# Thread: Partial Order Relation help

1. ## Partial Order Relation help

Let $I_{a}$ denote the identity relation on a set a

Show that if R is a partial order relation on a then $R\backslash I_{a}$ is a strict partial order relation on a

Show that if S is a strict partial order relation on a then $S \cup I_{a}$ is a partial order relation on a

I'm pretty sure I have these down I just want to make sure

For the first one, this one is just straight forward... I mean if you have a partial order relation and then you take out the identity it becomes irreflexive but I'm not sure how to "show" this. Do I just say:

$x \epsilon R \backslash I_{a}$ means that x cannot be of the form $(x,x)$ making $R \backslash I_{a}$ irreflexive and thus a strict partial order?

The second is basicly the same but the opposite way.

2. Often, the easiest and shortest way for both the writer and the reader is to write things down formally.

Suppose $R$ is a partial order on $a$. We need to prove that $R\setminus I_a$ is a strict partial order.

Irreflexivity. Let $x\in a$. Then $(x,x)\in I_a$, so $(x,x)\notin R\setminus I_a$. That was easy.

Transitivity. Suppose $(x,y),(y,z)\in R\setminus I_a$, i.e., $(x,y),(y,z)\in R$, $x\ne y$ and $y\ne z$. Then $(x,z)\in R$. It's left to show that $x\ne z$. Suppose $x=z$; then by antisymmetry of R we have x = y, a contradiction.

Thinking back, it is right that we had to use antisymmetry. If every reflexive and transitive (but not necessarily antisymmetric) relation would produce a strict partial order when one subtracts $I_a$, then it is unlikely that joining a strict order with $I_a$ would produce an antisymmetric relation as the second part of the problem says.

I recommend similarly writing out the second part in every detail.