Equivalence Relation On Set Let $A$ be a nonempty set and $$ R_1 \text{ and } R_2 $$ be two equivalence relations on $A.$ Investigate whether the following assertions are true or not.
$$R _1 \cup R_2 $$ is an equivalence relation.
$$R_1 \cap R_2$$ is also equivalence relation.
How can i show that following assertions are true or not ? A little bit of help would be enough I guess.
 A: You need to check whether the two relations are reflexive, symmetric, and transitive.  For reflexive, every equivalence relation has to include all the pairs of the sort $(a,a)$, which means both the union and intersection will as well.  For symmetric and transitive, if you have some kind of pair(s) you have to have other pairs.  Think about whether this requirement can be broken when you take the union or intersection.
A: It is called the closure property of equivalence relations.
Equivalence relations are closed under intersection.
For a relation to be equivalence it should satisfy Reflexive, Symmetric and Transitive property of relations.
Intersection is closed under all the three mentioned properties so the result of intersection of two equivalence relation will always be an equivalence relation.
.
See the proof here.
Equivalence relations are not closed under union.
Union is closed under Reflexive and Symmetric but not on the Transitive property so the union of two equivalence relation will not always be an equivalent relaion.See the proof here
A: Suppose:


*

*Both are relations on the set $\{a,b,c\}.$

*The equivalence classes for $R_1$ are $\{a\}$ and $\{b,c\},$ i.e. $b$ and $c$ are $R_1$-equivalent to each other, but not $R_1$-equivalent to $a.$

*The equivalence classes for $R_2$ are $\{a,b\}$ and $\{c\},$ i.e. $a$ and $b$ are $R_2$-equivalent to each other, but not $R_1$ equivalent to $b.$


That means $a$ is $R_1\cup R_2$-related to $b$ (i.e. either $R_1$-related or $R_2$-related) and $b$ is $R_1\cup R_2$-related to $c,$ but $a$ is not $R_1\cup R_2$-related to $c.$ Thus $R_1\cup R_2$ is not transitive.
But it's easy to show $R_1\cap R_2$ is an equivalence relation whenever $R_1$ and $R_2$ are equivalence relations on any set:
Suppose $x \mathrel{(R_1\cap R_2)} y$ and $y\mathrel {(R_1\cap R_2)} z.$
Then $x\mathrel{R_1} y$ and $y\mathrel{R_1} z,$ so $x\mathrel{R_1} z.$ And similarly for $R_2.$ Therefore $x \mathrel{(R_1\cap R_2)} z.$ Thus $R_1\cap R_2$ is transitive. And you can do the same with reflexiveness and symmetry.
