Let $\mathbb{N}$ be the set of all natural numbers and let $P(\mathbb{N})$ be the power set of $\mathbb{N}$. On the set $P(\mathbb{N})$ we define a relation $\sim$ by the formula: $$X\sim Y \iff |X\setminus Y|<\infty \wedge |Y\setminus X|<\infty.$$ It is easy to check that $\sim$ is an equivalence relation on $P(\mathbb{N})$.
My question (problem) is to give a brief description of the set of all equivalence classes of $\sim$. Precisely, I want to find a family of sets $\{A_i\in P(\mathbb{N}): i\in T\}$ such that for every $B\in P(\mathbb{N})$ there exists exactly one $i_0\in T$ such that $B\sim A_{t_0}$.
Well you can't.
I mean, you can, if you assume the axiom of choice. If you have that in your hand, then it's easy to just choose one element from every equivalence class. But that won't give you an explicit family like that. And you can't write in explicit details such family either.
The reason that you can't is that this set of representatives would amount of a non-measurable set. It is consistent that the axiom of choice fails, and all sets are measurable. So in such model where the axiom of choice fails, you cannot have a set of representatives.
If we could have defined such family explicitly, it would have existed in every model of set theory, including the one mentioned above. Since that's not happening (without set theory being inconsistent), it can't be done.