How do I show there exist $k \in [0,n]$ such that $\| f_k(y)-x\|< \epsilon$? Let $X$ be a bounded and closed subset of $\mathbb{R}^4$. Let $f: X \rightarrow X $ be a homeomorphism. Write $f_n$ for the nth iterate of $f$ if $n>0$, for the $-nth$ iterate of $f^{-1}$ if $n<0$, and the identity map if $n=0$. Thus, $f_{n+1}(x)=f(f_n(x))$ for all $n \in \mathbb{Z}$.
Write $A(x)=\{f_n(x):n \in \mathbb{Z}\}$ for $x \in X$. Suppose that $A(x)$ is dense in $X$ for all $x \in X$.
Show that for each $x \in X$ and all $\epsilon >0$, there exist $n>0$ such that for all $y \in X$, there exists $k  \in [0,n]$ such that $\| f_k(y)-x\|< \epsilon$.
Anyone suggest some direction to solve this question?
 A: Too long for a comment, requested by OP.
Assume that $f$ is an isometry, and that $X$ is bounded (but not necessarily closed).
Let $\epsilon >0$. Then, by a simple compactness of closure argument, we can find $F=\{y_1,...,y_k \} \subseteq X$ so that for all $z \in X$ there exists some $1 \leq j \leq k$ with
$$
\|z-y_j \|_\infty < \frac{\epsilon}{2} \qquad (*)
$$
Since by assumption $A(x)$ is dense in $X$, for each $1 \leq j \leq k$ there exists some $n_j \in \mathbb Z$ so that
$$
\|f_{n_j}(x)-y_j\| \leq \frac{\epsilon}{2}
$$
Since $f$ is an isometry, this is equivalent to
$$
\|x-f_{-n_j}(y_j)\| \leq \frac{\epsilon}{2} \qquad (**)
$$
Now, pick $M,N \in \mathbb N$ so that $-M \leq n_j \leq N$ for all $1 \leq j \leq k$.
Set $n=N+M$. We claim that this $n$ works.
Let $y \in X$ be arbitrary. Since $f_{M}(y) \in X$, by (*) there exists some $1 \leq j \leq k$ so that
$$
\|f_{M}(y)-y_j \| \leq \frac{\epsilon}{2}
$$
Since $f$ is an isometry we get
$$
\|f_{M+n_j}(y)-f_{n_j}(y_j) \| \leq \frac{\epsilon}{2}
$$
and hence by (**)
$$
\|f_{M+n_j}(y)-x \| <\epsilon
$$
Since $0=M+(-M) \leq M+n_j \leq M+N=n$ this proves the claim.
