Short Prove or Disprove: $R_1 \cap R_2$ is an equivalence relation Suppose $R_1, R_2$ are both equivalence relations defined on nonempty set $A$. Prove or disprove: $R_1 \cap R_2$ is an equivalence relation. 
What method (if any) would you take to prove this in as few sentences as possible?

Let $R=R_1\cap R_2.$ You must show the following:


*

*For any $a\in A,$ $\langle a,a\rangle\in R.$ ($R$ is reflexive on $A$.)

*For any $a,b\in A$, if $\langle a,b\rangle \in R,$ then $\langle b,a\rangle\in R.$ ($R$ is symmetric on $A$.)

*For any $a,b,c\in A$, if $\langle a,b\rangle,\langle b,c\rangle\in R,$ then $\langle a,c\rangle\in R$. ($R$ is transitive on $A$.)


Each step can be proved very directly and straightforwardly, since $R_1,R_2$ are equivalence relations on $A$. Let me prove symmetry to give you a taste of it.

Suppose that $a,b\in A$ such that $\langle a,b\rangle\in R.$ Then $\langle a,b\rangle\in R_1,$ so since $R_1$ is symmetric on $A$, then $\langle b,a\rangle\in R_1.$ Likewise, $\langle b,a\rangle\in R_2,$ so $\langle b,a\rangle\in R_1\cap R_2=R.$


Alternately, if you're not used to thinking of binary relations as sets of ordered pairs, we can proceed as follows.
We still let $R=R_1\cap R_2.$ By this, we mean that $a\:R\:b$ if and only if $a\:R_1\:b$ and $a\:R_2\:b.$ Then we need to show the following:


*

*For any $a\in A,$ $a\:R\:a.$

*For any $a,b\in A$, if $a\:R\:b,$ then $b\: R\: a.$

*For any $a,b,c\in A$, if $a\: R\: b$ and $b\:R\:c,$ then $a\:R\:c$.


Each step is again straightforward, and very similar to the approach that would be taken above. Once again, I'll prove symmetry.

Suppose that $a,b\in A$ such that $a\:R\:b.$ Then $a\:R_1\:b,$ so since $R_1$ is symmetric, then $b\:R_1\:a.$ Likewise, $b\:R_2\:a,$ so $b\:R\:a$ by definition of $R=R_1\cap R_2.$

