# Math Help - Preorders - Posets

1. ## 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!

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

The questions wants you to prove that for x,yE $R^+$ 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?

3. 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 $\Rightarrow$ x=y. The questions wants you to prove that for x,yE $R^+$ 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 $\left( {x,y} \right) \in \mathcal{R}\, \Leftrightarrow \,y - x > T$.
This means that $\mathcal{R}$ is trivially antisymmetric because it is always false that $\left( {x,y} \right) \in \mathcal{R} \wedge \left( {y,x} \right) \in \mathcal{R}$.

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

For part c, note that $\mathcal{S}=\mathcal{R} \cup \Delta_A$, uniting the diagonal, gives a new relation which is reflexive, transitive, and antisymmetri