What is the equivalence relation on the Set $A=\{1,2,3,4,5\}$ I have a set $A=\{1,2,3,4,5\}$, where its partition set is $E=\{\{1,5\}, \{2,3,4\}\}$
I am not sure what kind of equivalence relation gives a rise to above partition. 
 A: Whenever you have a partition $P$ on a set $S$, you can define the binary relation $\sim$ by $x\sim y$ if and only if $x$ and $y$ belong to the same element of $P$. Then:


*

*$\sim$ is an equivalence relation;

*the partition that $\sim$ induces is $P$.

A: $x \sim y $ if $x\in \{1,5\} \wedge y\in \{1,5\}$ and $x \sim y $ if $x\in \{2,3,4\} \wedge y\in \{2,3,4\}$ will work. It doesn't have to be a nice natural relation.
A: Simply define $x\sim y$ iff $x$ and $y$ lie in the same element of the partition. 
A: The equivalence relation on A, is given, explicitly, by $$R = \{(1, 1), (5, 5), (1,5), (5, 1), (2, 2), (2, 3),(2, 4), (3, 2), (3, 3), (3, 4),  (4, 2), (4, 3), (4, 4),\}.$$  You'll see that the sets comprising the equivalence classes ($E$) of the partition are indeed $\;\{1, 5\}, \;\;\{2, 3, 4\}$
The ordered pairs $(a, b) \in R$ indicate $a\,R\,b$, that is "$a$ is related to $b$ under $R$"  Here we have that the elements in each equivalence class are mutually related in $R$, but no element from $\{1, 5\}$ is related to any element in $\{2, 3, 4\}$, and vice versa.
The relation $R$ on $A$ is given by 


*

*reflexive: for every element $a \in A,\; (a, a) \in R$,

*symmetric: For every element $a, b$ in $A,\;$ if $(a, b) \in R$ then $(b,a) \in R$

*transitive: For every element $a, b, c \in A,\;$ if $(a, b)\in R$ and $(b, c)\in R,$ then $\;(a, c) \in R$.
It might be a good idea for you to confirm the properties above, just to be assured that the relation is indeed an equivalence relation. 
