Thread: Preorders - Posets

Hey,

So I have another question,

A relation that is transitive and antisymmetric is called a preorder. Define the "threshold relation" on the positive real numbers in the following way: Fix a positive real number T. For x, y contained in the positive real numbers, we say that y - x > T.

a) Show that R is a preorder.
b) Why is R called a threshold relation?
c) In general, if R is an arbitrary preorder on a set A, show that S is a partial order on A where S is defined in the following way: for x,y in A, xSy if x=y or xRy.

So for a) I prove transitivity quite easily but for antisymmetry can I assume that this is vacuously true since the assumptions for antisymmetry (y -x > T and x - y > T) are clearly impossible?
For b) I belive it's a threshold since any difference less than T is not contained within the set?
And for c) x=y is reflexivity, right? We're adding the reflexivity property.

Thanks for the help!

i have the same question. and i'm wondering how you can show antisymmetry as well. by definition, antisymmetry is: if xRy and yRx x=y.

The questions wants you to prove that for x,yE we say that xRy if y-x > T, where T is a fixed value = 5

but if we use antisymmetry on this, and we prove that y=x, then y-x won't be greater than T. so how can this be possilbe?

am i wrong in thinking this?

Originally Posted by scottie.mcdonald

i have the same question. and i'm wondering how you can show antisymmetry as well. by definition, antisymmetry is: if xRy and yRx x=y. The questions wants you to prove that for x,yE we say that xRy if y-x > T, where T is a fixed value = 5 but if we use antisymmetry on this, and we prove that y=x, then y-x won't be greater than T. so how can this be possilbe?

As was noted in the OP. The relation .
This means that is trivially antisymmetric because it is always false that .

As for part b, I have never seen that term before, so no comment.

For part c, note that , uniting the diagonal, gives a new relation which is reflexive, transitive, and antisymmetri