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Homework Help: Relations and Subsets

  1. Jul 3, 2011 #1
    Hi, this is my first time posting here, and I am trying to prove the following proofs and I do not know how to start:

    Suppose R and S are relations on set A

    1. If R is reflexive and R is a subset of S, then S is reflexive.
    2. If R is symmetric and R is a subset of S, then S is symmetric.
    3. If R is transitive and R is a subset of S, then S is transitive.

    I understand the definitions well enough, but I do not know how to specify these conditions.
    I am not certain that any of these are true. Any help would certainly be appreciated.

    Attempt on #1:

    Assume R is reflexive
    Assume R is a subset of S.

    Since R is reflexive, for all x in R, xRx ---> xRx
    and since R is a subset of S, all x in R is in S as well.

    Thus x in S, xSx ---> xSx (this is the step I am not sure of, nor do I know if I have anything else to work off of.)

    Therefore S is reflexive.


    The other two questions I set up the same way, but I use the other definitions. It definitely is not enough, and in the original problem it states (prove or show counterexample). So I am wondering if this is suitable.

    1. let R, S be relations on A
    R= { (1,1) (2,2) (1,2) (2,1)}
    S= { (1,1) (2,2) (1,2) (2,1) (1,3) (2,3) (3,2) (3,1)}

    Clearly R is reflexive and R is in S. But S is not reflexive as it does not contain (3,3).

    So for reflexivity at least this does not seem to be true. I am however trying to figure out if that is the case for symmetric and transitive.
     
    Last edited: Jul 3, 2011
  2. jcsd
  3. Jul 3, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    You are missing the sense of the 'subset'. R is a subset of AxA, i.e. ordered pairs (x,y) where x is in A and y is in A. If A={1,2,3} then your example for R isn't reflexive either. It's not reflexive on A={1,2,3}. It's only reflexive on the subset of A, {1,2}. That's not A. Make R reflexive on {1,2,3} and then say why if R is a subset of S as a set of ordered pairs, then S is reflexive.
     
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