Smallest open, dense, G-invariant subset of a metric space

Question

Let $X$ be a metric space and $G$ be a topological group acting continuously on $X$. Let $ \mathcal S $ be the set of open, dense and $G$-invariant subsets of $X$.

I need to take inverse limit (of some sets depending on $U$, say $A_U$) over $U \in \mathcal S$ where $\mathcal S$ is partially ordered by inclusion.

However, I dont really know how to find the mimimum (or infimum) of $\mathcal U$.

Is there a way to find the minimal element of $\mathcal S$? And if it doesn't exist, is there a way to find the infimum (maybe infimum is not the correct term here) of $\mathcal S$?

Or/and at least can you give me some intuition about the elements of $\mathcal S$?

Answer 1

For each $x \in X$, consider the orbit $Gx = \{gx: g \in G\}$ and its closure $\overline{Gx}$. Let $C$ be the set of $x$ such that $Gx$ is nowhere dense. If $x \in C$ then $X \backslash \overline{Gx}$ is a member of $\mathcal S$ that does not contain $x$. On the other hand, if $x \notin C$ then every member of $\mathcal S$ must contain $x$.

Given $A \in \mathcal S$, if $x \in A \cap C$, $A \backslash \overline{Gx}$ is a member of $\mathcal S$ that does not contain $x$. So the only way for $\mathcal S$ to have a minimal element $A$ is for that $A = X \backslash C$. This can occur (e.g. $C$ might be finite). But if $\overline{C}$ has nonempty interior, there is no minimal element.

Answer 2

In general there isn’t one. Let the multiplicative group $\{-1,1\}$ act on $S^1$ by multiplication. For rational $q\in(0,1)$ let $G_q=S^1\setminus\{e^{2\pi qi},-e^{2\pi qi}\}$. Each $G_q$ is dense, open, and $G$-invariant, and their intersection has empty interior and therefore contains no dense, open, $G$-invariant subset of $S^1$.