xεΑ & xεB & xεX) v (xεA & xεB & xεC & xεX & xεY ) v ( xεA & xεX & ~xεA)

    That is how far i could go .

    Please continue the problem for me.

    For the 2nd problem it is the same sticky situation.

  6. Oct 17, 2008 #5


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    There seems to be something missing in the first problem.
    It is to show

    \left(A \cap B \cap X\right) \cup \left(A \cap B \cap C \cap X \cap Y\right) \cup \left(A \cap X \cap A'\right) = A \cap B \cap X

    , correct? To shorten my typing I'll refer to the RHS (right hand side) and LHS (left hand side) of this statement.

    To prove RHS is contained in LHS.
    Suppose [tex] w \in A \cap B \cap X [/tex]. Then (with all the ugly glory)

    w \in A \cap B \cap X \subseteq (A \cap B \cap X) \cup (A \cap B \cap C \cap X \cap Y) \cup (A \cap B \cap A')

    so we know that LHS is a subset of RHS.

    Now to the other inclusion. Suppose [tex] w \in LHS [/tex].

    Case 1: If [tex] w \in A \cap B \cap X [/tex] we are done.
    Case 2: If

    w \in A \cap B \cap C \cap X \cap Y

    then [tex] w \in A \cap B \cap X [/tex] and again we are done.

    Case 3: (This is where I believe the problem lives). Suppose

    w \in A \cap X \cap A'

    We know that both [tex] w \in A [/tex] and [tex] w \in X [/tex], and [tex] x \in A' [/tex], but with the information given we cannot conclude that [tex] w \in B [/tex] (and now the latex preview is acting up) - we simply don't have enough knowledge of the relationships of the individual sets to make this conclusion.
  7. Oct 17, 2008 #6
    While looking through the book; Set Theory And Logic by ROBERT R. STOLL, i met this problem on page 22 ,exercise 5.3 (a) and not doing homework.

    The other problem is on the same page ,exercise 5.3 (b)

    So i do not think the problem 5.3(a) is not wrong
  8. Oct 17, 2008 #7
    Are you sure you typed the problem correctly? If so, all of Statdad's deductions are correct so far. Now, when we look at case 3: [tex]w\in A\cap X\cap A'[/tex]. This says that w is an element of A and w is an element of A' (the complement of A). What is wrong with this statement??

    Another way to think of it: We know, in general, that [tex]A\cap B=B\cap A[/tex] and that [tex](A\cap B)\cap C=A\cap(B\cap C)[/tex]. Using this, what is [tex]A\cap X\cap A'[/tex]? More specifically, what is [tex]A\cap A'[/tex] for any set?
  9. Oct 18, 2008 #8
    Where are you getting at??? I typed the problem straight from the book.

    Thanks for the help.
  10. Oct 18, 2008 #9
    If that is indeed the problem in the book, fine. Then answer my question:

    What is [tex]A\cap A'[/tex] for any set A? In other words, what is the intersection of a set and its complement?
  11. Oct 20, 2008 #10
    The empty set:Φ
  12. Oct 20, 2008 #11
    There you go. Can you do the rest of the problem now?
  13. Oct 21, 2008 #12
    NO because i cannot follow why all the above cases and any added should lead us to the desired result.
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