# Quotient set univocally defined

I know that what I am going to ask is pretty basic, borderline stupid, nevertheless it is bugging me. By definition I know that given a set $A$ and a equivalence relation $\rho$, then the items $[\alpha]_{\rho}$ and $A/\rho$ are defined as follows:

$[\alpha]_{\rho} = \{x \in A : \alpha\;\rho\; x\} \\A/\rho = \{[\alpha]_{\rho}:\alpha \in A\}$

let's consider the relation $\sim$ so defined in $J=\{0,1,2,3,4,5,6,7,8,9\}$:

\begin{align} a\sim b \Leftrightarrow a^2-1 \equiv b^2-1 \mod 3\end{align}

then given that if I had to determine $[5]_{\sim}$ and $[9]_{\sim}$, I would write

$[5]_{\sim} = \{1,2,4,5,7,8\} \\ [9]_{\sim} = \{0,3,9\}$

is it correct stating:

$J/\sim=\{[0], [1]\} = \{[2],[3]\}=\{[6],[8]\} =\; ...$

and so basically $J/\sim$ isn't univocally defined?

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$J/\sim = \{[0], [1]\}$ is unambigiously defined. It doens't change the definition of $J/\sim$ at all, if you write its elements in another form. Note, that $[0]$ and $[3]$ are the same set, namely $$[0] = \{x \mid x \sim 0\} = \{0,3,9\} = \{x \mid x \sim 3\} = [3]$$ an the same for $[1]$ and $[2]$. So the sets $\{[0], [1]\}$ and $\{[2], [3]\}$ have the very same elements and hence are equal, they both equal moreover the set $$J/\sim = \bigl\{\{0,3,9\}, \{1,2,4,5,6,7,8\}\bigr\}$$ To give another example note that $\{\frac12\} = \{0.5\}$ also its elements are written in another form. But this doesn't change the elements, and hence not the set.