How big is the lattice of all functions?

Question

Define the lattice $(\mathcal{L},\prec)$ as the set of all function $f:\mathbb{N}\rightarrow\mathbb{N}$ satisfying $f(n)\leq f(n+1)\leq f(n)+1$, where two functions are considered equal if they differ at finitely many points, and $f\prec g$ is defined as $f(n)\leq g(n)$ with finitely many exceptions.

This lattice is large, e.g. you can construct chains and antichains of size continuum, chains of antichains of the same cardinality and so on. Although these constructions are structurally simple, defining them in a precise way gets quite complicated. Therefore I am interested into references dealing with this lattice or sublattices, making the intuitive "largeness" precise.

Background: Y. Barnea and I constructed a set of groups parametrized by elements of $\mathcal{L}$, and want a way to convince group theorists that we really have "many" groups, and don't want to waste a lot of space on constructions which yield results far inferior to what every lattice theorist would immediately see.

Edit: As every function in $\mathcal{L}$ is equivalent to either a function $f$ satisfying $f(0)=0$ or a function $f(n)=n+c$, the interesting part is the sublattice of functions satisfying $f(0)=0$. Also I always assume AC and would not mind too much about assuming CH.

Answer 1

One partial answer is the following. The large lattice ${\cal P}(\omega)/fin$ (which is defined here) can be embedded into ${\cal L}$ as follows:

For $A\subseteq \omega$ define $f_A:\omega \to \omega$ by $$f_A(n) = |\{0,\ldots,n\}\cap A|$$ for all $n\in \omega$. It is not hard to verify that the map $\varphi:{\cal P}(\omega)/fin \to {\cal L}$ sending $[A]_{fin}$ to $[f_A]$ is well-defined and injective. (It is also a lattice homomorphism.)

Could it be that ${\cal L}$ is some kind of power of ${\cal P}(\omega)/fin$? I'll ask that in a separate question.

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EDIT: Andreas Blass pointed out an error in the above construction, but it seems like Jan-Christoph Schlage-Puchta was able to fix the error -- see comments below.

Answer 2

Here are a few tricks to play with. For every $A\subseteq\omega$ define $f_A$ by $f_A(0)=0$, $f_A(n+1)=f_A(n)+1$ if $n\in A$, and $f_A(n+1)=f_A(n)$ if $n\notin A$. If $A\subset^*B$ (so $B\setminus A$ is infinite) then $f_A\prec f_B$: fix $m$ such that $A\setminus B\subseteq m$; in fact, $f_B(n)-f_A(n)$ will diverge to infinity. Now take a tower $\langle T_\alpha:\alpha<\mathfrak{t}\rangle$ such that $T_{\alpha+1}\setminus T_\alpha$ is always infinite. Then $\langle f_{T_\alpha}:\alpha<\mathfrak{t}\rangle$ is a $\mathfrak{t}$-chain in $\mathcal{L}$. One can also create a copy of the unit interval in $\mathcal{L}$ in this way.

If $\mathcal{A}$ is an almost disjoint family then we can modify it as follows: for every $A$ let $X_A=\bigcup_{n\in\omega}[3^n,3^{n+1})$. Then the set $\{f_{X_A}:A\in\mathcal{A}\}$ is an antichain in $\mathcal{L}$. To see this consider two infinite sets $A$ and $B$ with finite intersection, say $A\cap B\subseteq m$, and write $g_A=f_{X_A}$ and $g_B=g_{X_B}$. For every $n$ we have $g_A(3^n), g_B(3^n)\le \sum_{k<n}3^k=\frac12(3^n-1)$. Now, take $l\ge m$ and let $n\le l$ be the first member of $(A\cup B)\setminus l$; now if $n\in A$ then $g_A(3^{n+1})=g_A(3^n)+2\cdot 3^n$ and $g_B(3^{n+1})=g_B(3^n)$, so that $g_A(3^{n+1})>g_B(3^{n+1})$. If $n\in B$ then this reverses. It follows that $g_A$ and $g_B$ are $\prec$-incomparable.