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Suppose $X$ is a compact metric space, and that $\varphi: X\to X$ is a homeomorphism of $X$.

We say a subset $A$ of $X$ is $\varphi$-invariant if $\varphi(A) = A$. A $\varphi$-invariant set is minimal if it is closed, $\varphi$-invariant, nonempty and the smallest of all such sets. We say $(X,\varphi)$ is essentially minimal if $X$ contains a unique minimal $\varphi$-invariant set.

An orbit of $x \in X$ is the set $O_\varphi(x)= \{ \varphi^n(x) \;|\; n\in\mathbb{Z} \}$.

My Question

Now suppose that $\varphi: X\to X$ is a homeomorphism such that $X$ contains exactly two minimal $\varphi$-invariant subsets $M_1,M_2\subset X$. Moreover, $\varphi$ has the property that for all $x\in X$ either $M_1\subset\overline{O_\varphi(x)}$ or $M_2\subset\overline{O_\varphi(x)}$ but not both.

Does it follow that $X=X_1\dot{\cup} X_2$ splits into two clopen $\varphi$-invariant subsets?

If not, I would be thankful for a counter example.

More generally, I would like to answer the following more general question. Suppose $\varphi: X\to X$ is a homeomorphism and let $\mathcal{M}$ be the set of all minimal $\varphi$-invariant sets in $X$. Suppose also that $\varphi$ satisfies the following property: $$\forall\; x\in X: \exists !\; M\in\mathcal{M}: M\subset\overline{O_\varphi(x)}. $$

For $M_1,M_2\in\mathcal{M}$, can one find $\varphi$-invariant open subsets $U_1,U_2\subset X$ such that $M_i\subset U_i$ for $i=1,2$ and $U_1\cap U_2=\emptyset$ ?

My motivation is, among other things, to show that certain homeomorphisms with this property can be decomposed into essentially minimal systems, i.e. $X=\bigcup E_\alpha$ and the $E_\alpha$ are closed, $\varphi$-invariant, essentially minimal and pairwise disjoint. For the case I am interested in, a positive answer to the above (more general) question would be sufficient. But since I cannot even answer the more specific question, an answer to that would already be very helpful.

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In the second paragraph, when you define the minimal $\varphi$-invariant sets, you say that they should be "smallest", but I think you mean merely that they should be minimal here. Right? (Smallest = contained in all other instances; minimal = no strictly smaller instance) –  Joel David Hamkins Nov 26 '12 at 0:50
Yes, that is exactly what I mean. Thank you –  Gabor Szabo Nov 26 '12 at 12:19
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2 Answers

up vote 2 down vote accepted

The answer is no.

For each pair $(n,k) \in \mathbb{Z}^+ \times \mathbb{Z}$ we define a point $p(n,k) \in \mathbb{R^2}$ as: $$p(n,k)=\left(1+\frac{1}{n},\frac{k}{n^2} \right)$$ if $|k| \leq n$, and $$p(n,k)=\left( \frac{1}{k}\cos(1/n), \frac{1}{k} \sin(1/n) \right)$$ otherwise.

We let $$X= \left\{ p(n,k): (n,k) \in \mathbb{Z}^+ \times \mathbb{Z} \right\} \cup \left\{(0,0), (1,0) \right\}$$ viewed as a (compact) subspace of $\mathbb{R}^2$, and define $\varphi:X \to X$ by $\varphi(0,0)=(0,0)$, $\varphi(1,0)=(1,0)$ and $\varphi(p(n,k))=p(n,k+1)$.

It is easy to check that all the requirements are met (note that the two minimal $\varphi$-invariant sets are $M_1=\{(0,0)\} $ and $M_2=\{(1,0)\}$). However, if $U$ is an open subset of $X$ containing $(0,0)$ then for each $n$ there is a $k$ such that $p(n,k) \in U$; so if $U$ is also $\varphi$-invariant then it must contain every $p(n,k)$; therefore, if $U$ is also closed, it must contain $(1,0)$. In other words, the only $\varphi$-invariant clopen is $X$.

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No. Let $X$ be a closed interval and let $\phi:X\to X $ be an order-preserving bijection that fixes only the endpoints.

Edit: This is an answer to a trivial question which is not the intended one.

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Don't you need still to have only the two fixed points (the endpoints), as in your original answer? Otherwise, this won't satisfy the hypothesis of having exactly two minimal invariant sets. –  Joel David Hamkins Nov 26 '12 at 1:33
Yes, sorry. In correcting a typo I also messed it up. Fixed now. –  Tom Goodwillie Nov 26 '12 at 4:08
I don't see how these maps satisfy the given property. Without loss of gerenality, we can work with $X=[0,1]$. If the endpoints are the only fixed points, either $\varphi(x)>x$ for all $x\neq 0,1$ or $\varphi(x)< x$ for all $x\neq 0,1$. Assume the latter. So for a given $x_0\in (0,1)$, the sequence $x_n=\varphi^n(x_0)$ is decreasing and bounded, hence it converges. Since the limit will be a fixed point, the limit is 0. Similarly, the sequence $y_n=\varphi^{-n}(x_0)$ will converge to 1. In particular, almost all orbit closures contain both minimal sets, which is ruled out from the beginning. –  Gabor Szabo Nov 26 '12 at 12:38
I see. I misunderstood the question because you used "or" to mean exclusive "or". I have taken the liberty of editing the end of your "My Question" for clarity. –  Tom Goodwillie Nov 26 '12 at 13:17
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