# Partial Order Relation and Equivalence Relation

• MHB
Yankel
Hello all,

If R is a partial order relation, is it true to say that

$R\cup R^{-1}$

$R^{2}$

$R\cap R^{-1}$

Are equivalence relations ?

Regarding the first one, I think that the answer is yes. If

$xRx$

then it remains after the union. Asymmetry means that $xRy$ without $yRx$ but when I apply the union both are in, so it becomes symmetric, and there is no reason why transitive won't work. Am I correct, or not even close ? What about the other two ?

Thank you

Gold Member
MHB
there is no reason why transitive won't work.
There is, actually.

Try to come up with proofs using precise statements and formulas rather than words. If a universal statement is false, this has to be shown by producing a counterexample.[/QUOTE]

Yankel
Yes, I found an example now, and I did solve the last one.

The only thing I am stuck with is R^2.

Can I say that if R is a partial order relation it's composite R^2=R ? I tried one example which worked.

Gold Member
MHB
Can I say that if R is a partial order relation it's composite R^2=R?
What do you mean by "composite"? It is true that $R$ is transitive iff $R^2\subseteq R$. But does the fact that $R$ is a partial order imply that $R^2$ is symmetric?

Yankel
By composite I Mean xRRy.

Gold Member
MHB
By composite I Mean xRRy.
It's important to say things correctly. First, $R\circ R$ is called composition (I have not seen the word "composite" used for this). Second, it is not clear what $x$ and $y$ are in $xRRy$. For given $x$ and $y$, $xR^2y$ is true or false, while $R^2$ is a relation, not something true or false.

Can I say that if R is a partial order relation it's composite R^2=R?
Yes. Since $R$ is transitive, $R^2\subseteq R$. For converse inclusion, if $(x, y)\in R$, then $(y, y)\in R$ due to reflexivity, so $(x, y)\in R^2$.