Relationship of ${\cal P}(\omega)/fin$ and ${\cal L}$

Question

Define ${\cal L}$ as in this question: the set of functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(n)\leq f(n+1)\leq f(n)+1$, where two functions are considered equal if they differ at at most finitely many points, and $f\prec g$ is defined to mean $f(n)\leq g(n)$ for all but at most finitely many exceptions $n$.

Let ${\cal P}(\omega)/fin$ be defined as in this post. In another post, it was established that there is some kind of "natural" (not in the categorical sense) lattice embedding from ${\cal P}(\omega)/fin$ into ${\cal L}$.

My question is: Can the relationship between ${\cal P}(\omega)/fin$ and ${\cal L}$ be made more precise? Isn't ${\cal L}$ essentially ${\cal P}(\omega)/fin$ "stacked on top of itself $\omega$ times"? Or is ${\cal L}$ (a quotient of) a product of ${\cal P}(\omega)/fin$?

Answer 1

Look at your and my second answer to the original question. Take your map from $\mathcal{P}(\omega)$ into $\mathcal{L}$ (or rather the set of functions before identifying almost equal elements). That is for $A\subseteq\omega$ define $f_A$ as in your answer: $f_A(n)=|n\cap A|$. What I showed is this: if $A\subset^*B$ then $f_B(n)-f_A(n)$ diverges to infinity. What you can also show is: if $A=^*B$ then $f_A-f_B$ is constant on a tail. Indeed, if $A\setminus m=B\setminus m$ then for $n\ge m$ we have $f_A(n)-f_B(n)=f_A(m)-f_B(m)$. On the other hand, if $A\neq^*B$ then $f_A-f_B$ is not constant on a tail: if $n\in A\setminus B$ then $f_A(n)<f_A(n+1)$ but $f_B(n)=f_B(n+1)$ and vice versa.

This shows that $A\mapsto f_A$ induces an injective map from $\mathcal{P}(\omega)/\mathit{fin}$ into $\mathcal{L}/{\equiv}$ where $f\equiv g$ means that $f-g$ is constant on a tail. The map is also onto: given $f\in\mathcal{L}$ look at $g$ defined by $g(n)=f(n)-f(0)$; then $g=f_A$, where $A=\{n:f(n+1)=f(n)+1\}$.

As shown in the other answer the map is order-preserving in the sense that $A\subset^*B$ implies that $f_B(n)-f_A(n)$ diverges to infinity.

On the other hand disjoint sets can map to comparable functions: let $A$ consist of the even numbers and let $B=\{4n+1:n\in\omega\}$. Then $f_A$ grows like $n/2$ and $f_B$ grows like $n/4$ so that $f_A(n)-f_B(n)$ will diverge to infinity.