Let $X$ be a compact metric space and $f:X\to X$ a homeomorphism. Assume that $X$ is minimal, i.e. $\{f^i(x):i\in \mathbb{N}_0\}$ is dense in $X$ for all $x\in X$. Prove that for any $\epsilon>0$, there exists an integer $N>0$, which only depends on $\epsilon$, such that $\{x,f(x),\dots,f^N(x)\}$ is $\epsilon$-dense for all $x\in X$.
So far I've only been able to prove this proposition for any finite subset $\{x_1,\dots,x_k\}$ of $X$.
Fix $1\leq j\leq k$. Since $\{f^i(x_j):i\in \mathbb{N}_0\}$ is dense, it is $\epsilon$-dense. So the open balls $\{B(f^i(x_j),\epsilon):i\in \mathbb{N}_0\}$ form a cover of $X$. By compactness of $X$, we can find a subcover $\{B(f^{i_1}(x_j),\epsilon),\dots,B(f^{i_{m(j)}}(x_j),\epsilon)\}$. Then $\{x_j,f(x_j),\dots,f^{i_{m(j)}}(x_j)\}$ is $\epsilon$-dense.
So we can repeat this argument for $x_2,\dots,x_k$ and set $N=\max\{i_{m(1)},\dots,i_{m(k)}\}$.
But how can I prove the general case, when $X$ is infinite? Should I use uniform continuity of $f$? I really have no idea. (The answer presented in this thread is apparently wrong.)
Any help will be appreciated. Thanks!
Let $d$ be the metric of the space $X$. Suppose to the contrary that there exists $\epsilon>0$ such that for any natural $n$ there exists $x_n,y_n\in X$ such that $d(f^i(x_n),y_n)\ge\epsilon$ for each $0\le i\le n$. Taking subsequences, we can assume that sequences $\{x_n\}$ and $\{y_n\}$ converges to points $x$ and $y$ of $X$, respectively. The minimality implies that there exists $m$ such that $d(f^m(x), y)<\epsilon/2$ that is $x\in f^{-m}(B(y,\varepsilon/2))$. Since the latter set is open there exists a number $M’$ such that $x_n\in f^{-m}(B(y,\epsilon/2))$ for each $n>M’$. There is a number $M\ge M’$ such that $d(y_n,y)< \varepsilon/2$ for each $n>M$. It follows that $$d(f^{m}(x_n),y_n)\le d(f^{m}(x_n),y)+d(y,y_n)< \epsilon/2+\epsilon/2=\epsilon$$ for each $n>M$, a contradiction.