Let $T$ be a topological semigroup. A subset $A$ of $T$ is called syndetic if there is compact subset $K$ of $T$ such that $K^{-1}A= T$. This means that for every $t\in T$, $Kt\cap A\neq \emptyset$. Also a subset $R$ of $T$ is called a thick set if for every compact subset $K$ of $T$, there is $t\in T$ such that $Kt\subseteq R$. I interest to study relation between these notions.
If for every compact set $K\subseteq T$, we have $S\cap (\bigcap_{k\in K} k^{-1}S)\neq \emptyset$, then $S$ is thick set.
Q.What can say about the converse of inclusion?
Please help me to know it.
Let $T$ be a topological semigroup. Then we show that the following condition are equivalent.
$(1)$ $A\subseteq T$ is a thick set,
$(2)$ for every compact set $K$ there is some $t\in A$ such that $Kt\subseteq A$.
$(3)$ for every compact set $K$ we have
$A\cap_{k\in K}k^{-1}A\neq \emptyset$.
$(1)\Rightarrow (2)$. Let $r\in T$ be arbitrary. For compact set $K\subseteq T$, we have $Kr\cup{r}$ is a compact set in $T$. Since $A$ is thick set, for compact set $Kr\cup{r}$, there is $s\in T$ with $(Kr\cup{r})s\subseteq A$. Now, let $t=rs$, then $t\in A$ and $Kt\subseteq A$.
$(2)\Rightarrow (3)$. Let $K\subseteq T$ be a compact set in $T$. There is $t\in A$, with $Kt\subseteq A$. This means that $t\in K^{-1}A$. Hence, $t\in A\cap_{k\in K}k^{-1}A$, i.e. $A\cap_{k\in K}k^{-1}A\neq \emptyset$.
$(3)\Rightarrow (1)$ Let $K\subseteq T$ be a compact set. Since $A\cap_{k\in K}k^{-1}A\neq \emptyset$, hence assume that $t\in A\cap_{k\in K}k^{-1}A$. Then $Kt\subseteq A$.