The set of all equivalence classes of an equivalence relation on a uncountable set may be countable or uncountable. 
The set of all equivalence classes of an equivalence relation on a
   an uncountable set may be countable or uncountable.

Can you please give examples of each. $x$~$y$:x is connected via a path. The equivalence class is entire $\mathbb R$ is an equivalence class. right? I couldn't find an example of The set of all equivalence classes of an equivalence relation on an uncountable set be uncountable. Please help me. I have found one example. I don't know, revising the question with an attempt is legal or not. sorry for an incorrect attempt.
 A: On any set you have two very simple equivalence relations (smallest and larges, if you look at ordering by inclusion). 
If each element is equivalent to each other, you have single equivalence class $\{X\}$. (This corresponds to equivalence relation $R=X\times X$, i.e. $x\sim y$ for every $x,y\in X$.)
If each element is equivalent only to itself, the equivalence classes are singletons. So the partition is $\{\{x\}; x\in X\}$ and it has the same cardinality as $X$. (In the case $X=\mathbb R$ it is uncountable.) This is the relation $R=\{(x,y); x=y\}$, i.e. $x\sim y$ iff $x=y$.
If you want an equivalence relation on $\mathbb R$ such that there are countably many equivalence classes, simply take any partition of $\mathbb R$ into countably many subsets and the corresponding equivalence relation. For example, you could take the partition $\mathcal P=\{[x,x+1); x\in\mathbb Z\}$.
A: On $\mathbb R$ if a number is equivalent only to itself, there is one equivalence class for each number $x\in R$. 
If all numbers are equivalent to each other, there is one equivalence class, and one is countable. 
Or say two numbers are equivalent iff they have the same floor (ie for each $x\in \mathbb R$ there is an integer $n_x$ with $n_x\le x \le n_x+1$, set $x\equiv y$ iff $n_x=n_y$). That gives a countably infinite set of equivalent classes.

For a different uncountable example, make two real numbers equivalent iff they differ by a rational number. That requires a little more thought, and partitions the reals as an uncountable collection of countably infinite sets.
A: *

*On $\mathbb R$, define the equivalence relation $x \sim y$ iff $x-y \in \mathbb Z$. The set of equivalence class representative here is an uncountable set e.g. $[0,1)$.

*On $\mathbb R - \{0\}$, define the equivalence relation $x \sim y$ iff $xy > 0$. This partitions into two equivalence classes, positive and negative reals.
A: For a countable set of equivalence classes of an equivalence relation $R$ on an uncountable set $A$, how about using  $R= \{ \langle x,y\rangle| x \in A \land y \in A \}$
In other words, every element from the uncountable set stands in the $R$ relation to every element from the uncountable set.
That should give you exactly one equivalence class, namely $A$ itself, and so the set of equivalence classes in this case contains $1$ element ... making it a countable set.
A: relation $x$ ~$y$ iff $x=y$ 
$\{\overline{x}=\{y| y=x \}\}$ is uncountable (The singletons of R).
