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Any ideas, hints on the following would be great.

Suppose that $T\colon X \to X$ is continuous, and there exist at least two distinct periodic orbits. Show that if one of the periodic orbits is attracting then $T$ is not topologically transitive.

Note that the definition of attracting is: If $\Gamma = O(x)$ is the periodic orbit then there is and open set $U$ containing $\Gamma$ such that $\omega(x') = \Gamma$ for every $x' \in U$.

It may be assumed that $X$ is compact.

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What's $\omega$? – Gerry Myerson Mar 14 '13 at 11:43
Let the attracting orbit be $A$, the other periodic orbit, $B$. Choose an open set $U$ containing $A$ and "small enough" that $T(U)\subseteq U$ (I'm imagining $X$ to be a metric space; I'm not quite sure how to phrase then in a more general setting). Choose an open set $V$ containing the other periodic orbit and disjoint from $U$. Then $T^k(U)\cap V$ is empty for all $U$. I'm a bit out of my depth here and not confident enough to leave this as an answer. – Gerry Myerson Mar 14 '13 at 11:50
up vote 1 down vote accepted

Well, let's see. Let $U$ is a non-empty open set with the property that every point in $U$ converges to the attractive orbit. If the transformation is topologically transitive, then $$\bigcup_{k=1}^{\infty} T^k(U)$$ covers the space. Must be a contradiction right about there, seeing as how one of the $T^k(U)$s must contain a point in the other orbit.

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@Gerry Ah yes, \union - very clever of me. I typeset with Mathematica more than LaTeX these days and it gets so confusing. – Mark McClure Mar 14 '13 at 12:02
I think there's something better than what I used --- maybe \bigcup --- but I don't have my TeX book handy. You could experiment. I note that your answer bears a resemblance to my comment, but is much cleaner. – Gerry Myerson Mar 14 '13 at 12:09
Yep - I think your comment appeared while I was typing. – Mark McClure Mar 14 '13 at 12:13

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