Necessary condition for minimal dynamical systems Let $f:X\to X$ be an homeomorphism on a compact metric space.
I want to prove that if $X$ is minimal then for all $\epsilon>0$ there exists an $N\in \mathbb{N}$ such that $\{x,f(x),...,f^N(x)\}$ is $\epsilon$-dense in $X$ for all $x\in X$.
My attempt:
Assume that $f$ is minimal. We know that in a compact topological space, every point in a minimal set is recurrent. In fact, the $\omega-$limit of a point is (non empty in compact spaces) clearly closed and $f-$invariant and therefore contains the minimal set of $f$. So, in this case, we also have that the $\omega-$limit of a point $x$ coincide with $X$ for all $x\in X$; i.e.
$$\omega(x):=\bigcap_{n\in\mathbb N}\overline{\bigcup_{i\geq n}f^i(x)}=X \qquad \forall x\in X$$
This means in particular that, for all $n\in \mathbb N$ and all $x\in X$, the set $\{f^i(x)\}_{i\geq n}$ is dense in $X$.\
Now we get that for any $\epsilon>0$ and all $x,n$ there exists
an $N=N(\epsilon, x, n)\in \mathbb N$ such that $\{f^{n}(x),...,f^{n+N}(x)\}$ is $\epsilon-$dense.
How can I eliminate the fact that $N$ depend on $x$? I still have to use that it's a homeo. Maybe should I do this also with the inverse of $f$ that has the same properties?
 A: Let $\varepsilon >0$.
First, fix a $x \in X$. Because $f$ is minimal, the set $\lbrace f^k(x), k \in \mathbb{Z} \rbrace$ is dense in $X$. In particular, the open balls $B(f^k(x), \varepsilon/2)$ cover $X$, so by compacity of $X$, there exists $N_x \in \mathbb{N}$ such that
$$X = \bigcup_{k=-N_x}^{N_x} B(f^k(x), \varepsilon/2)$$
Now, for each $x \in X$, there exists an open neighbourhood $V_x$ of $x$ such that if $y \in V_x$, then $$X = \bigcup_{k=-N_x}^{N_x} B(f^k(y), \varepsilon)$$
(indeed, for $y$ sufficiently near $x$, the ball $B(f^k(y), \varepsilon/2)$ is near the ball $B(f^k(x), \varepsilon/2)$, so the ball $B(f^k(y), \varepsilon)$ contains the ball $B(f^k(x), \varepsilon/2)$, so the balls $B(f^k(y), \varepsilon)$ cover $X$).
Again by compacity of $X$, there exists $x_1, ..., x_p \in X$ such that
$$X= \bigcup_{i=1}^p V_{x_i}$$
Let $N=\max \lbrace N_{x_i}|\text{ } i=1, ..., p \rbrace$. By construction, for all $x \in X$, you have
$$X = \bigcup_{k=-N}^{N} B(f^k(x), \varepsilon)$$
i.e. the family $\lbrace f^{-N}(x), ..., x, f(x), ..., f^{N}(x) \rbrace$ is $\varepsilon-$dense. Because this is true for all $x \in X$, you can apply it to all the $f^N(x)$ to get that for all $x$, the family $\lbrace x, ..., f^{2N}(x) \rbrace$ is $\varepsilon-$dense for all $x \in X$.
