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QUESTION: Let $X$ be a topologically complete metric space and $T:X\to X$ a continuous map. Let $x\in X$ be a point whose orbit is a closed set. Show that either $x$ has an iterated that is periodic or $\omega_T(x)=\emptyset$.

I tried to do it by exclusion: if $\omega_T(x)=\emptyset$, then no iterated of $x$ is periodic and, if $\omega_T(x)\neq\emptyset$, I want to prove that an iterated of $x$ is periodic, but I couldn't do it.

Can someone help me?

Gus
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1 Answers1

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We have $$ \omega(x)=\bigcap_{n\in\mathbb N}\overline{\{T^k(x):k\ge n\}}=\bigcap_{n\in\mathbb N}\{T^k(x):k\ge n\}, $$ since the forward orbit of $x$ is closed. If $y\in\omega(x)$, then there exists $k$ such that $T^k(x)=y$. But one can replace the intersection by $\bigcap_{n>k}$ and so there exists $m>k$ such that $T^m(x)=y$. Hence, $T^m(x)=T^k(x)$ and so $T^{m-k}(z)=z$, where $z=T^k(x)$. In other words, $z$ is periodic.

John B
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  • @Anna Completeness is not used. Incidentally, I found out that a much simpler argument. I will edit my answer soon. – John B Apr 03 '16 at 12:23
  • In both answers you've used that a closed set minus a finite set of points is closed. Can you please explain why ${ 1/n , : , n \in \mathbf{N} } \cup { 0 } $ remains closed if you remove $0$? Maybe your argument actually needs somd extra feature from the problem? – ulilaka Apr 10 '16 at 23:07
  • @ulilaka No I haven't. Do you know what is an $\omega$-limit set? – John B Apr 10 '16 at 23:11
  • Yes, i do. In the first solution you said explicitely that the orbit of $y$ is also closed because it is the orbit of $x$ minus a finite number of points. In the second, you erased the bar from the sets ${ T^k x , : , k \geq n }$ for the same reason, I guess. – ulilaka Apr 11 '16 at 00:50
  • @ulilaka OK, then you know that your example with the sequence $1/n$ does not satisfy the properties of the problem. For the same reason, I could not use what you are saying that I used, and indeed something else is used. What? – John B Apr 11 '16 at 00:54
  • Probably the fact that the space is topologically complete. This is a property that is inherited by closed subsets, and thus may be applied to conclude something stronger about the limit points of the orbit of $x$. It is just that I can't see why the orbit of $y$ or any of the sets figuring in the intersection that defines the $\omega$-limit set should be indeed closed. – ulilaka Apr 11 '16 at 01:02
  • I think not, but I may be wrong. Think like this: Each point in the $\omega$-limit set is the limit of some suborbit, but erasing finitely many elements $T^{k_n}(x)$ does not change the limit. For that reason the closure doesn't change, and there is no need to use completeness. – John B Apr 11 '16 at 01:15
  • That proves $\omega (y) = \omega (x)$, with which I agree. Claiming that the orbit remains closed after removing finitely many points is a lot stronger. It is saying that none of the removed points is recurrent. – ulilaka Apr 11 '16 at 01:24
  • I now understood that I need to correct my answer (since what I wrote in my former comment is wrong). I regret that now is time to sleep (here, regards to my friends from USP), but let me comment that we are really only talking about $A_n={T^k(x):k\ge n}$, whose closure is the set plus the limit of sequences in the set; erasing finitely many points doesn't change the limits, but of course the closure may change. One the other hand, we need to show that $\bigcap_n\overline A_n=\bigcap_nA_n$, and not that $\overline A_n=A_n$. – John B Apr 11 '16 at 01:37
  • PS: I will look for a cleaner argument to show that indeed $\bigcap_n\overline A_n=\bigcap_nA_n$, as a clarification of what I wrote right now. – John B Apr 11 '16 at 01:40