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Symmetric relation on ordered pairs

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    For sets A and B, define a relation [tex]\mathcal{R}[/tex] on A∪B by:
    [tex]\forall A, B \in A \cup B, x\mathcal{R}y[/tex] if and only if [tex](x,y) \in A \times B[/tex]

    For all sets A and B, if R is symmetric, then A = B


    2. Relevant equations



    3. The attempt at a solution

    I tried doing this, and I heard it's supposed to be false. BUT I can't see why it can be. The cartesian product should emply that x and y are only related when x is in A and y is in B. So, if it's symmetric, then it's already assumed that all elements in A are related to an element in B, and similarly, all elements in B are related to that same element in A. Which implies that they are equal. No?

    Is my logic messed up? I don't see how this could be false. If either of A or B were the empty set, there'd be nothing to prove, since the cartesian product of anything with an empty set is empty, then the if part can never be assumed, and it should be vacuously true.

    Can anyone point me in the right direction if I'm doing this wrong? Thanks!
     
  2. jcsd
  3. Apr 9, 2010 #2

    LCKurtz

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    Perhaps so, but I don't find your argument convincing. Look at an example:

    A = {1,2,3} B = {2,3,4} AUB = (1,2,3,4}

    A X B = {(1,2)(1,3)(1,4)(2,2)(2,3)(2,4)(3,2)(3,2)(3,4)}

    Here 2R3 and 3R2 because (2,3) and (3,2) are both in A X B.

    And 1R2 since (1,2) is in A X B, but (2,1) isn't so we don't have 2R1. So we don't have a symmetric relation. Of course, we also don't have A = B in this example. What you need to show is if A = B this can't happen and conversely, if the relation is symmetric that A = B.

    You haven't shown either yet.
     
  4. Apr 9, 2010 #3
    Actually, I figured it out (I think). Your example wouldn't work because it isn't symmetric which means symmetry (if part) fails so it's vacuously true either way, but I'm pretty sure this is right.

    Proof:

    Symmetry (in this context) means for all x,y in AUB, if (x, y) is in A x B, then (y, x) is also in A x B.

    So the "if" part of the statement I'm trying to prove (For all sets A and B, if R is symmetric, then A = B) is an if statement itself.

    I pick A = {1} and B = ∅
    Thus AUB = {1}. A x B = ∅. So there never exists a relation to satisfy the "if" part of the symmetry statement. Because there are no values in A x B. Thus, it is symmetric vacuously. So it is true that it is symmetric, but A != B.

    QED

    This makes sense in my crazy mind... Can anyone see anything wrong with it?
     
  5. Apr 9, 2010 #4

    LCKurtz

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    I think you are missing the point of this problem. Let's consider two non-empty sets A and B and a relation R on AUB. That is, R is just a subset of (AUB) X (AUB).

    R has the property that if [itex](x,y) \in R \rightleftarrows (x,y) \in A \times B[/itex].

    In other words [itex](x,y) \in R[/itex] if and only if [itex]x\in A\hbox{ and }y \in B[/itex]

    What you are asked to prove is: R is symmetric if and only if A = B. There are two parts; I will show you the easy part and get you started on the hard part.

    Part a. Given A = B show R is symmetric.
    Proof: Suppose [itex](x,y)\in R[/itex]. Then [itex]x\in A\hbox{ and }y \in A[/itex]. This implies [itex](y,x)\in R[/itex], so R is symmetric.

    Part b. Given R is symmetric, show A = B.
    So you need to show [itex]A \subset B\hbox{ and }B \subset A[/itex]. You might begin by observing that [itex]A \times B \subset R[/itex].
     
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