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Set Algebra

  1. Jan 23, 2012 #1
    How would I prove [itex]A \subseteq B \Leftrightarrow A \cap B^{c} = \emptyset[/itex] ?
     
    Last edited: Jan 23, 2012
  2. jcsd
  3. Jan 23, 2012 #2
    This is what I have so far:

    Suppose [itex]A \cap B^{c} = \emptyset[/itex]. Let [itex]x \in A[/itex]. We want to show [itex]x \in B[/itex]. Since [itex]A \cap B^{c} = \emptyset[/itex] and [itex]x \in A[/itex] then [itex]x \notin B^{c}[/itex]. Hence [itex]x \in B[/itex].
     
  4. Jan 23, 2012 #3

    PAllen

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    Why not start by assuming either side of the two way implication, rather than assuming something completely different?
     
  5. Jan 23, 2012 #4
    It is supposed to be [itex]A \subseteq B \Leftrightarrow A \cap B^{c} = \emptyset[/itex].
     
  6. Jan 23, 2012 #5

    PAllen

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    Ok, then what you've got is a good start. Now work it the other direction.
     
  7. Jan 23, 2012 #6
    So ([itex]\Rightarrow[/itex]) is correct? First, I suppose [itex]A \subseteq B[/itex]. Then I let [itex]x \in A[/itex] and therefore [itex]x \in B[/itex] by supposition, but then it does not lead me anywhere.
     
  8. Jan 23, 2012 #7

    PAllen

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    Can something be in B and B complement?
     
  9. Jan 23, 2012 #8
    [itex]\emptyset[/itex] ?
     
  10. Jan 23, 2012 #9
    Also, we know [itex]\emptyset \subseteq A \cap B^{c}[/itex] and we want to show [itex]A \cap B^{c} \subseteq \emptyset[/itex], but I do not know how to do it.
     
  11. Jan 23, 2012 #10

    PAllen

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    Can there be any element in B and B complement?
     
  12. Jan 23, 2012 #11
    No, that would be a contradiction.
     
  13. Jan 23, 2012 #12

    PAllen

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    The rest should follow. Can't think of another hint that isn't the complete answer.
     
  14. Jan 24, 2012 #13
    Is this correct?

    Suppose [itex]A \subseteq B[/itex]. Let [itex]x \in A[/itex]. Then [itex]x \in B \because A \subseteq B[/itex]. [itex]A \cap B^{c} \Rightarrow x \in A[/itex] and [itex]x \in B^{c}[/itex] by definition of intersection. Since [itex]x \in B[/itex], [itex]A \cap B^{c} = \emptyset[/itex].
     
  15. Jan 24, 2012 #14

    PAllen

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    Ok, but you can tighten up the the argument. If x in A, then x in B, then x not in B complement; then A intersect B complement is empty.
     
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