Complexity of a fixed point 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
 A: 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. 
A: 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.
