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Given an ultrafilter on $\omega$ and a function $f:\omega\longrightarrow\omega$, what does it mean that $f$ is unbounded modulo $D$?


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For $n\in\omega$ let $b_n=\{k\in\omega:f(k)\le n\}$. If $b_n\in D$, we say that $f$ is bounded by $n$ modulo $D$. Thus, $f$ is unbounded modulo $D$ iff $\forall n\in\omega\,(b_n\notin D)$.

Roughly speaking, one says that something is true modulo $D$ if it is true for all $n$ in some member of $D$. For example, given $f,g:\omega\to\omega$, we say that $f=g$ modulo $D$ if $\{n\in\omega:f(n)=g(n)\}\in D$. Similarly, we might say that $f\le g$ if $\{n\in\omega:f(n)\le g(n)\}\in D$. The statement that $f$ is bounded by $n$ modulo $D$ is a minor extension of this idea: it’s equivalent to saying that $f\le g$, where $g(k)=n$ for all $k\in\omega$.

Note that if $f=g$ modulo $D$, and $d\in D$ is arbitrary, then $\{n\in d:f(n)=g(n)\}\in D$. Similarly, if $f$ is bounded by $n$ modulo $D$, and $d\in D$ is arbitrary, then $\{k\in d:f(k)\le n\}=d\cap b_n\in D$. If you think of members of $D$ as ‘large’ sets and their complements as ‘small’ sets, then a function that is bounded by $n$ modulo $D$ is bounded by $n$ on a large subset of any large set. Alternatively, even if you lump together all of the ‘bad’ points at which $f$ is not bounded by $n$ to form $\{k\in\omega:f(k)>n\}$, you have only a small set: $\{k\in\omega:f(k)>n\}\notin D$.

Similarly, if $f=g$ modulo $D$, then $f=g$ on a large subset of any large set: $\{k\in\omega:f(k)\ne g(k)\}$ is a small set, meaning that it’s not in $D$.

You might think that ‘$f$ is unbounded modulo $D$’ ought to mean that there is a $d\in D$ such that $f\upharpoonright d$ is unbounded, but that turns out to be unsatisfactory: it’s entirely possible that there is a $d_0\subseteq d$ such that $d_0\in D$ and $f$ is bounded by some $n$ on $d_0$, i.e., that $\{k\in d:f(k)>n_0\}\notin D$. In that case $f$ is bounded by $n$ modulo $D$, so we really don’t want to say that $f$ is unbounded modulo $D$. Things work out nicely, though, if we define ‘unbounded modulo $D$’ to mean ‘not bounded by any $n\in\omega$ modulo $D$’, as I did above. Then for each $n\in\omega$ the set $b_n\notin D$. This means that for any $n\in\omega$, $f$ is actually larger than $n$ on a ‘large’ set.

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thank you very much for your complete answer, it was very useful! – rush fan Dec 27 '12 at 18:47
@rushfan: I’m glad to hear it; you’re very welcome. – Brian M. Scott Dec 27 '12 at 18:47

It is useful to think of an ultrafilter $D$ on $I$ as a $\{0,1\}$-valued finitely additive "measure" on the powerset of $I$. The subsets of $I$ in the ultrafilter have "measure" $1$, and their complements have measure $0$.

A function $f: I\to \omega$ is unbounded modulo $D$ if for any $N\in \omega$, $f(i)\gt N$ "almost everywhere." The almost everywhere bounded functions are uninteresting. For if $I$ is split into finitely many pairwise disjoint sets, then precisely one of the sets lies in the ultrafilter. So an almost everywhere bounded function is almost everywhere constant.

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