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Intersection proof

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove A intersects B=empty set if and only if B is a subset of (X/A)

    2. Relevant equations

    3. The attempt at a solution

    Would I prove the contrapositive in this case?

    If B is not a subset of (X/A), then the intersection of A at B is not the empty set

    Could someone please show me what to do?

    Thank you very much
  2. jcsd
  3. Mar 7, 2008 #2
    "if and only if" means that you must prove the statement both ways. Proving by contradiction seems the easiest way for each. The approach you suggest will only get you half the proof.
  4. Mar 7, 2008 #3
    Thank you very much

    Could you show me where to go from there?

    Thank you
  5. Mar 7, 2008 #4
    Well since you wanted to use the contrapositive for A intersects B = empty -> B is a subset of X\A I'll explain that way. Let [itex]B=\{b_1,b_2,...\}[/itex]. If B is empty what we want is vacuously true since the empty set is a subset of every set (and thus B can never not be a subset of [itex]X \setminus A[/itex]). If [itex]B\not\subset X\setminus A[/itex] then at least one [itex]b_i \in B[/itex] is [itex]\in X\setminus (X \setminus A) = A[/itex] which implies A and B have these elements in common.

    You can prove the other way in a similar fashion. My personal suggestion is contradiction for the other way.

    On this note, you have another question that's pretty similar. Your questions really boil down to choosing elements of certain sets and then showing by logic that they must/must not exist in other sets. Try to proceed like this in your other question also.
    Last edited: Mar 7, 2008
  6. Mar 7, 2008 #5
    Thank you very much

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