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Proofs involving relations

  1. Nov 4, 2013 #1
    I'm currently reading the section on relations in Velleman's "How to prove it" and I have found a statement somewhere that I want to prove, but I'm not sure whether what I have come up with is reasonable and I also have some questions on the logic used in these type of proofs.

    The theorem is: Let R be a relation on set A. Then it holds that:
    [itex]R^2 \subseteq R <=>[/itex] R is transitive [itex]<=> R^i \subseteq R ,\forall i \geq 1[/itex]

    I'll start with [itex]R^2 \subseteq R[/itex] <=> R is transitive:

    => Assume that [itex]R^2 \subseteq R[/itex]. R^2 means that for an arbitrary ordered pair (x, z)[itex]\in[/itex]A there is a y[itex]\in[/itex]A such that xRy and yRz. The goal is to show that R is transitive, which is basically a conditional statement. If xRy and yRz hold, then we can infer that xRz. Therefore I can assume xRy and yRz, and now should prove that xRz.But because R^2[itex]\subseteq[/itex]R, I know that xRy and yRz is true for an arbitrary (x, z)[itex]\in[/itex]A. It follows that (x, z)[itex]\in[/itex]R.

    <= Assume that R is transitive. Therefore if for any (x, z)[itex]\in[/itex]A it is true that there is a y[itex]\in[/itex]A such that xRy and yRz, I know that xRz. Now since R^2[itex]\subseteq[/itex]R is also a conditional statement of the form [itex]\forall[/itex](x,y)[itex]\ni[/itex]AxA((x,y)[itex]\ni[/itex]R^2 then (x,y)[itex]\ni[/itex]R, I can assume that there is an ordered pair, call it (x,z)[itex]\ni[/itex]R^2. Since we assumed that R is transitive, it follows by modus ponens that (x,z)[itex]\ni[/itex]R.

    Now I'll try to prove: R is transitive <=> R^i [itex]\subseteq[/itex]R, [itex]\forall[/itex]i[itex]\geq[/itex]1.

    => Assume R is transitive. To prove that R^i[itex]\subseteq[/itex]R, for all i [itex]\geq[/itex]1 we can use induction.

    Base case: i=1 -> R[itex]\subseteq[/itex]R obviously holds.

    Now I'm not really sure whether I actually have to use induction here. I think what needs to be shown here is pretty straightforward. If there is a y[itex]\in[/itex]A such that x1Ry and yRx2 hold, then because R is transitive it follows that x1Rx2. And then you can continue this way with x2 and x3 and in general any value i. But I don't really know how to write this in a formally correct way. Can anybody help me with that? And please give some feedback on the first part of the proof.
     
  2. jcsd
  3. Nov 6, 2013 #2

    Stephen Tashi

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    It that a typo?

    [itex] R^2 [/itex] is a set, so you shouldn't say it "means" something. The definition of a set tells you what it means for an element to be in that set. So you may say "To say an element [itex] x [/itex] is in [itex] R^2 [/itex] means that ...".
     
  4. Nov 6, 2013 #3
    ^Agreed.

    In particular, it's the set [itex]R^2 = \{(x,z)\in A^2:\enspace \exists y\in A \text{ such that } xRy \text{ and } yRz\}[/itex], and we use the notation [itex]xR^2z[/itex] as a shorthand for [itex](x,z)\in R^2[/itex].
     
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