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How do you prove things with null

  1. Jul 16, 2007 #1
    For instance, the following seems obvious but I don't know how to state the proof formally (and directly):

    Show [tex]A \cap (B-A) = \{\} [/tex]

    Here is a try:
    For any [tex]x \in U \ if \ x \in A[/tex] then [tex] x \notin (B-A)[/tex]
    therefore [tex] A \cap (B-A) = \{\} [/tex]

    there is something missing....
  2. jcsd
  3. Jul 16, 2007 #2
    Well, by definition (of set exclusion),
    [tex]x \in B-A \leftrightarrow x \in B \wedge x \notin A[/tex]
    Also by definition (of set intersection),
    [tex]x \in A \cap (B-A) \leftrightarrow x \in A \wedge x\in B-A[/tex]
    Substituting the first statement into the second, you get
    [tex]x \in A \cap (B-A) \leftrightarrow x \in A \wedge x \in B \wedge x \notin A[/tex]
    But the RHS is clearly false (x is not simultaneously in A and not in A), meaning that the LHS is also false. Therefore, for all x,
    [tex]x \notin A \cap (B-A)[/tex]

    Therefore, [itex]A \cap (B-A)[/itex] satisfies the defining property of the empty set.
  4. Jul 17, 2007 #3


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    You have shown that if x is in A then it is not in [tex]A \cap (B-A)[/tex] What if x is not in A? That's what's missing. (Yes, it's trivial but you should say it.)
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