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Let $\varphi:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ be a homeomorphism of the plane with fixed point $p$, i.e. $\varphi(p)=p$, and no other periodic points. Let $r$ be a fixed natural number. My question is:

Is it possible to partition the plane into a finite number of closed sets $A_{i}$, $i=1,...,k$ ($\bigcup_{i=1}^{k}A_{i}=\mathbb{R}^{2}$), such that $\varphi^{j}(A_{i})\cap A_{i}\subset\{p\}$ for any $j=1,...,r$, $i=1,...,k$. (This condition means that the intersection $\varphi^{j}(A_{i})\cap A_{i}$ is either empty, or the point $p$). The problem here is the finiteness of the family $\{A_{i}\}$, as the answer is clearly affirmative for a countable family of $A_{i}$'s.

[I came across this problem while considering some concrete systems in the plane with a finite number of periodic points. Then it is possible to formulate an analoguous question, but I am asking the most simple variant here, since I cannot imagine neither a counterexample, nor a proof even in this case...]

s::l

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    $\begingroup$ What do you understand by a partition? If $\{A_i\}_{i=1}^n$ is a partition of $\mathbb{R}^2$ (in the usual sense that $A_i\cap A_j$ is empty for $i\ne j$ and $\cup_{i=1}^n A_i=\mathbb{R}^2$) and $A_1,\dots A_{n-1}$ are closed, then $A_n$ is open. $\endgroup$
    – Julián Aguirre
    Commented May 16, 2011 at 15:27
  • $\begingroup$ Sorry, I meant only that the union of the A_i is the whole plane. They are allowed to intersect. $\endgroup$
    – t22
    Commented May 16, 2011 at 15:32
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    $\begingroup$ Then you should say "to cover" rather than "to partition". Also, I think in line 4 you mean the singleton {p}. $\endgroup$
    – Pietro Majer
    Commented May 16, 2011 at 15:50
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    $\begingroup$ In case it is not obvious, you can easily find a partition (I mean partition) with $r+1$ sets with the required intersection property. The problem is to make the sets closed. $\endgroup$
    – Emil Jeřábek
    Commented May 16, 2011 at 16:06
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    $\begingroup$ Extending Emil Jeřábek's comment, it is not too hard to find such a partition (I mean it too) into something like $2r+1$ Borel sets. The problem is still to make them closed. $\endgroup$
    – Clinton Conley
    Commented May 16, 2011 at 16:29

2 Answers 2

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You can extend your homeomorphism to the sphere with two fixed points and no other periodic points, and if it preserves a probability measure with total support it is called an irrational pseudo-rotation in this paper (see proposition 0.2 and recall Oxtoby-Ulam's theorem stating that if a homeomorphism preserves a probability measure with total support then it is conjugated to a conservative one).

There, it is proved that an irrational pseudo-rotation has, for every $n \geq 0$ a curve joining the fixed points such that it is disjoint from its first $n$ iterates and ordered exactly as in the irrational rotation. This allows to construct the desired $A_i$ if $n$ is sufficiently large compared with $r$.

When it does not preserve a measure with total support, the result does not apply, but many of the tools there may be useful, in particular, the Brower translation theorem.

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Let $A_i$ be the sector $\theta \in \left[ \frac{i}{2r}2\pi, \frac{(i+1)}{2r}2\pi \right]$ with $i = 0, \ldots, (2r-1)$.

Think of $\mathbb{R}^2$ as $\mathbb{C}$, and let $\varphi: z \mapsto \frac{1}{2}{\rm e}^{2\pi{\rm i}/r}z$. This rotates sectors two places counterclockwise, thus ensuring that $\varphi^j(A_i) \cap A_i = \{ 0 \}$ (the only fixed point of $\varphi$) for all $j$ from $1$ to $r$, while the factor $\frac{1}{2}$ ensures no other periodic points.

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