MHB Partial Order Relation and Equivalence Relation

[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

 

[Evgeny.Makarov]

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.

 

[Evgeny.Makarov]

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.

 

[Evgeny.Makarov]

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 $(x, y)\in R^2$.