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Let $X$ be a compact metric space and $f:X\to X$ be a continuous function.

Show that if $\omega(x,f)$ is a finite set, then there exists a periodic point $y$ in $X$ such that: $d(f^n(x),f^n(y))\to 0$.

I can already use that there exists a periodic point $y$ such that $\omega(x,f)=o(y,f)$ where $o(y,f)$ denotes the orbit of $y$ under $f$.

But from here I don't know what else to do, because I want the orbits of $y$ and $x$ to approach at the same time, so I don't know how to proceed.

Thanks for your help!!

Robert Z
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user392559
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1 Answers1

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Hint: Show that the distance of $f^n(x)$ to the $\omega$-limit set of $x$ tends to zero when $n\to+\infty$.

Otherwise it would exist $\delta>0$ and a subsequence $n_k\nearrow+\infty$ such that ${\rm dist}(f^{n_k}(x),\omega(x))\ge\delta$ for all $k$. Taking a converging subsequence of $f^{n_k}(x)$ say to $z\in\omega(x)$ you get ${\rm dist}(z,\omega(x))\ge\delta$.

Contradiction!

Now use the fact that $\omega(x)$ is the orbit of some point $y'$ and use continuity to obtain what you want. Note that $y$ and $y'$ may be different.

John B
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