Let $G$ be an infinite finitely generated discrete group. I call an infinite set $S$ irregular iff for every $g\in G$, $g\neq 1$, we have that $S\cap gS$ is finite. For example $\{z^3|z\in\mathbb{Z}\}$ is irregular in $\mathbb{Z}$. Now my easy to state question: Does every free ultrafilter on $G$ contain at least one irregular subset?

The following is true and might probably be useful: $G$ acts freely on the space of all ultrafilters (the Stone-Cech compactification of $G$ as a discrete space).

free ultrafilters: http://en.wikipedia.org/wiki/Ultrafilter

  • $\begingroup$ In the definition of irregular, I guess you want $g$ to be non-identity? $\endgroup$ – Joel David Hamkins Nov 17 '12 at 11:31
  • $\begingroup$ To save people searching through the wikipedia article: free = non-principal. $\endgroup$ – HJRW Nov 17 '12 at 12:03
  • $\begingroup$ @Joel: Yeah, I'll correct that, thx. $\endgroup$ – Werner Thumann Nov 17 '12 at 14:05

No, a free ultrafilter on the additive group of integers need not contain an irregular set. The Galvin-Glazer proof of Hindman's theorem (which is nowadays the standard proof of that theorem) begins by showing the existence of idempotent ultrafilters $U$ on $\mathbb N$. I won't bother to define "idempotent" here, since what I need is not the definition but the following consequence of it. If $X$ is any set in $U$, then there is an infinite set $\{x_0<x_1<\dots\}$ of positive integers such that any finite sum of distinct $x_i$'s is in $X$. In particular, the intersection of $X$ and $x_0+X$ contains $x_0+x_i$ for all $i>0$, and therefore $X$ is not irregular.

  • $\begingroup$ Thx for the answer, but are you sure this argument carries over to the group Z? Does Elli-Numakura really give an idempotent besides the trivial one around 0? Im a layman in this field, so plz dont blame me if this is a stupid question. ;) $\endgroup$ – Werner Thumann Nov 18 '12 at 18:32
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    $\begingroup$ Any idempotent ultrafilter on $\mathbb N$ generates (by closing it under supersets) an idempotent ultrafilter on $\mathbb Z$. More generally, the canonical embedding $\beta\mathbb N\to\beta\mathbb Z$ is a homomorphism of semigroups. $\endgroup$ – Andreas Blass Nov 19 '12 at 1:37

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