How can I prove that there is a bijective function? Let $A$ be a nonempty set. Prove that there is a bijective function
$$ F \colon \{ \text{Equivalence relations on } A\} \rightarrow \{\text{Partitions of }A\}. $$
I am completely lost on where to proceed with this question. 
 A: Given an equivalence relation $\sim$ on $A$, the equivalence class of an element $a \in A$ is defined by $[a] := \{ b \in A \, | a \, \sim \, b \}$. Check using the definition that given $a,b \in A$ we have either $[a] = [b]$ or $[a] \cap [b] = \emptyset$ and thus $A = \bigcup_{a \in A} [a]$ is a partition of $A$ into disjoint non-empty subsets.
In the other direction, given a partition $A = \bigcup A_i$, show that you can define an equivalence relation whose equivalence classes are precisely the subsets $A_i$. You will need to check that the binary relation you define satisfies the properties of an equivalence relation.
A: An equivalence relation $R$ on $A$, is one which is reflexive, symmetric, and transitive.
$R\subseteq A\times A $ where


*

*$\forall a\in A:\big( (a,a)\in R\big)$, 

*$\forall a\in A,\forall b \in A : \big((a,b)\in R\to (b,a)\in R\big)$, 

*$\forall a\in A,\forall b\in A,\forall c\in A:\Big(\big( (a,b)\in R\wedge (b,c)\in R\big) \to (a,c)\in R\Big)$


A partition $P$ of $A$ is a set of non-empty subsets of $A$ where every element of $A$ occurs in one and only one of the subsets.
$P\subseteq \mathcal P(A)$ where $\forall a\in A \exists B\in P, \forall C\in P:\Big(x\in B\wedge \big(x\in C\to B=C\big)\Big)$
In short, show that you can map any partition $P$ of $A$, to an equivalence relation $R$, with a one to one correspondence.


 Hint 1: Consider the relation: "$x$ is in the same partition subset as $y$". That is: $(x,y)\in R \iff \exists B\in P \big(x\in B\wedge y\in B\big)$   Is this an equivalence relation?  



 Hint 2: Consider the definition of equivalence class $[x]= \{y: (x,y)\in R\}$   Do the sets of equivalence classes of the elements of $A$ form a partition?

