I have to check if sum and intersection of family of equivalence relations is equivalence relation.
Here is the exercise:
Let $\mathcal{R}$ be a family of equivalence relations defined on some set $X$. Check if $\bigcup \mathcal{R}$ and $\bigcap \mathcal{R}$ are equivalence relations on $X$.
I am doing this exercise and, so far, I have proved $\bigcup \mathcal{R}$ is reflexive and $\bigcap \mathcal{R}$ is reflexive and symmetric but I have no idea how to prove $\bigcup \mathcal{R}$ is symmetric.
Could you please give me some hints how to finish the proof (or how to build counter - example)?
The intersection of a family of equivalence relations will be an equivalence relation. The proofs are the "follow your nose" variety - there is no real problem solving, you just have to write them out.
For unions, it is true that the union of a family of symmetric relations will be symmetric. The key point is that if $(a,b) \in \bigcup \mathcal{R}$, to use the notation from the question, then $(a,b) \in R$ for some $R \in \mathcal{R}$. Thus $(b,a) \in R$, because $R$ is symmetric, so $(b,a) \in \bigcup \mathcal{R}$ as well. This is the key point of working with this kind of union: the only way to be in the union of a family is to be in one of the sets of the family.
The union of a family of equivalence relations does not need to be transitive, though. You can find a counterexample with appropriate equivalence relations on a 3-element set $\{a,b,c\}$.