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Math Help - Cancellation of sets

  1. #1
    Forum Admin topsquark's Avatar
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    Cancellation of sets

    I'd appreciate it if someone would look this over for me. This is probably easier than I am making it.

    Let A, B, and C be subsets of a set X. I wish to show
    \displaystyle B = C \Leftrightarrow A \cup B = A \cup C

    First: \displaystyle B = C \implies A \cup B = A \cup C

    If x \in B then x \in A \cup B, but x \in B \implies x \in C, thus \forall x \in B \text{, } x \in A \cup C

    Thus A \cup B \subset A \cup C. Since the roles of B and C are interchangable in the above by symmetry we also have that A \cup C \subset A \cup B. Thus A \cup B = A \cup C.

    I can't figure out how to start the other half: A \cup B = A \cup C \implies B = C. If x \in A I can't seem to go anywhere and if x \in B - A I get down to a statement that says B - A = C - A, which seems to be a rather circular result. (I think this means B = C since x \not \in A, but I'm a little sketchy on the logic here.)

    Thanks in advance for the help.

    -Dan
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  2. #2
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    Quote Originally Posted by topsquark View Post
    I can't figure out how to start the other half: A \cup B = A \cup C \implies B = C. If x \in A
    Do you really think that it is true?
    Let A=\{1,2,3\},~B=\{1,2\},~\&~C=\{2,3\}.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    Do you really think that it is true?
    Let A=\{1,2,3\},~B=\{1,2\},~\&~C=\{2,3\}.
    (sighs) I got bogged down in the proof and didn't even look at the obvious stuff. Thanks for the catch.

    (I'm definately going to hit my Set Theory text.)

    -Dan
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