Prove that $[a]_X \subseteq [a]_Y$ . Let $X,Y$  be two equivalence relations  defined on a set $A.$ Show that $X \cup Y$ be the equivalence relation  iff $[a]_X \subseteq [a]_Y$ (or $[a]_Y \subseteq [a]_X$). Here $[a]_{X(Y)}$ denotes the eqiuvalence class of elemenent $a \in A.$ My idea was consider an element $b \in [a]_X \setminus [a]_Y]$ and verify   the intersection:
$$
[a]_{X \, \cup \, Y} \cap [b]_{X \, \cup \, Y}.
$$
Since $[a]_{X \, \cup \, Y}=[a]_{X } \cup [a]_{ Y}$  we  get 
\begin{gather*}
[a]_{X \, \cup \, Y} \cap [b]_{X \, \cup \, Y}=([a]_{X } \cup [a]_{ Y}) \cap ([b]_{X} \cup [b]_{ Y})=\\ \\=([a]_{X } \cap [b]_{ X}) \cup ([a]_{X } \cap [b]_{ Y}) \cup ([a]_{Y } \cap [b]_{ X}) \cup ([a]_{Y } \cap [b]_{ Y}).
\end{gather*}
Since $b \in [a]_X$ then $[b]_X=[a]_X$ and  $[a]_{X } \cap [b]_{ X}=[a]_X$. Thus 
$[a]_{X \, \cup \, Y} \cap [b]_{X \, \cup \, Y}$  is  not empty set. It follows  that
$[a]_{X \, \cup \, Y} = [b]_{X \, \cup \, Y}.$
Is  this enougth?
 A: I’m afraid that I can’t follow your reasoning: the fact that $[a]_{X\cup Y}\cap[b]_{X\cup Y}\ne\varnothing$ does not imply that $[a]_{X\cup Y}=[b]_{X\cup Y}$. Moreover, the statement that you’re to prove is an iff statement, so you have two implications to prove, not just one: 


*

*if $[a]_X\subseteq[a]_Y$ for each $a\in A$, then $X\cup Y$ is an equivalence relation; and  

*if $X\cup Y$ is an equivalence relation, then either $[a]_X\subseteq[a]_Y$ for each $a\in A$, or $[a]_Y\subseteq[a]_X$ for each $a\in A$.


I would start by observing that $X\cup Y$ is reflexive: if $a\in A$, then $\langle a,a\rangle\in X\subseteq X\cup Y$. Moreover, $X\cup Y$ is symmetric: if $\langle a,b\rangle\in X\cup Y$, then either $\langle a,b\rangle\in X$ or $\langle a,b\rangle\in Y$. In the first case $\langle b,a\rangle\in X\subseteq X\cup Y$, and in the second case $\langle b,a\rangle\in Y\subseteq X\cup Y$, so $\langle b,a\rangle\in X\cup Y$. Thus, $X\cup Y$ is an equivalence relation iff it’s transitive. I’ll get you started on both parts; see if you can finish them.


*

*Suppose that $[a]_X\subseteq[a]_Y$ for each $a\in A$. We want to show that $X\cup Y$ is transitive, so suppose that $\langle a,b\rangle,\langle b,c\rangle\in X\cup Y$. Then $b\in[a]_X\cup[a]_Y=[a]_Y$, and ... 

*Now suppose that there is some $a\in A$ such that $[a]_X\nsubseteq[a]_Y$ and $[a]_Y\nsubseteq[a]_X$; we want to show that $X\cup Y$ is not transitive. Let $b\in[a]_X\setminus[a]_Y$ and $c\in[a]_Y\setminus[a]_X$. Then $\langle b,a\rangle\in X\cup Y$ and $\langle a,c\rangle\in X\cup Y$; why? Now show that $\langle b,c\rangle\notin X\cup Y$.
