inverse on topological semigroup

Question

Assume $(G,\cdot)$ denotes a topological semigroup (no id, non-commutative). Let $V\subset G$ be open and take some arbitrary $g\in G$.

Define $$gV^{-1}:= \bigcup_{x\in V}gx^{-1} = \bigcup_{x\in V}\{h\in G: hx=g \} = \{h\in G: g\in hV\}.$$

Then my question is: Is $gV^{-1}$ open?

Thanks for the help!

Answer 1

There are many counterexamples. Consider the case that $G$ is a topological monoid, such that the unit $e$ is the only left or right invertible element, and $e$ is not isolated. Take any open $e \in V \subseteq G$. Then $e V^{-1}=\{e\}$ is not open. An example is therefore $G=\mathbb{R}_{\geq 0}$.