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Set theory proof help?

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A and B be sets. Prove that A U B=(A-B) U (B-A)U (A∩B)


    2. Relevant equations



    3. The attempt at a solution
    If I want to prove the foward direction: A U B⊆(A-B) U (B-A)U (A∩B) then
    from my understanding I know that xεA or xεB. And I can assume wolog that xεA. But since I assume that x is an element from A, its already equal to (A-B) U (B-A)U (A∩B)? Since xεA and x is not an element of B or xεb and not an element of A or x is both an element of A and B? is that all i have to show for the foward direction?
     
    Last edited: Oct 4, 2012
  2. jcsd
  3. Oct 4, 2012 #2
    ok i was thinking this for a while. Lets take two sets A={1,2,3,4,5} and {3,4,5,6,7}
    so I can divide this proof into 3 cases:
    case 1: xεA
    case 2: xεB
    case 3: xεA and xεB.

    am i right?
     
  4. Oct 4, 2012 #3
    Im not sure what you've covered so far. Do you know the logical definition of a subset?
     
  5. Oct 4, 2012 #4
    Foward Direction:A U B⊆(A-B) U (B-A)U (A∩B)
    Case 1: xεA
    Since xεA then xε(A-B). xε(A-B) means that xεA and x is not an element of B.
    Case 2: xεB
    Since xεB then xε(B-A) which means that xεB and x is not an element of A.
    Case 3:xεA and xεB
    Since xεA and xεB then xε(A∩B) means that xεA and xεB.
    Hence A U B⊆(A-B) U (B-A)U (A∩B).

    Reverse direction
    (A-B) U (B-A) U (A∩B)⊆A U B
    Background: I have to show that xε(A-B) or xε(B-A) or xε(A∩B) is a subset of A U B.

    Case 1:xε(A-B)
    Since xε(A-B) then this means that xεA and x is not a element of B. Which means that xε(AUB). xε(AUB) means that xεA or xεB. Hence (A-B)⊆A U B.

    Case 2: xε(B-A)
    Since xε(A-B) then this means that xεB and x is not an element of A. Hence xε(AUB) which is similar as the above case. Hence (B-A)⊆A U B.

    Case 3: xε(A∩B)
    Since xε(A∩B) which implies that xεA and xεB. This means that xε(AUB). Since xεA or xεB or both xεA and xεB. Hence (A∩B)⊆A U B.
    Therefore (A-B) U (B-A) U (A∩B)⊆A U B
    Im not entirely sure if this proof is right. It seems verbose but then again im taking an introduction class.
     
    Last edited: Oct 4, 2012
  6. Oct 5, 2012 #5
    Not quite. Let me go through what you did.


    Case 1: xεA

    "Since xεA then xε(A-B)" (This is not true consider x in both A and B. Then x wont be in A-B but it is in A). xε(A-B) means that xεA and x is not an element of B.
    Case 2: xεB
    "Since xεB then xε(B-A)" (Again this is not true) which means that xεB and x is not an element of A.
    Case 3:xεA and xεB
    Since xεA and xεB then xε(A∩B) "means that xεA and xεB (redundant)".
    Hence A U B⊆(A-B) U (B-A)U (A∩B).

    Reverse direction
    (A-B) U (B-A) U (A∩B)⊆A U B
    Background: I have to show that xε(A-B) or xε(B-A) or xε(A∩B) is a subset of A U B.

    Case 1:xε(A-B)
    Since xε(A-B) then this means that xεA "and x is not a element of B (useless info)". Which means that xε(AUB). xε(AUB) "means that xεA or xεB (This step should be before you say x is in AUB). Hence (A-B)⊆A U B.

    Case 2: xε(B-A)
    Since xε(A-B) then this means that xεB "and x is not an element of A (useless info)". Hence xε(AUB) which is similar as the above case. Hence (B-A)⊆A U B.

    Case 3: xε(A∩B)
    Since xε(A∩B) which implies that xεA and xεB. This means that xε(AUB). Since xεA or xεB or both xεA and xεB (this step should be before concluding xε(AUB)) . Hence (A∩B)⊆A U B.
    Therefore (A-B) U (B-A) U (A∩B)⊆A U B
    Im not entirely sure if this proof is right. It seems verbose but then again im taking an

    The first direction needs more work. The second just needs a little fix. The first direction is harder and it might require showing that A = (A-B)U(A∩B). Also, I'm not sure if your class covered this but, xε(A-B) <=> xε(A∩B[itex]^{c}[/itex]) (B[itex]^{c}[/itex]) is the complement of B)
     
  7. Oct 5, 2012 #6
    since the definition of AUB is xεA or xεB then I thought that x could be in either set A or set B or both. That was the reason behind my three cases. Oh maybe I should have written it like this for the foward direction

    Case 1: xεA and x is not element of B
    Case 2: xεA and xεB
    Case 3: xεB and x is not an element of A but that is the same as case 1.
     
  8. Oct 6, 2012 #7
    I'm not sure what you are trying to justify.
    Here is the definition of a subset:
    A[itex]\subset[/itex]B [itex]\Leftrightarrow[/itex] for any x, x [itex]\in[/itex]A [itex]\Rightarrow [/itex] x [itex]\in[/itex]B
     
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