Is there a compact set $S$ in topological semigroup $T$ with $t^{-1}K\subseteq S$, where $K$ is compact set in $T$?

Question

Let $T$ be a topological semigroup with $e\in T$ and for $t\in T$, $A\subseteq T$, define $t^{-1}A=\{r: tr\in A\}$. If $K\subseteq T$ is a compact set then $tK$ is a compact set in $T$. It seems that $t^{-1}K$ is not a compact set, in general.

Can we say that there is compact set $S$ in $T$ such that $t^{-1}K\subseteq S$? Please help me to know it.

Answer 1

Not if $T$ is Hausdorff.

As it is compact in a Hausdorff space, $K$ is necessarily closed. Since the action of $t$ is continuous (definition of topological semigroup), $t^{-1}K$, the inverse image of $K$ under $t$, would also be closed. A closed subset of a compact set is necessarily compact itself. If you could guarantee such an $S$, you could prove that $t^{-1}K$ is always compact.

Answer 2

Let $T = ({\Bbb N} \cup \{+ \infty\}, \max, 0)$ equipped with the discrete topology. Let $K = \{+\infty\}$ and $t = +\infty$. Then $K$ is compact, but $t^{-1}K = T$ is not contained in any compact set.