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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 $A_{i}$, $i=1,...,k$ ($\cup_{i=1}^{k}A_{i}=\mathbb{R}^{2}$), $\bigcup_{i=1}^{k}A_{i}=\mathbb{R}^{2}$), such that $\varphi^{j}(A_{i})\cap A_{i}\subset{p}$ 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}$'s. [I came across this problem while considering some concrete systems in the plane with a finite number of periodic points. Then it is possible to formulate an analoguous question, but I am asking the most simple variant here, since I cannot imagine neither a counterexample, nor a proof even in this case...] s::l 1 # Complexity of a fixed point Let$\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$($\cup_{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