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Let $f:X\rightarrow X$ be continuous on a Hausdorff compact space with no isolated point. Suppose there is a $x\in X$ such that:

  • $\exists_{p\geq1} f^p(x)=x$ ($f$ composed $p$ times)

  • There is a neighborhood U of x such that $f^p(\bar{U})\subset {U}$

  • $\cap_{n\geq0}f^{pn}(U)=\{x\}$

I would like to deduce from this that there is no point $y\in X$ with dense orbit, i.e such that $\cup_{n\geq0}f^{n}(y)$ is dense in $X$.

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1 Answer 1

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HINT: Suppose that $y$ has a dense orbit. Then $f^n(y)\in U$ for some $n\ge 0$, and it follows that

$$\left\langle f^{n+kp}(y):k\in\Bbb Z^+\right\rangle\to x\;.$$

Show that for $i=0,\dots,p-1$ we must have

$$\left\langle f^{n+kp+i}(y):k\in\Bbb Z^+\right\rangle\to f^i(x)\;.$$

Thus, the orbit of $y$ is eventually trapped in any open nbhd of $\{f^i(x):i=0,\dots,p-1\}$.

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I can see how to obtain these steps, but I can't find the right argument to conclude. I'd like to deduce from there that the orbit of $x$ is itself dense (hence a contradiction), but is that the right way to conclude? –  Lynn Feb 27 '13 at 10:50
    
@Lynn: Take an open nbhd $V$ of $\{f^i(x):i=0,\dots,p-1\}$ small enough that $W=X\setminus\operatorname{cl}V\ne\varnothing$. $W$ contains only finitely many points of the orbit of $y$, and $W$ itself is infinite (since $X$ has no isolated points), so $W$ contains a non-empty open set disjoint from the orbit of $y$. –  Brian M. Scott Feb 27 '13 at 10:56
    
What if for all $V$ we have cl $V=X$? How to exclude that case? –  Lynn Feb 27 '13 at 11:07
    
@Lynn: Pick any point $z\notin\{f^i(x):i=0,\dots,p-1\}$. $X$ is compact Hausdorff, so it’s regular, and there is therefore an open nbhd $V$ of $\{f^i(x):i=0,\dots,p-1\}$ such that $z\notin\operatorname{cl}V$. –  Brian M. Scott Feb 27 '13 at 11:13

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