Let $X$ be a compact Hausdorff space without isolated points and $f:X\rightarrow X$ be a homeomorphism such that the $\mathbb{Z}$-action on $X$ that is defined by $f$ is minimal. Namely, for each $x\in X$, the orbit of $x$ defined below:
$$ [x]_f = \big\{ f^n(x): n\in \mathbb{Z} \big\} $$ is dense in $X$. For each subset $A\subseteq X$, we define the orbit of $A$ the same way:
$$ [A]_f = \big\{ f^n(a): a\in A,\, n\in\mathbb{Z} \big\} $$ In this case, is it always true that, or is there a necessary condition that implies that there exists a proper subset $A\subseteq X$ such that $[A]_f=X$?
This is a rather naive example. In a compact metric space $(X, d)$, suppose there exists $r>1$ such that for any $x, y\in X$, $d\big( f(x), f(y) \big) \geq rd(x, y)$, or we call $f$ is $r$-expansive. Then clearly for any (proper) open ball $B$, the diameter of $f^n(B)$ will increase as $n$ increases, and hence $[B]_f$ must be the entire $X$ since the diameter of $X$ is finite in this case. However, in this example, I am considering a compact metric space $(X, d)$ that has an $r$-expansive homeomorphism, but in general do not know if any compact metric space can have an $r$-expansive homeomorphism.
Counter-examples of a compact metric space that does not have any $r$-expansive (or even $1$-expansive) homeomorhpism, and counter-examples to my question above are both welcomed.
Updates: This post shows that in the set-up above an $f$-invariant subset always exists even when $f$ is not invertible.
