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Question: Let $\Gamma$ be a countable group. Is the intersection of two thickly syndetic sets still thickly syndetic?

I've only seen the proof for the group $\mathbb{Z}$ (and I believe this method generalises to amenable groups). (there is an "extra" question below)

Background:

  • a set $S \subset \Gamma$ is syndetic if there is a finite $A \subset \Gamma$ to that $AS = \Gamma$.
  • $T$ is thick if for all finite $F \subset \Gamma$, $\cap_{f \in F} fT \neq \emptyset$.
  • $U$ is thickly syndetic, if for any finite $F \subset \Gamma$, $U' = \cap_{f \in F} fL$ is syndetic.

Partial answer: It's an exercise to see that if $S$ is syndetic and $T$ is thick, then their intersection is non-empty. Indeed, if $A$ is the set for which $S$ is syndetic, then $$ \begin{array}{rl} \displaystyle \bigcap_{a' \in A} a'T &= \displaystyle \Gamma \cap \big( \bigcap_{a' \in A} a'T \big) \\ &= \displaystyle \big( \cup_{a \in A} aS \big) \cap \big( \bigcap_{a' \in A} a'T \big)\\ &= \displaystyle \bigcup_{a \in A} \bigg( aS \cap \big( \bigcap_{a' \in A} a'T \big) \bigg)\\ &= \displaystyle \bigcup_{a \in A} a \bigg( S \cap \big( \bigcap_{a' \in A} a^{-1}a'T \big) \bigg)\\ & \subseteq \displaystyle \bigcup_{a \in A} a \big( S \cap T \big) \end{array} $$ Since $T$ is thick the $S \cap T$ is non-empty. Now if $T$ is thickly syndetic, then $S \cap T$ is even syndetic. However this argument does not yield thickly syndetic.

extra question: it's another exercise to show that a set $S$ is syndetic iff its intersection with any thick is non-empty. Likewise a set is syndetic iff its intersection with any thick is non-empty. Is a set thickly syndetic iff its intersection with any thickly syndetic set is thickly syndetic?

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    $\begingroup$ If the first question is true then the extra question is also true since any set contains a thickly syndetic set is also thickly syndetic.I remember a fact I learned in a course on topological dynamics that a set is thickly syndetic iff its intersection with any picewise syndetic set (the intersection of a syndetic set and a thick set) is non-empty. Maybe this result is helpful. $\endgroup$
    – Siming Tu
    Commented Jul 20, 2017 at 23:22
  • $\begingroup$ Thanks! Actually I just realised the answer is obvious... one needs only to replace $S$ by $S' = \displaystyle \cap_{f \in F} S$ and $T$ by $T' = \displaystyle \cap_{f \in F} T$. Now if $S$ and $T$ are thickly syndetic, then $S'$ is syndetic and $T'$ is thick (so the same argument follows). Since $S' \cap T' = \displaystyle \bigcap_{f \in F} (S \cap T)$, one gets the conclusion... $\endgroup$
    – ARG
    Commented Jul 21, 2017 at 7:34
  • $\begingroup$ I am sorry that I did not quite understand your argument. It seems that you have proved that $\displaystyle \bigcap_{f \in F} f(S \cap T)$ is a piecewise syndetic set and it seems this can not imply the intersection is thickly syndetic as piecewise syndetic is weaker than syndetic. $\endgroup$
    – Siming Tu
    Commented Jul 24, 2017 at 15:17
  • $\begingroup$ Perhaps because I did not put emphasis on the following fact: If $T$ is thickly syndetic then, for any finite set $F$, $\displaystyle \bigcap_{f \in F} fT$ is still thickly syndetic. Indeed, if $E$ is another finite set, then $\displaystyle \bigcap_{e \in E} e \big( \bigcap_{f \in F} fT \big) = \bigcap_{e \in E} \big( \bigcap_{f \in F} ef T \big) = \bigcap_{d \in EF} d T$. Since $EF$ is still finite, the fact follows. $\endgroup$
    – ARG
    Commented Jul 24, 2017 at 18:03
  • $\begingroup$ So basically the argument I wrote was to show that the intersection of a thick set $T$ and a syndetic set $S$ is non-empty. Because of the fact, if $T$ is thickly syndetic this argument shows that the intersection is actually syndetic. Now if $S$ is also thickly syndetic, the argument in my first comment uses the fact that for any $F$, $S'$ and $T'$ are still thickly syndetic. In particular, $S'$ is syndetic and $T'$ is thickly syndetic. So their intersection $\displaystyle \bigcap_{f \in F} f(S \cap T)$ is syndetic. Since this holds for any $F$, this means that $S \cap T$ is thickly syndetic $\endgroup$
    – ARG
    Commented Jul 24, 2017 at 18:12

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