Partial Order Relation help

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• Feb 10th 2011, 05:57 PM
gutnedawg
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.
• Feb 11th 2011, 01:29 AM
emakarov
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.