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Let $T: {\bf R}^n \rightarrow {\bf R}^n$ be an homeomorphism and $x$ a point in ${\bf R}^n$. The positive orbit of $x$ is the set $\{T^n(x) \mid n \in {\bf N}\}$ and its $\omega$-limit set is the set of accumulation points of its orbit. $$\omega(x) = \{y \mid \exists \, n_k \rightarrow +\infty \hbox{ such that } T^{n_k}(x) \rightarrow y\}$$

If the $\omega$-limit set of $x$ is bounded, does it imply that the positive orbit of $x$ is also bounded?

For flows, it is easy by a connectedness argument, but I am not sure that it holds for homeomorphisms.

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We can modify the connectedness argument to show it also holds for homeomorphisms: suppose $\omega(x)$ is contained in an open ball $B(0,R)$, with $R>0$, and let $N$ be so big that $T(B(0,R))\subseteq B(0,N)$.

Then if the positive orbit of $x$ is not bounded, it must contain infinite points inside $B(0,N)\setminus B(0,R)$, which contradicts the fact that $\omega(x)\subseteq B(0,R)$.

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