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Relations, power sets and the empty/null set.

  1. Sep 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose R is a relation on A, and define a relation S on P (A) as follows:
    S = {(X, Y ) ∈ P (A) × P (A) | ∀x ∈ X∃y ∈ Y (xRy)}.
    For each part, give either a proof or a counterexample to justify your
    answer.
    (a) If R is reflexive, must S be reflexive?
    (b) If R is symmetric, must S be symmetric?
    (c) if R is transitive, must S be transitive?


    2. Relevant equations

    P(A) = power set of A
    Uppercase letters = sets
    Lowercase = element(s) of the sets

    3. The attempt at a solution
    For parts a and b, and possibly c, the relation S is undefined when I take one of the X, Y to be the empty set, since there are no x,y in X,Y when X or Y are the empty set xRy is undefined.

    a)Assume R is reflexive.
    Want to prove ∀X∈P(A)((X,X)∈S)
    Let x be an arbitrary element of P(A)
    then x is an arbitrary subset of A;
    take A to be the empty set, but then the empty set has no elements therefore the requirement ∀x ∈ X∃y ∈ Y (xRy) for (X,X) to be an element of S is invalid, that is undefined.

    I get the same result for part b, and I'm assuming it's the same way for c) as well. Did this question mean to exclude the empty set as a possibility for the domain of S? Or am I possibly under some misconception.
     
  2. jcsd
  3. Sep 6, 2014 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    A statement like "for all x in E, x has property P" (i.e. ##\forall x\in E~P(x)##) is true when ##E=\varnothing##, regardless of what P is. To say that it's false is to say that there's an x in E that doesn't have property P. So the statement isn't false, and must therefore be true.

    This is probably less confusing if we recall that the statement ##\forall x\in E~P(x)## is actually an abbreviation for
    $$\forall x\ \big(x\in E\ \Rightarrow\ P(x)\big)$$ If E is empty, the implication is clearly true for all x.
     
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