# Homeomorphisms on a compact set

Let $X\subset {\Bbb R}^4$ be compact and $f:X\to X$ be a homeomorphism. Define $f^n$ as $$f^0=id_X, \ f^{(n+1)}=f^nf\ (n>0),\ f^n=(f^{-1})^{-n}\ (n<0).$$ Assume that $\{f^n(x)\}_{n\in {\Bbb Z}}$ is dense in $X$ for every $x\in X$. Show that for every $x\in X$ and every $\epsilon>0$, there exists $n>0$ such that for all $y\in X$ we have $$f_k(y)\in B(x,\epsilon)$$ for some $k\in[0,n]$, where $B(x,\epsilon)$ is the open ball centered at $x$ with radius $\epsilon$.

I tried to prove the statement by contradiction. Assume the following is true: $$\exists x_0\in X\ \exists\epsilon>0 \forall n>0\ \exists y_n\in X\ \forall k\in[0,n]\\ \|f_k(y_n)-x\|>\epsilon.$$ Since $X$ is compact in ${\Bbb R}^4$, $\{f_n(y_n)\}_{n=1}^{\infty}$ has a convergent subsequent $\{f_{n_k}(y_{n_k})\}_{k=1}^{\infty}$. Denote the limit as $y_{*}$. Then there exists $m\in{\Bbb Z}$ such that $$\|f_m(y_{*})-x_0\|\leq\epsilon/2$$ by the dense assumption. One the other hand $f_m$ is a homeomorphism which implies that $$\|f_m(y_{*})-f_m(f_{n_k}(y_{n_k}))\|\leq\epsilon/2$$ for large enough $k$. By triangle inequality we have $$\|f_m(f_{n_k}(y_{n_k}))-x_0\|\leq\epsilon$$ I just get stuck here: there is no way to guarantee that $m+n_k\in[0,n_k]$ in order to reach a contradiction.

How can I rescue the proof or does any one have other ideas?

Assume the following is true: $$\exists x_0\in X\ \exists\epsilon>0 \ \forall 2n>0\ \exists y_n\in X\ \forall k\in[0,2n]\\ \|f_k(y_n)-x\|>\epsilon.$$ Since $X$ is compact in $R^4$, $\{f_n(y_n)\}_{n=1}^{\infty}$ has a convergent subsequence $\{f_{n_k}(y_{n_k})\}_{k=1}^{\infty}$. Denote the limit as $y_{*}$. Then there exists $m\in{\Bbb Z}$ such that $$\|f_m(y_{*})-x_0\|\leq\epsilon/2$$ by the dense assumption. One the other hand $f_m$ is a homeomorphism which implies that $$\|f_m(y_{*})-f_m(f_{n_k}(y_{n_k}))\|\leq\epsilon/2$$ for large enough $k$. By triangle inequality we have $$\|f_m(f_{n_k}(y_{n_k}))-x_0\|\leq\epsilon$$ Note that for large enough $k$ we have $m+n_k\in[0,2n_k]$, which implies that we have got the contradiction.
For each $y$, at least one $f^k(y)$ is in $B(x,\epsilon)$. This means that $$\bigcup_{k\in\mathbb Z}f^{-k}[B(x,\epsilon)]=X.$$ Since this is a union of open sets, by compactness of $X$ there are $m_0,m_1\in\mathbb Z$ such that $$\bigcup_{k\colon m_0\le k\le m_1}f^{-k}[B(x,\epsilon)]=X.$$ Set $n:=m_1-m_0$. For $y\in X$ there is an $\ell\in[m_0,m_1]$ such that $f^{-m_0}(y)\in f^{-\ell}[B(x,\epsilon)]$, i.e. for $k:=\ell-m_0\in[0,n]$ we have $$f^k(y)=f^\ell(f^{-m_0}(y))\in B(x,\epsilon).$$