Let $R$ be an equivalence relation on set $A$. Prove that for any $x,y \in A$ either $[x] = [y]$ or $[x] \cap [y] = \varnothing$. For any $x$ and any $y$
$(x,x) \in R$ and $(y,y) \in R$ (since $R$ is reflexive)
So there exist $[x]$ and $[y]$ equivalence classes for $R$.
When $[x]$ and $[y]$ are equivalence classes of $R$ then,
there are possible two status which is either $[x] = [y]$ or $[x] \neq [y]$,
Assume if $[x] \neq [y]$ then $[x] \cap [y] \neq \varnothing$.
Let $z$ be an arbitrary element for $[x] \cap [y]$.
then $z \in [x] \cap [y] \implies z \in [x]$ and $z \in [y]$
but $[x] \neq [y]$
Therefore, This is contradiction!
Therefore, if $[x] \neq [y]$ then $[x] \cap [y] =  \varnothing$.
if either $[x] = [y]$ or $[x] \neq [y]$,
$\implies$ either $[x] = [y]$ or $[x] \cap [y] =  \varnothing$. (since if $[x] \neq [y]$ then $[x] \cap [y] =  \varnothing$)
Is this correct? Is there a fault? Is there another way to prove it?
 A: There are only two possibilities: either $[x]\cap[y]=\emptyset$ or $[x]\cap[y]\neq\emptyset$. Now $[x]\cap[y]\neq\emptyset$ means that there exists an element $z\in[x]\cap[y]$.
If $z\in[x]\cap[y]$, then $xRz$ and $yRz$, which implies by transitivity and symmetry ($yRz\implies zRy$) that $xRy$. In that case you can show $[x]=[y]$. This is because for any $a\in A$ with $aRx$,  $aRx$ and $xRy$ implies by transitivity that $aRy$, and hence $[x]\subseteq[y]$. The reverse inclusion $[y]\subseteq[x]$ follows similalry.
A: Suppose that $[x] \cap [y] \neq \emptyset$ and $[x] \neq [y]$.
Since the intersection is nonempty, we may pick $z \in [x] \cap [y]$. Since the classes are not equal, we may pick an element $w \in R$ which is in one class but not in the other. WLOG, assume that $w \in [x]\setminus[y]$.
Since $w, z \in [x],$ we have that $x \sim w$ and $x \sim z$. Using symmetry and transitivity, we get that $z \sim w$. However, we also have $y \sim z$. Transitivity gives $y \sim w$ which is a contradiction since $w \notin [y]$.
A: Let $x,y\in A$.
Either $x R y$ of $x \not R y$.
Case 1:  $x R y$.
$[x]=\{z\in A| z R x\}$, $[y]=\{z\in A|z R y\}$.
If $m\in [x]$ then $m Rx$ and as $R$ is transitive and $xR y$ then $m Ry$ so $m \in [y]$ so $[x]\subset [y]$ but if $n \in[y]$ then $n Ry$ but as $R$ is symmetric and $xRy$ the $yR x$ and as $R$ is transitive $nRx$ so $n\in [x]$ and $[y]\subset [x]$ so $[x]=[y]$.
Case 2:  $x \not R y$.
If $n \in [x]\cap [y]$ then $n Rx$ and $nR y$.  As $R$ is symmetric than $xRn$ and $nRy$.  As $R$ is transitive $x R y$ which is a contradiction.  So $[x]\cap [y] = \emptyset$.
So $x Ry \implies [x]=[y]$
And $x \not R y \implies [x]\cap [y] = \emptyset$.
Those are the only two options.
A: It suffices to prove that, for all $x,y\in A$, if $[x]\cap[y]\neq \varnothing$, then $[x]=[y]$.
Suppose $x,y$ are arbitrary in $A$ and let $z\in [x]\cap[y]$. Then $zRx$ and $zRy$ by definition. The former implies $xRz$, which, together with the latter, gives $xRy$. So $[x]\subseteq [y]$. But by symmetry, $[y]\subseteq [x]$. Hence $[x]=[y]$.
