Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$. For $f,g\in\omega^\omega$ we say $f\simeq_{\text{fin}} g$ if there is $n\in \omega$ such that $f(k) = g(k)$ for all $k\geq n$.

We say that $A\subseteq \omega$ has measure 1 if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n+1\}|}{n+1} = 1.$$We define another equivalence relation on $\omega^\omega$ by setting $f\simeq_1 g$ if $$\{n\in\omega:f(n)=g(n)\} \text{ has measure } 1.$$

When I look at the posets $(\omega^\omega/\simeq_{\text{fin}}, \leq_{\text{fin}})$ and $(\omega^\omega/\simeq_{1}, \leq_{1})$ (where the ordering relations are defined "pointwise modulo the equivalence relation"), I get the feeling that they are not isomorphic and am quite certain this is correct. However a proof has eluded me. Any help is appreciated!

  • $\begingroup$ Measure 1 sets are not preserved by arbitrary permutations. Does this help? Gerhard "Still Working On 'Pointwise Modulo'" Paseman, 2016.02.13. $\endgroup$ – Gerhard Paseman Feb 13 '16 at 20:36
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    $\begingroup$ I find it likely that you can distinguish $P(\omega)/\text{Fin}$ and $P(\omega)/I_0$, where $I_0$ is the ideal of density zero sets, as algebras, using ideas from forcing or cardinal characteristics, and this would correspond to your question with $2^\omega$ in place of $\omega^\omega$. $\endgroup$ – Joel David Hamkins Feb 13 '16 at 20:43
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    $\begingroup$ I think perhaps Ilijas Farah's paper will help: math.yorku.ca/~ifarah/Ftp/n2003d08-s.pdf. $\endgroup$ – Joel David Hamkins Feb 13 '16 at 20:46

Very nice question!

They are not isomorphic.

What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, but when we take the quotient modulo finite, there is no such antichain. This is a property that distinguishes the isomorphism types.

To see this, I shall use the characteristic of the size of the smallest infinite maximal almost disjoint family of sets in two quotients. It turns out that these cardinals are different for the two quotients.

Namely, on the one hand, the size of the smallest infinite maximal almost-disjoint family modulo the finite sets is well-known to be uncountable. This is the cardinal characteristic known as $\frak a$.

On the other hand, when we take the quotient of $P(\omega)$ modulo the ideal of asymptotic density zero sets, usually denoted $Z_0$, then it turns out that there is a countably infinite maximal almost disjoint family. Simply let $A_0$ be half the numbers (the evens, say); let $A_1$ be half of what is left; let $A_2$ be half of what is left after that, and so on. These sets are disjoint, union to the whole space and $A_n$ has density $1/2^{n+1}$; so for any set $A$ with positive density, there is large enough $N$ such that it does not concentrate on the tail sets $\bigcup_{k\geq N}A_k$, since this union is too sparse, and so it must have $A\cap A_n$ with positive density for some $n<N$. (See additional detail in lemma 2.5 in Barnabás Farkas, Lajos Soukup, The zero density ideal, cardinal invariants and related forcing problems.)

If $f_n$ is the characteristic function of $A_n$, then it follows that $0<f_n<1$ for all $n$, modulo density, and the $f_n$ are incomparable and have meet $0$ and the constant-$1$ function is a minimal upper bound of the $f_n$, since if $h<f$ modulo density, then the set where it dropped would have positive density on some $A_n$, and in this case, we wouldn't have $f_n\leq h$ modulo density. So in the density quotient $\omega^\omega/Z_0$, we have the countable antichain above $0$ with a minimal upper bound.

But this situation cannot arise in the other quotient $\omega^\omega/\text{Fin}$. Namely, if we have $0<g_n<h$ modulo finite and $g_n\wedge g_m=0$ for all $n\neq m$, then the support of the $g_n$'s must form an almost-disjoint family. But since it is countable, there is another set almost disjoint from all these supports, and contained in the support of $h$. So we can reduce $h$ on that set to find a strictly smaller $h'<h$, which is still an upper bound of the $g_n$. So $h$ is not a minimal upper bound of the $g_n$ modulo finite.

So the two quotient order structures are not isomorphic.

  • $\begingroup$ It was a good question--thanks! I wonder what other ways we can see it? $\endgroup$ – Joel David Hamkins Feb 14 '16 at 21:18

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