Let $\{A_i\}$ be a partition of $A$, where each $A_i$ is a set. Define $a \sim b$ if and only if $a$ and $b$ lie in the same $A_i$. Show that this is an equivalence relation.
Define $f(\{ A_i\}) = \ \sim$. We will find an inverse function.
Let $ \# $ be an equivalence relation on $A$. I claim produces a partition on $A$. For $a \in A$, let $[a] = \{ b \in A : a \# b\}$. I claim that for any $[a]$ and $[b]$, $[a]$ and $[b]$ are either disjoint or equal.
For the proof, suppose that $a \# b$ are related. Then, if $x \in [a]$, then $ x \# a$, and transitivity and reflexivity implies $x \# b$,, hence $x \in [b]$. Onthe other hand, if $y \in [b]$, then $ y \# b$, and transitivity and reflexivity implies $y \# a$,, hence $y \in [a]$. Together, this proves that $[a] = [b]$.
If $[a]$ and $[b]$ are not related, then they are disjoint. For if $c \in [a]$ and $c \in [b]$,then $c \# a$ and $c \# b$, which by transitivity and reflexivity implies that $a \# b$, which is a contradiction. Hence, $[a]$ is disjoint from $[b]$.
Now, letting $\{[a]\}$ be the set of $[a]$,we observe this is a partition of $A$, because it is a disjoint collection, and every element $e$ of $A$, is in it's own equivalence class $[e]$. Define $g(\#) = \{[a]\}$
That is, every equivalence relation gives a partition and every partition gives an equivalence relation. I leave you to prove that $f$ and $g$ are inverses of each other. This establishes a bijection between the set of partitions and equivalence relations.
Best Answer
The equivalence relation would be "belongs to the same set of the partition". Nobody said that it had to be expressed without referring to the partition.
(And where is $9$ in your partition?)