Please check and critique the following proof:

Let $\displaystyle R_{1}$and $\displaystyle R_{2}$ be equivalence relations on a set A.

a) Is $\displaystyle R_{1}\cup R_{2}$ an equivalance relation on A? Prove or disprove.

First we need to test the Reflexive property: Let $\displaystyle a_{1}\in$ A such that $\displaystyle (a_{1},a_{1})\in R_{1}$ and $\displaystyle a_{2}\in A $ such that $\displaystyle (a_{2},a_{2})\in R_{2}$. Since $\displaystyle (a_{1},a_{1})\in R_{1}$, then $\displaystyle (a_{1},a_{1})\in R_{1}\cup R_{2}$. Similarly, $\displaystyle (a_{2},a_{2})\in R_{2}$, then $\displaystyle (a_{2},a_{2})\in R_{1}\cup R_{2}$. This proves that $\displaystyle R_{1}\cup R_{2}$ is Reflexive.

Second we need to test the Symmetric property: Let $\displaystyle a_{1},b_{1}\in$ A such that $\displaystyle (a_{1},b_{1}),(b_{1},a_{1})\in R_{1}$ and $\displaystyle a_{2},b_{2}\in$ A such that $\displaystyle (a_{2},b_{2}),(b_{2},a_{2})\in R_{2}$. Clearly if $\displaystyle (a_{1},b_{1}),(b_{1},a_{1})\in R_{1}$ then $\displaystyle (a_{1},b_{1}),(b_{1},a_{1})\in R_{1}\cup R_{2}$. Also if $\displaystyle (a_{2},b_{2}),(b_{2},a_{2})\in R_{2}$ then $\displaystyle (a_{2},b_{2})$,$\displaystyle (b_{2},a_{2})\in R_{1}\cup R_{2}$. This proves that $\displaystyle R_{1}\cup R_{2}$ is Symmetric.

Third we need to test the Transitive property: Let $\displaystyle a_{1},b_{1},c_{1}\in$ A such that $\displaystyle (a_{1},b_{1}),(b_{1},c_{1})(a_{1},c_{1})\in R_{1}$ and $\displaystyle a_{2},b_{2},c_{2}\in$ A such that $\displaystyle (a_{2},b_{2}),(b_{2},c_{2})(a_{2},c_{2})\in R_{2}$. Clearly if $\displaystyle (a_{1},b_{1}),(b_{1},c_{1})(a_{1},c_{1})\in R_{1}$ then $\displaystyle (a_{1},b_{1}),(b_{1},c_{1})(a_{1},c_{1})\in R_{1}\cup R_{2}$. Similarly if $\displaystyle (a_{2},b_{2}),(b_{2},c_{2})(a_{2},c_{2})\in R_{2}$ then $\displaystyle (a_{2},b_{2}),(b_{2},c_{2})(a_{2},c_{2})\in R_{1}\cup R_{2}$. Therefore $\displaystyle R_{1}\cup R_{2}$ is Transitive. Since $\displaystyle R_{1}\cup R_{2}$ is Reflexive, Symmetric, and Transitive, it is an equivalence relation on A.

b) Is $\displaystyle R_{1}\cap R_{2}$ an equivalence relation on A? Prove or disprove. Suppose $\displaystyle a_{1}\in A$ such that $\displaystyle (a_{1},a_{1})\in R_{1}$ and $\displaystyle (a_{1},a_{1})\notin R_{2}$. Then clearly $\displaystyle (a_{1},a_{1})\notin R_{1}\cap R_{2}$. Since $\displaystyle R_{1}\cap R_{2}$ is not Reflexive, it is not an equivalence relation on A.