Function $f\colon 2^{\mathbb{N}}\to 2^{\mathbb{N}}$ preserving intersections and mapping sets to sets which differs only by finite number of elements Define on $2^{\mathbb{N}}$ equivalence relation 
$$
X\sim Y\Leftrightarrow \text{Card}((X\setminus Y)\cup(Y\setminus X))<\aleph_0
$$
Is there exist a function $f\colon 2^{\mathbb{N}}\to 2^{\mathbb{N}}$ such that
$$
f(X)\sim X
$$
$$
X\sim Y \Rightarrow f(X)=f(Y)
$$
$$
f(X\cap Y)=f(X)\cap f(Y)
$$
 A: Let $\{X_\alpha : \alpha\in\mathcal{A}\}\subset\mathbb{N}$ be an uncountable family of sets such that 
$$
\alpha,\beta\in\mathcal{A},\quad\alpha\neq\beta\Rightarrow \text{Card}(X_\alpha\cap X_\beta)<\aleph_0
$$
Such a family does exist. Indeed for each irrational number $x\in\mathbb{I}$ consider sequence of rational numbers $\{x_n\}_{n=1}^{\infty}\subset\mathbb{Q}$ tending to $x$. Let $\varphi(x)=\{x_n:n\in\mathbb{N}\}$ be the set of this rational numbers. Obviously for $x,y\in\mathbb{I}$ such that $x\neq y$ we have $\text{Card}(\varphi(x)\cap\varphi(y))<\aleph_0$. Also obviously for all $x\in\mathbb{I}$ we have $\text{Card}(\varphi(x))=\aleph_0$. Let $i\colon 2^\mathbb{Q}\to 2^\mathbb{N}$ be some bijection between $2^\mathbb{Q}$ and $2^\mathbb{N}$ then we may take by definition $\{X_\alpha : \alpha\in\mathcal{A}\}=\{i(\varphi(x)):x\in\mathbb{I}\}$. This will be desired family.
Let $\alpha,\beta\in\mathcal{A},\alpha\neq\beta$. Then $X_\alpha\cap X_\beta\sim\varnothing$. And from the second and third properties we obtain $f(X_\alpha)\cap f(X_\beta)=f(X_\alpha\cap X_\beta)=f(\varnothing)$.
Now for each $\alpha\in\mathcal{A}$ consider $Y_\alpha=f(X_\alpha)\setminus f(\varnothing)$. By construction $X_\alpha$ is infinite, so does $f(X_\alpha)$, and as the consequence $Y_\alpha\neq\varnothing$. Now for all $\alpha,\beta\in\mathcal{A},\alpha\neq\beta$ we have 
$$
Y_\alpha\cap Y_\beta=f(X_\alpha\cap X_\beta)\setminus f(\varnothing)=\varnothing
$$
Thus we built an uncountable family of disjoint subsets $\{Y_\alpha : \alpha\in\mathcal{A}\}$ in countable set $\mathbb{N}$. Contradiction, hence such a function doesn't exist.
