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Let $X$ be a metric space with metric $d$ and let $f:X\rightarrow X$ be dynamical system.

We say a sequence $\{y_i\} \subset X$ is a $\delta$-pseudo-orbit of $f$ if $d(f(y_k),y_{k+1}) < \delta,$ for each $k\in \mathbb{Z}$.

Let $\epsilon > 0.$ We say that a point $x\in X$ $\epsilon$-traces the $\delta$-pseudo-orbit $\{y_i\}$ if $d(f^k(x),y_k) < \epsilon,$ for each $k\in \mathbb{Z}$.

We say that the dynamical system $f$ has the "POTP" (the "pseudo-orbit tracing property") if given $\epsilon > 0$, there exists $\delta > 0$ such that for any $\delta$-pseudo-orbit $\{y_i\}$ there is a point $x$ that $\epsilon$-traces $\{y_i\}$.

I can show that a circle rotation does not have the POTP, however I am having trouble generalizing this to show that an isometry of a manifold does not have the POTP.

To see that the circle rotation does not have the POTP let $f:[0,1)\rightarrow [0,1)$ be the cirlce rotation map $f(x) = x + a\mod 1$. Consider the $\delta$-pseudo-orbit given by $y_0 = 0,$ $y_1 = a + \frac{\delta}{2} \mod1,$ and in general $y_{k+1} = y_k + a + \frac{\delta}{2}\mod1.$

I will leave out the "$\mod1$" below.

$\{y_i\}$ is a $\delta$-pseudo-orbit, since $$d(f(y_k),y_{k+1}) = d(y_k + a,y_k + a + \frac{\delta}{2}) < \delta.$$

Now, note that $d(f^k(x),y_k) = d(x + ka, k(a + \frac{\delta}{2})) = d(x,k\frac{\delta}{2})$ for all $k.$ For $\delta < \frac{2}{3}$ we see that for some $k,$ $k\frac{\delta}{2}$ is not in a $\frac{1}{3}$-ball around $x$.

Thus, picking $\epsilon = \frac{1}{3}$ the $\delta$-pseudo-orbit $\{y_i\}$ is not $\epsilon$-traced by $x.$

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Could you share some ideas of your proof and where you get stuck extending it ? – nonlinearism Mar 30 '13 at 19:06
Good idea... I edited this into my question. In the case of the circle map I constructed a specific $\delta$-pseudo-orbit such that for small enough $\delta$ the orbit is not $\epsilon$-traced by any $x$. – Bohring Mar 30 '13 at 20:43
up vote 2 down vote accepted

It suffices to assume that $X$ is a metric space with at least one nontrivial connected component $C$. Pick two distinct points $a,b\in C$ and let $\epsilon=d(a,b)/3$. Given any $\delta>0$, let $z_0=a,\dots, z_n=b$ be a chain of points from $a$ to $b$ so that $d(z_{k+1},z_k)<\delta$ for all $k=0,\dots,n-1$. (Such a chain exists by the connectedness of $C$). Let $$y_k= \begin{cases} f^k(a) ,\quad &k<0 \\ f^k(z_k), \quad & 0\le k\le n \\ f^k(b) \quad & k>n \end{cases}$$ This is a $\delta$-pseudo-orbit. Indeed, for $k=0,\dots,n-1$ $$d(y_{k+1}, f(y_k)) = d(f^{k+1}(z_{k+1}), f^{k+1}(z_{k}))=d(z_{k+1}, z_{k})<\delta$$ and for the other values of $k$ we have $y_{k+1}=f(y_k)$.

Suppose that a point $x$ satisfies $d(f^k(x),y_k)<\epsilon$ for all $k$. Then $d(x,a)<\epsilon$ and $d(f^n(x),f^n(b))<\epsilon$. Since $f$ is an isometry, the latter inequality is equivalent to $d(x,b)<\epsilon$. Now we have a contradiction: $d(x,a)<\epsilon$ and $d(x,b)<\epsilon$, where $\epsilon=d(a,b)/3$.

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Very nice proof. – Bohring Apr 6 '13 at 3:38

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