Consider the following equivalence relation on $\{0,1\}^\omega$:

$x\simeq y$ iff there is $n\in\omega$ such that $x(k)=y(k)$ for all $k\in\omega$ with $k\geq n$.

It is easy to see that the following is a well-defined ordering relation on $\{0,1\}^\omega/\simeq$ :

$[x]_\simeq \leq [y]_\simeq$ iff there is $n\in \omega$ such that $x(k)\leq y(k)$ for all $k\in\omega$ with $k\geq n$.

Let us order $\{0,1\}^\omega$ componentwise. Is there an order-preserving surjective function $f:(\{0,1\}^\omega/\simeq)\to \{0,1\}^\omega$?


Yes, in fact, there is a surjective lattice-homomorphism. Your two posets are better known as the Boolean algebras $P(\omega)/fin$ and $P(\omega)$ ($fin$ being the ideal of finite sets). Let $\omega=\bigcup A_n$ be a partition of $\omega$ into infinitely many infinite sets and let $U_n$ be a nonprincipal ultrafilter containing $A_n$ for each $n$. Define $f:P(\omega)/fin\to P(\omega)$ by $f(A)=\{n:A\in U_n\}$ (under Stone duality, this corresponds to the map $\beta\omega\to\beta\omega\setminus\omega$ that extends the map on $\omega$ sending $n$ to $U_n$). This is a homomorphism of Boolean algebras, and furthermore it is surjective: for any $B\subseteq \omega$, $f\left(\bigcup_{n\in B} A_n\right)=B$.

If you don't care about having a lattice-homomorphism, you don't need to use ultrafilters: just let $U_n$ be principal filter generated by $A_n$, and define $f$ as before. Then $f$ will still be surjective and order-preserving.


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