(When) does a group action generate a basis of a topological space by refinements of open covers? Let $G$ be a discrete amenable group (maybe also finitely generated, if required to assume) acting on a compact, metrisable, finite dimensional topological space $X$. Let $\{U_\alpha\}_{\alpha\in I}$ be a finite open cover of $X$. Is it true that if the action is free and / or minimal, then the refinement of $\{U_\alpha\}_{\alpha\in I}$ with the open covers of the form $\{gU_\alpha\}_{\alpha\in I}$ for $g\in G$ generates a basis for the topology of $X$? If so, how to prove this? If not, what further assumptions on the action are required?
My attempts in this direction have principally been in lines motivated by the Poincare recurrence. Unfortunately most references I have come across as yet discuss principally the measure picture (ergodic theory rather than topological dynamics), as of course one has many more tools in that setting. References that might contain some ideas in this direction too would be very helpful.
 A: Alp's comment to the question helped me get to the solution, I think. Assuming $G$ to be finitely generated, the necessary and sufficient condition for this to happen is for the action to be expansive. We call a continuous action $\alpha$ of a group $G$ on a compact metric space $X$, $\alpha:G\curvearrowright X$, to be expansive if there exists a constant $c>0$ such that for all $x,y\in X$ and $x\neq y$, there exists $g\in G$ satisfying $d(\alpha_g(x), \alpha_g(y))>c$. This $c$ is referred to as the expansive constant for the action.
The following result is Proposition 2.4 from a recent article by A. Barzanouni. But first we set up the following notation: For two covers $\mathcal{U}$ and $\mathcal{V}$, we say $\mathcal{U}\prec\mathcal{V}$ if $\mathcal{U}$ is a refinement of $\mathcal{V}$, and their join $\mathcal{U}\wedge \mathcal{V}$ is given by the cover $\{U\cap V\mid U\in\mathcal{U}, V\in\mathcal{V}\}$.
The following are equivalent:

*

*$\alpha:G\curvearrowright X$ is an expansive group action.

*there exists an open cover $\mathcal{U}$ of $X$ such that for any other open cover $\mathcal{V}$, we have $\bigwedge_{g\in F} \alpha_{g^{-1}}\mathcal{U} \prec \mathcal{V}$ for some finite set $F\Subset G$.

$(1)\implies (2)$: Let $c>0$ be an expansive constant for the action $\alpha$. Take a finite open cover $\mathcal{U} = \{U_i\}_{i=1}^n$ such that $diam(U_i) < c$. Since $G$ is a finitely generated group, we can take $G = \{g_i\}_{i=0}^\infty$. Let $F_N = \{g_0, g_1, \dots, g_N\}$. If $(2)$ is not true, there is an open cover $\mathcal{V}$ such that for every finite set $F_N$, $$\bigwedge_{g\in F_N} \alpha_{g^{-1}}\mathcal{U}\nprec\mathcal{V}.$$
Since
$$\bigwedge_{i=0}^N \alpha_{g_i^{-1}}\mathcal{U}=\left\{\bigcap_{i=0}^{N} \alpha_{g_i^{-1}} U_{k_i} \mid k_i \in \{0, 1, \dots, n\} \right\},$$
$\left\{\bigcap_{i=0}^{N} \alpha_{g_i^{-1}} U_{k_i} \right\}$ is not contained in any $V\in\mathcal{V}$
Now consider $x_N,y_N$ in some open set in this cover $\bigwedge_{i=0}^N \alpha_{g_i^{-1}}\mathcal{U}$. We must have $d(x_N , y_N ) >c'$ where $c'$
is Lebesgue number for the cover $\mathcal{V}$. Also, since for each $i$ such that $0\leq i \leq N$ we must have $x_N,y_N\in \alpha_{g_i^{-1}}U_{k_i}$ for some $k_i$ and consequently $\alpha_{g_i}(x_N), \alpha_{g_i}(y_N) \in U_{k_i}$ and $diam(U_{k_i})<c$, we have $d(\alpha_{g_i}(x_N), \alpha_{g_i}(y_N)) < c$ for all $0 \leq i \leq N$. Since $X$ is compact, we can assume that there exist convergent subsequences $x_{N_k} \longrightarrow x$ and $y_{N_k} \longrightarrow y$. This implies that $d(x, y) > c'$ and $d(\alpha_g(x), \alpha_g(y)) < c$ for all $g \in G$.
This contradicts that $c$ is an expansive constant.
$(2) \implies (1)$: Assume there exists an open cover $\mathcal{U}$ such that for any other open cover $\mathcal{V}$, there exists a finite set $F \Subset G$ such that,
$$\bigwedge_{g\in F} \alpha_{g^{-1}}\mathcal{U}\prec\mathcal{V}.$$
Let $c>0$ be the Lebesgue number of the open cover $\mathcal{U}$. We claim that $c$ is an expansive constant. If it is not true, then there are $x,y\in X$ such that $x \neq y$ and $d(\alpha_g(x), \alpha_g(y)) < c$ for all $g \in G$. Take an open cover $\mathcal{V}$ such that $diam(V) < \frac{d(x,y)}{2}$ for all $V \in \mathcal{V}$. Then there is a finite set $F \Subset G$ such that $\bigwedge_{g\in F} \alpha_{g^{-1}}\mathcal{U}\prec\mathcal{V}$. Since for every $g \in F$, $d(\alpha_g(x), \alpha_g(y)) < c$, hence for every $g \in F$, there is $U \in \mathcal{U}$ such that $x, y \in \alpha_{g^{-1}}U$. This implies that $$x,y \in \bigwedge_{g\in F} \alpha_{g^{-1}} \mathcal{U} \prec \mathcal{V}.$$
Hence, there is $V \in \mathcal{V}$ such that $x, y \in V$. So $d(x, y) \leq diam(V) < \frac{d(x,y)}{2}$. That is a contradiction.
