Let $k$ be a field, $X=$Spec$A$ be an affine scheme with A a f.g. $k$-algebra. $G=$Spec$R$ is a linearly reductive group acting rationally on A. (i.e. every element of $A$ is contained in a finite dimensional $G$-invariant linear subspace of $A$.) By Nagata's theorem, $A^G$ is a f.g. $k$-algebra. We have the affine GIT quotiont $X \rightarrow X//G :=$Spec$A^G$ induced by the inclusion $A^G \rightarrow A$ of $k$-algebras.
Question: is the affine GIT quotient, viewed as a map of the underlying topological spaces, necessarily an open map? (it doesn't need to be an open immersion of schemes.) If not, any counterexample?
I know a ridiculous counterexample. $R= \mathbb{k} [x, y] / (xy=0)$. Consider the action of $G=\mathbb{G}_m$ given by $(x, y) \rightarrow (tx, y)$. Then $R^G = \mathbb{k}[y]$. Consider $U = \{ x \neq 0 \}$. It is an open subset, but the image is $\{ y=0 \}$ (not open).