I have the following question :
$A \subseteq P(\mathbb{N} \times \mathbb{N}) $ group of equivalence relation above $\mathbb{N}$, let $E \subseteq A^2$ a relation that is defined in the following manner :
$(R_1,R_2)\in E$ if exists function $f : \mathbb{N} \rightarrow \mathbb{N}$ surjective and injective such all $x,y \in \mathbb{N}$, $(x,y)\in R_1$ if and only if $(f(x),f(y))\in R_2$
$E$ is a equivalence relation, let $R \in A$ such that $|\mathbb{N}/R|=2$, Proof if $R$ has finite equivalence class so $[R]_E$ is countable.
What I did :
I don't really know how to approach this, This is what I managed to concluded:
Since $|\mathbb{N}/R|=2$ so there is $[C]_R,[D]_R \in$ $\mathbb{N}$ $/R$ such that $N/R=\{[C]_R,[D]_R\}$ its clear that $[C]_R \neq [D]_R$ and as well that $(C,D)\not \in R$, but I don't understand how the equivalence class of $\mathbb{N}/R$ is connected with $R/E=[R]_E$?
Any ideas/suggestions?
Any help will be appreciated.
Let $R$ has a finite equivalence class. This means there exists a finite subset $F$ of $\Bbb N$ such that for each $x,y\in\Bbb N$ hold $(x,y)\in R$ iff both $x,y$ in $F$ or both $x,y$ not in $F$. Let $(R,R’)\in E$. Then there exists a bijection $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $x,y \in \mathbb{N}$, $(x,y)\in R$ iff $(f(x),f(y))\in R’$. Thus $(f(x),f(y))\in R’$ iff both $f(x),f(y)$ in $f(F)$ or both $f(x),f(y)$ not in $f(F)$. Therefore the relation $R’$ is completely determined by image of $f(F)$. Since the set $F$ is finite, there are countably many different images $f(F)$, so there are at most countably many different relations $R’$ such that $(R,R’)\in E$.