# Transitive relation

• Feb 18th 2009, 08:54 PM
namelessguy
Transitive relation
1)Prove or disprove: If A is a set for which each $\displaystyle R \in A$ is a transitive relation, then $\displaystyle \bigcup A$ is a transitive relation.

I think this is true. So I tried to prove it. Let $\displaystyle (x,y) \in \bigcup A$ and $\displaystyle (y,z) \in \bigcup A$. Then by definition of union, there exists $\displaystyle R$ such that $\displaystyle (x,y) \in R \in A$ and $\displaystyle (y,z) \in R \in A$. Since R is transitive by definition of set A, we have $\displaystyle (x,z) \in R \in A$. Hence, $\displaystyle (x,z) \in \bigcup A$
I am not sure about my argument here. Is it possible that $\displaystyle (x,y) \in R \in A$ while $\displaystyle (y,z) \in R' \in A$?Any help is appreciated.

2) Prove or disprove: If A is a nonempty set each of whose elements is an equivalence relation on the set B, then $\displaystyle \bigcup A$ is an equivalence relation on B.
I don't think this is correct, but I can't come up with a counterexample. I tried the problem but with $\displaystyle \bigcap A$, and it seems that three properties of equivalence relation do hold for this case, but for union, somehow I don't see it follows.
• Feb 19th 2009, 03:02 AM
clic-clac
Hi
1) Check $\displaystyle A=\{\{(a,b),(b,c)\},\{(b,d)\}\}$
Take care, as you said after, the "first $\displaystyle R$" has no to be equal to the "second one"!

2) Here again, think of a particular but simple case where $\displaystyle A$ has only two elements. What if there is one class in the first relation of equivalence intersects partially a class of the second relation, for instance: (the idea is to lose the transitivity)

$\displaystyle B=\{a,b,c\},\ A=\{x,y\},\$ $\displaystyle x=\{(a,a),(b,b),(c,c),(a,b),(b,a)\},\ y=\{(a,a),(b,b),(c,c),(b,c),(c,b)\}$