For $A, B\subseteq \mathbb{R}^+$, we denote $$A+B= \{a+b: a\in A, b\in B\},$$ and $$A^{-1} +B= \{t: \text{ there is } a\in A \text{ such that } a+t\in B\}.$$
A subset $B\subseteq \mathbb{R}^+$ is called syndetic, if there is a compact set $K$ of $\mathbb{R}^+$ such that $\mathbb{R}^+= K^{-1}+B$
A subset $B\subseteq \mathbb{R}^+$ is called $g$- syndetic, if there is a compact set $K$ of $\mathbb{R}^+$ such that $\mathbb{R}^+= K+B$
One can check that every non-empty subset $B\subseteq \mathbb{R}^+$ with $0\in B$ is syndetic if and only it is $g$-syndetic.
Assume $\mathcal{T}=\{T(t)\}_{t\geq 0}$ is $C_0$-semigroup of bounded operators on Banach space $X$.
There exist a strictly increasing sequence $\{t_n\}\subseteq [0, \infty)$ with $t_n\to \infty$ and a sequence $\{x_n\}$ in $X$ such that $\lim_{n\to \infty}T(t_n)x_n=y$.
Assume that $A\subseteq [0, \infty)$ is syndetic. Is there a sequence $\{r_n\}\subseteq A$ with $r_n\to \infty$ and $\lim_{n\to \infty}T(r_n)x_n= \lim_{n\to \infty}T(t_n)x_n=y$?