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Simple proof involving sets

  1. Sep 3, 2008 #1
    1. The problem statement, all variables and given/known data
    For any sets A and B, prove that

    A[tex]\cap[/tex](A[tex]\cup[/tex]B) = A

    2. The attempt at a solution
    Now keep in mind I dont have any experience with proofs(and Im looking for a nudge in the right direction not a full proof).

    Here was my first instinct(and don't yell at me too much for it):
    Suppose x [tex]\in[/tex] A [tex]\cup[/tex] B
    Then x [tex]\in[/tex] A or x [tex]\in[/tex] B
    If x [tex]\in[/tex] A
    then x [tex]\in[/tex] A [tex]\cap[/tex] A
    so A = A
    IF x [tex]\in[/tex] B

    Now after writing that I felt that this is not a good way to prove the problem(or a way to do it at all). So any hints would be appreciated.
  2. jcsd
  3. Sep 3, 2008 #2


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    Homework Helper

    The only cases you really need to worry about are i) x is in A and ii) x is not in A. Can you handle those two?
  4. Sep 4, 2008 #3


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    Homework Helper

    You need to show

    A \cap \left( A \cup B \right) = A,

    correct? This means you must show that each set is a subset of the other.
    1. Start with [tex] x \in A [/tex] and show that it has to follow that

    x \in A \cap \left(A \cup B \right)

    This will give that [tex] A \cap \left(A \cup B \right) \supseteq A [/tex]

    2. Now pick [tex] x \in A \cap \left(A \cup B\right)[/tex]. You need to show that this means [tex] x \in A [/tex] (this should be rather easy). This will show that

    A \cap \left(A \cup B \right) \subseteq A

    and you will be done.
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