Results 1 to 8 of 8

Math Help - Simple Set Question

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    239

    Simple Set Question

    To prove that there exists a unique set  N \in \mathcal{P}(X) such that  A \Delta N = A for all  A \in \mathcal{P}(X) is it as simple as choosing  N = \emptyset and so:

     (A \cup \emptyset) - (A \cap \emptyset) = A - \emptyset = A ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2007
    Posts
    239
    I guess its right then?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Not quite. You have shown existence but not uniqueness.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2007
    Posts
    239
    To show uniqueness: We want to find a set  N \in \mathcal{P}(X) such that  (A \cup N) - (A \cap N) = A . This means that  A \cup N = A and  A \cap N = \emptyset so that  A - \emptyset = A .

    By definition,  A \cup N = A \Rightarrow N = \emptyset (which is unique). And so it follows that  A \cap \emptyset = \emptyset and  N = \emptyset .

    Would this show uniqueness?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    This what must be done.
    A\Delta B = A\quad  \Rightarrow \quad B = \emptyset
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    239
    Suppose that  x \in A \Delta B and  x \in A . Then  x \in (A \cup B) - (A \cap B)   \Rightarrow  x \in A and  x \in A \Rightarrow x \in (A \cup B) - (A \cap B) . It follows that  B = \emptyset .

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    No that does not work.

    I think it is safe to say that most of us use this equivalence:
    A\Delta B = \left( {A - B} \right) \cup \left( {B - A} \right).

    With that we can say if A\Delta B = A then that means \left( {A - B} \right) \cup \left( {B - A} \right) = A.
    From which we can conclude that \left( {B - A} \right) \subseteq A\quad  \Rightarrow \quad B = \emptyset .

    QED
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Aug 2007
    Posts
    239
    Why doesn't it work?

     x \in  A \Rightarrow x \in A ?

    Oh ok: Its apparent that if  B-A \subseteq A then   B = \emptyset . Sort of like a contradiction if we didn't have the empty set.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Really simple question too hard for a simple mind
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 5th 2010, 08:03 AM
  2. Easy Question - Simple Interest, Simple Discount
    Posted in the Business Math Forum
    Replies: 0
    Last Post: September 21st 2010, 08:22 PM
  3. Replies: 4
    Last Post: May 4th 2010, 09:08 AM
  4. Simple Integral Question (I think it's simple)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 13th 2010, 02:37 PM
  5. Replies: 1
    Last Post: May 30th 2009, 02:59 PM

Search Tags


/mathhelpforum @mathhelpforum