Why do we have $E_0(\Bbb N^\Bbb N)\sim_c E_0$?

Question

Let $E_0$ and $E_0(\Bbb N^\Bbb N)$ be the relations of eventual agreement on $2^\Bbb N$ and $\Bbb N^\Bbb N$, so $$xE_0y\iff\exists m\forall n\geq m x(n)=y(n)$$ and similarly for $E_0(\Bbb N^\Bbb N)$. Clearly $2^\Bbb N\hookrightarrow\Bbb N^\Bbb N$ is a continuous reduction of $E_0$ to $E_0(\Bbb N^\Bbb N)$. Why do we have a continuous reduction in the opposite direction as well?

Answer 1

Here's one way to do it:

Let $(A_i)_{i\in\mathbb{N}}$ be a partition of $\mathbb{N}$ into infinitely many infinite pieces, and let $a_{i,j}$ be the $j$th smallest element of $A_i$. Given $f\in\mathbb{N}^\mathbb{N}$, we'll build a set $S_f\in 2^\mathbb{N}$ by coding $f(i)$ into $S_f\upharpoonright A_i$: $$S_f(\langle i,j\rangle)=\begin{cases} 1 & \mbox{ if $f(i)<j$,}\\ 0 & \mbox{ otherwise.}\\ \end{cases}$$

Here $\langle\cdot,\cdot\rangle$ is your favorite pairing function. The infinite binary sequence $S_f$ is built by interweaving infinitely many infinite binary sequences $S(\langle i, -\rangle)$, each of which is eventually zero; this latter point guarantees $(S_f,S_g)\in E_0$ whenever $(f,g)\in E_0(\mathbb{N}^\mathbb{N})$, while the other direction follows since for all $m$ there is some $n$ such that for all $k>n$ we have $A_k\cap\{0,1,...,m\}=\emptyset$.