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Prove that (A-B)-C=(A-C)-(B-C)

  1. Feb 23, 2012 #1
    Let A,B and C be sets. Prove that
    (A-B)-C=(A-C)-(B-C).

    Attempted solution:

    Suppose [itex]x \in (A-B)-C[/itex]. Since [itex]x \in (A-B)-C[/itex] this means that [itex]x \in A[/itex] but [itex]x \notin B[/itex] and [itex]x \notin C[/itex].

    I'm not sure how to show how these two statements are equal.
     
  2. jcsd
  3. Feb 23, 2012 #2

    jbunniii

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    So far so good. You have established that x is in A, but not in B and not in C.

    Next, answer these questions:

    * Is x in A - C?
    * Is x in B - C?

    and see what you can conclude.
     
  4. Feb 23, 2012 #3
    Ok well you said [itex]x\in A[/itex] and [itex]x\notin C[/itex] What does that mean?
     
  5. Feb 24, 2012 #4
    Let A,B and C be sets. Prove that
    (A-B)-C=(A-C)-(B-C).

    Attempted solution:

    i.
    Suppose [itex]x \in (A-B)-C[/itex]. Since [itex]x \in (A-B)-C[/itex] this means that [itex]x \in A[/itex] but [itex]x \notin B[/itex] and [itex]x \notin C[/itex].

    ii.
    Suppose [itex]x \in (A-C)-(B-C)[/itex]. Since [itex]x \in (A-C)-(B-C)[/itex] it makes since that [itex]x \in A[/itex] and [itex]x \notin B[/itex] and [itex]x \notin C[/itex].

    Therefor these two statements are equal and (A-B)-C=(A-C)-(B-C).
     
  6. Feb 24, 2012 #5

    HallsofIvy

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    You need to finish this! [itex]x \notin B[/itex] and [itex]x \notin C[/itex] means what about x being in (A- C)- (B- C)?

    Why dfoes that make sense? And what does that tell you about x being in (A- B)- C?

     
  7. Feb 24, 2012 #6

    Deveno

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    you want to show the two sets are subsets of each other; that is, that they have precisely the same elements.

    if x is (A-B)-C, what does that mean?

    first of all, it means x is in A-B, but x is not in C.

    secondly, since x is in A-B, it means x is in A, but not in B.

    putting these two statements together, we have: x is in A, x is not in B, x is not in C.

    now if x is not in B, then it is not in B-C, since that is a subset of B.

    (x is not only NOT in the part of B that lies outside of C, it's totally not in B anywhere).

    but x IS in A, and x is NOT in C, so x IS in A-C.

    so x IS in A-C and x is NOT in B-C, so x IS in (A-C)-(B-C).

    that's "half" of the proof. the "other half" starts with assuming x is in (A-C)-(B-C).
     
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