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