Let $T$ be a topological semigroup. We say that $T$ satisfies $SFC$-condition, if for every finite set $H\subseteq T$ and every $\epsilon>0$, there is a finite set $K\subseteq T$ such that $|K\setminus tK|<\epsilon |K|$ for all $t\in H$.
Take $T=[0, \frac{1}{2}]\cup \{1\}$ with usual topology and multiple as operation. In my research, I need to know that whether $T$ satisfies $SFC$-condition or not?
I think, it does not have $SFC$-condition, because if finite set $H$ contains $0\in T$ and $\epsilon>0$ be sufficiently small, then there is no finite set $K$ such that $|K\setminus tK|<\epsilon |K|$ for all $t\in H$, but I do not know that my claim is true or not. Please help me to know it.
If the semigroup $T$ has a null element $0\in T$ then $SFC$-condition holds.
In fact, one set $K$ suffices independently of $H$ chosen. Indeed, take $K = \{0\}$, then $|K\setminus tK|<\varepsilon |K|$ is equivalent to $0<\varepsilon$.