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

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 ${\cal P}(\omega)/fin$?

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 $$f_A(n) but $$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.