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Let $X$ be a complete metric space and $T:X\rightarrow X$ continous. if exists a point $x$ with closed orbit, then $x$ is eventually periodic or $\omega(x)=\emptyset$

I've tried so far:

If $\omega(x)\neq \emptyset$, then $\displaystyle \exists y \in\omega(x)=\bigcap_{k\geq 0}\overline{\left\{T^{n}(x):n\geq k\right\}}.$

In particular, $y \in \overline{\left\{T^{n}(x):n\geq 0\right\}}=\left\{T^{n}(x):n\geq 0\right\}$, because $x$ has closed orbit. So $\exists k_{0}\in \mathbb{N}$ such that $y=T^{k_{0}}(x)$.

Hence, $T^{m}(\omega(x))\subset \omega(x)$ and $T^{m}(y)=T^{m+k_{0}}(x) \in \omega(x),$ for all $m\in \mathbb{N}$.

But, as $T^{m+k_{0}}(x) \in \omega(x)$, we have that $T^{m+k_{0}}(x) \in \overline{\left\{T^{n}(x):n\geq 0\right\}}=\left\{T^{n}(x):n\geq 0\right\}.$

So, exists $j_{0} \in \mathbb{N}$ such that $$T^{m+k_{0}}(x)=T^{j_{0}}(x),$$ let $z=T^{j_{0}}(x)$ and $m>j_{0}$ and we have $$T^{(m-j_{0})+k_{0}}(z)=z.$$

I don't know what is wrong with my answer and i din't use the hiphotesys of $X$ be complete space.

Lucas Galhego
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  • btw, i don't want direct answers, i just want a tip or if my answer is right. – Lucas Galhego Nov 29 '18 at 16:47
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    But, we don't know if $\left{ T^{n}(x):n \geq k \right}$ is closed, that's why i used the fact that $\omega(x)$ absorv $\omega(x)$ by $T$. – Lucas Galhego Nov 29 '18 at 17:13
  • Why are you sure that $m+k_0\neq j_0$? I think you need to show it, otherwise your last step is just $T^0(z)=z$. – Leo163 Nov 29 '18 at 17:47
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    Is for all $m > j_{0}$, so if $m+k_{0}=j_{0}$, take $m_{1}>m$. – Lucas Galhego Nov 29 '18 at 17:51
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    It is still unclear to me why you should be able to find $j_0$ such that $T^{m+k_0}(x)=T^{j_0}(x)$ for every $m$. – Leo163 Nov 29 '18 at 17:59
  • I thought a little and it's true, for each $m$ exists a $j_{0}$... i guess the problem is there. @Leo163, you have any tip for this problem? – Lucas Galhego Nov 29 '18 at 18:14
  • Of course such an $j_0$ exists, it is $j_0=m+k_0$. It is not at all assured just from the closedness of the orbit that there exists a non-trivial $j_0\ne m+k_0$. – Lutz Lehmann Nov 30 '18 at 19:08

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That's how I would do it (I do not claim the solution to be optimal in any sense). If for every $k$ $\overline{\{T^n(x):n>k\}}=\{T^n(x):n>k\}$, then you are done since this implies that some $y=T^m(x)$ must appear infinitely often, and in particular it appears twice, which is enough to say that $x$ is eventually periodical.

If this is not the case, it means that there is $k$ such that $\overline{\{T^n(x):n>k\}}\neq\{T^n(x):n>k\}$: this implies that there is $n\leq k$ such that $T^n(x)$ is an accumulation point for a sequence $\langle T^i(x):i\in I\rangle$, for some (infinite) $I\subseteq \omega$ (this follows from the fact that we are assuming that the orbit of $x$ is closed). By the continuity of the map $T$, we know that for every $j$ the sequence $\langle T^{i+j}(x):i\in I\rangle$ converges to $T^{j+n}(x)$. Now, we claim that, given this, we can construct a Cauchy sequence of points in the orbit that does not converge to any of the points of the orbits: by completeness of the space $X$, this sequence will have to converge to some point, and since this point will not be in the orbit we are contradicting the hypothesis that the orbit is closed (so we actually get to use all of the hypotheses).

SO let's build the sequence: let $d$ be the distance between $T^0(x)=x$ and $T^n(x)$, and let $\epsilon_0:=\min(d/2,1/2)$. Since there is a sequence of points converging to $T^n(x)$, let $a_0=T^p(x)$ be a point of the orbit such that $p>n$ such that $d(a_0,T^n(x))<\epsilon_0$. By what we said earlier, there is a sequence of points converging to $a_0$. Let $d_1:=d(a_0,T(x))$, and let $\epsilon_1:=\min(d_1/2,\epsilon/4)$. Again, we can find a point $a_1$ such that $d(a_0,a_1)<\epsilon_1$. We proceed like these for $\omega$ steps.

The sequence of the $a_n$ is clearly Cauchy. Moreover, it cannot converge to any point of the orbit, since at every step we make sure the distance between $T^s(x)$ and $a_s$ is always positive (e.g., if $s=0$, we said that $\epsilon_1<\epsilon_0/4$, so no $a_i$ will be closer than $\epsilon_0/2$ to $T^0(x)$). This gives the desired contradiction.

Leo163
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    Why do you demand $ϵ1\le 1/3$? You need $ϵ{k+1}\le ϵk/2$ for the Cauchy property $$d(a_m,a_n)\le ϵ_k\text{ for }m,n>k$$ and $ϵ_k\le d(T^k(x), a_k)/4$ for $$d(T^k(x),a_n)\ge d(T^k(x), a_k)-d(a_n,a_k)\ge d(T^k(x), a_k)-2ϵ_k\ge d(T^k(x), a_k)/2\text{ for }n>k$$ to avoid that $T^k(x)$ is the limit of the sequence $(a_k){k\in\Bbb N}$. Thus $ϵk=\min(ϵ{k-1}/2,d(T^k(x), a_k)/4)$ with $a_{k+1}\in B(a_k,ϵ_k)$ should be sufficient for your construction. – Lutz Lehmann Nov 30 '18 at 15:41
  • Yes, it is. I just wanted to make explicit why the sequence was Cauchy, but that is actually not the way to do it. Thank you for your remark. – Leo163 Nov 30 '18 at 16:16