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Proof subset?

  1. Mar 9, 2008 #1
    Proof subset????

    Given three sets A, B, and C, set X = (A-B) U (B-C) U (C-A) and
    Y = (A∩B∩C) complement C. Prove that X is subset of Y. Is Y necessarily a subset of X? If yes, prove it. If no, why???
    ---When I draw the two venn diagrams X and Y, they are the same, but I don't know how to prove it...

    Can someone help me out here...
    Thanks in advance!
  2. jcsd
  3. Mar 9, 2008 #2


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    The usual proof for such a statement is: Let x be an element from X and try to prove that it is also an element of Y. So if x is in X, you know that it is in A but not in B, and/or it is in B but not in C, and/or it is in C but not in A. You want to show that it must be in (A∩B∩C)C. If it is in (A∩B∩C) then it would be in A and B and C, so it being in the complement means it is at least not in A or not in B or not in C. So you could suppose it is both in A and B and show that it cannot be in C.

    So let me write this out in the right order:
    Let [itex]x \in X[/itex]. Suppose that [itex]x \in A, x \in B[/itex]. Then definitely, [itex]x \not\in A - B, x \not\in C - A[/itex], because if it is in A it cannot be in any set from which we remove all elements of A (and similarly for B). But it must be in one or more of (A-B), (B-C) and (C-A), so it must be in (B - C). That is, x is in B (which we knew) but not in C. So if x is not in C, it cannot be in the intersection of C with whatever set you make up. In particular, it is not in [itex]A \cap B \cap C[/itex]. Therefore, it must be in the complement of that set, which is called Y.

    Now try to do the same reasoning for [itex]Y \subset X[/itex]. You have already shown by your Venn diagram that if x lies in Y, it must lie in X. So try to prove it in the same way as I just did.
  4. Mar 9, 2008 #3
    Thanks alot !!!!!
  5. Mar 9, 2008 #4


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    There is, however, a technical problem with "Let [itex]x \in X[/itex]"- the proof collapses is X is empty. Far better to start "IF [itex]x \in X[/itex]". That way, if X is empty, the hypothesis is false and the theorem is trivially true.
  6. Mar 9, 2008 #5


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    You are right, obviously the empty set is a subset of any set S (vacuously, all its elements are also in S).
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