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I Question from set theory

  1. Aug 7, 2017 #1
    We can prove that
    When A and B are two sets(A≠B)
    (A-B) = (A∩B') = (A-(A∩B))
    {We can also confirm them using venn diagram}
    From first and third relation
    A-B = A - (A∩B)
    By cancelling A from both side
    I get B = (A∩B)
    Which is only possible when A and B are same set.
    What is wrong in my proof , is it not valid to cancell sets A from both side(if yes then why?)
     
  2. jcsd
  3. Aug 7, 2017 #2

    mfb

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    There is no such operation.
     
  4. Aug 7, 2017 #3
    So we can not cancell sets
    Can you give a reason for not to cancell sets from both side( or another similar example)
    Another question:
    what does addition of two sets means
    I know what does (A∪B) ,(A-B) meant
    But can you tell me what does A + B means
    A,B,C are sets
     
  5. Aug 7, 2017 #4

    mfb

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    A-B = A-C does not imply B=C.
    Similarly, A∪B = A∪C does not imply B=C either.
    From the context, probably the union, (A∪B). If in doubt, ask the person who wrote that.
     
  6. Aug 7, 2017 #5
    Ok, A∪B And A+B are same
    And n(A∪B)= n(A)+ n(B) -n(A∩B)
    Is just for cardinality of (A∪B)

    in my first question
    You are trying to say that, we can not apply operations on sets we can only apply operations to their cardinality(your example completely proved the reason why, thanks)

    A-B = A-(A∩B)
    A-A = B-(A∩B)
    Φ+(A∩B) = B
    Does it implyes any thing?
     
  7. Aug 7, 2017 #6
    I got a example
    A={1,2,3,4}
    B={2,3,7,8,9}
    C={2,3,7,8,11,13}
    A-B = A-C = {1,4}
    Clearly we can see B≠C
     
  8. Aug 7, 2017 #7

    mfb

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    That is not correct.
    What is Φ? The empty set? That equation is not correct either.

    You can see all this by drawing diagrams.
    Right. A shorter example:
    A={1}, B={2}, C={3}
     
  9. Aug 7, 2017 #8
    I.e what i am saying ,what i have learn from our conversation and from theory is that we can not apply every algebraic operations on set but we can apply them on cardinality of them.
     
  10. Aug 7, 2017 #9

    mfb

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    Cardinalities of finite sets are just integers, and you can add and subtract integers as usual.
     
  11. Aug 7, 2017 #10

    jbriggs444

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    Sets do not form an algebraic group when defining "addition" as taking the union. You are correct that we cannot perform algebra on them.

    However, it is not clear that cardinalities work either.

    A = {1}
    B = {2}
    A-B = {1}
    |A| - |B| is not equal to |A-B|
     
  12. Aug 7, 2017 #11
    Yes,algebraic operations do not work with the cardinality of sets , i just write it because i can write n(A∪B) = n(A) + n(B) - n(A∩B)
    But i cant write A = B+(C - D)
     
  13. Aug 7, 2017 #12

    FactChecker

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    In general, you are trying too many things with sets that are not valid. You should take a step back, pay strict attention to the operations that are defined for sets, and only use the basic operations and things proven from the basic operations. Otherwise, it is tempting to adopt bad habits.
     
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