# Prove that the union of relations is an equivalence relation

Let $\{\alpha_i \mid i\in \mathbb N\}$ is family of equivalence relations on the set $A$ such that for every $i \in N$ $\alpha_i\subseteq\alpha_{i+1}$. Prove that the union of all $\alpha_i$ is equivalence relation on $A$.

Whenever there are problems involving family of relations I'm clueless. I know that we obviously need to prove reflexivity,symmetry and transitivity but other than that I can't even begin.

• I guess you forgot to mention that all the $\alpha_i$ are equivalence relations. (Otherwise there are trivial counterexamples.) – Stefan Mesken Nov 16 '17 at 17:25
• I also thought that they are supposed to be. But nowhere in the question it is stated that they are also equivalence relations.That's why I was mostly clueless about how to begin(It's an old exam question). – DreaDk Nov 16 '17 at 17:30
• Well, that must be a mistake. Before looking at my hint: Can you proof this assuming that all the $\alpha_i$ are equivalence relations? – Stefan Mesken Nov 16 '17 at 17:31
• I also fixed the formatting of your post. For the future, here is a MathJax tutorial. – Stefan Mesken Nov 16 '17 at 17:33
• I would perhaps be able to prove some of it. But the whole idea of family of sets/relations is a bit confusing to me (mainly cause I haven't encountered it much) – DreaDk Nov 16 '17 at 17:33

Let's proof that $\alpha := \bigcup \{ \alpha_i \mid i \in \mathbb N \}$ is transitive - reflexivity and symmetry are even easier:
Let $x,y,z \in A$ such that $(x,y) \in \alpha$ and $(y,z) \in \alpha$. Then there are, by the definition of $\alpha$, $i,j \in \mathbb N$ such that $(x,y) \in \alpha_{i}$ and $(y,z) \in \alpha_j$. Let $k = \max\{i,j\}$. Then $\alpha_i, \alpha_j \subseteq \alpha_k \subseteq \alpha$ and thus $(x,y),(y,z) \in \alpha_k$.
Now we use the fact that $\alpha_k$ is an equivalence relation to conclue that $(x,z) \in \alpha_k$. Since $\alpha_k \subseteq \alpha$, it follows that $(x,z) \in \alpha$ and thus that $\alpha$ is transitive.