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.
The proof is really straightforward and one of those 'let use make sure you understand the definition' kind of exercises. I'll get you started and then you'll pick it up from there:
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.