Topological homeomorphism which takes convergent sequence to a specific convergent sequence For some reasons the sequence $x_n=1/n$ with $n\in\mathbb{N}$ is good for my porposes, and I have to deal with more general sequences (covergent ones) on $\mathbb{R}^2$. So let $y_n$ be a convergent sequence on $\mathbb{R}^2$ (let us consider $y_n\neq y_m$ if $m\neq n$). My question is: Is there a homeomorphism $\phi:\mathbb{R}^2\to\mathbb{R}^2$ such that $\phi(1/n,0)=y_n$ for all $n\in\mathbb{N}$ (or at least for $n\geq k$, for some $k\in\mathbb{N}$)?
 A: I prefer to write your sequence as $p_n =(x_n,y_n)$. If $p_n \to p$ and some $p_k = p$, it is impossible to find a homeomorphism $\phi$ such that $\phi(p_n) = (1/n,0)$ for all $n$ because $\phi(p_k) = \phi(p)  =\lim_{n \to \infty} \phi(p_n) = (0,0)$.
A positive answer to your question can be obtained if we either

*

*exclude convergent sequences $(p_n)$ such that $p_k =  \lim_{n \to \infty} p_n$ for some $k$ and prove that there exists a homeomorphism $\phi$ such that $\phi(p_n) = (1/n,0)$ for all $n$


*allow that $p_k =  \lim_{n \to \infty} p_n$ for some $k$ and prove that there exists a homeomorphism $\phi$ such that $\phi(p_n) = (1/n,0)$ for all $n \ne k$. Note that there exists at most one index $k$ such $p_k =  \lim_{n \to \infty} p_n$. This is a strong form of your "at least" variant.
We consider case 1. The proof is technically complicated.
W.l.o.g. we may assume that $p_n \to 0$ (if $p_n \to p$, then the translation $\tau(z) = z - p$ is a homeomorphism and $p_n' = \tau(p_n)$ is a sequence with this property).
We now need the following
Lemma. Let $R = [a,b] \times [c,d]$ be a closed rectangle. Then for any two points $p, q \in \operatorname{int} R = (a,b) \times (c,d)$ there exists a homeomorphism $h : R \to R$ keeping the boundary of $R$ fixed such that $h(p) = q$.
We do not give a proof of this intuitively clear lemma here (this would be worth another question).
We write $\lvert R \rvert = \max(b-a,d-c)$.
Step 1.
Since all $p_n \ne 0$ and $p_m \ne p_n$ for $m \ne n$, we can construct inductively a sequence of pairwise disjoint rectangles $R_n$ such that $p_n \in \operatorname{int} R_n$, $0 \notin R_n$ and $\lvert R_n \rvert < 1/n$.
Choose inductively boundary-fixing homeomorphisms $h_n$ on $R_n$ such that the $y$-coordinates of the $h_n(p_n)$ are pairwise disjoint and $\ne 0$. This is possible because all $[c_n,d_n]$ are infinite.
The $h_n$ and the identity on $E = (\mathbb R^2 \setminus \{0\} \setminus \bigcup_n \operatorname{int} R_n$ fit together to a homeomorphism $h$ on $\mathbb R^2 \setminus \{0\}$: The map is well-defined because $h_n = id$ on $R_n \cap E$; it is clearly a bijection. It is continuous because it is continuous on all $R_n$ and on $E$ and
the $R_n$ and $E$ form a locally finite closed cover of $\mathbb R^2 \setminus \{0\}$. The continuity of $h^{-1}$ is verified analogously. Defining $h(0) = 0$ we get a bijection $h$ on $\mathbb R^2$. The maps $h$ and $h^{-1}$ are continuous in all points $\ne 0$. Moreover, $h$ is also continuous in $0$: Given $\epsilon > 0$, choose $N$ such that $R_n \subset U_\epsilon(0)$ for $n  \ge N$. This is possible since $p_n \in U_{\epsilon/2}(0)$ for $n \ge N_1$ and $\lvert R_n \rvert < 1/n$. Then $U = U_\epsilon(0) \setminus \bigcup_{n=1}^{N-1}$ is an open neigborhood of $0$ such that $h(U) \subset U_\epsilon(0)$. The continuity of $h^{-1}$ in $0$ is verified analogously.
Step 2.
Using the above $h$, we can assume w.l.o.g. that the $y_n$ are  pairwise distinct and $\ne 0$.  We can now construct inductively a sequence of pairwise disjoint rectangles $R'_n$ such that $0 \notin R'_n$, $p_n =(x_n,y_n)$ and $(1/n,y_n)$  are contained in $\operatorname{int} R'_n$ and $\lvert R'_n \rvert \to 0$. The last condition can be achieved because $x_n \to 0$. Choose boundary-fixing homeomorphisms $h'_n$ on $R'_n$ such that the $h'_n(p_n) = (1/n,y_n)$. As in step 1 the $h'_n$ and the identity on  $E'' = \mathbb R^2 \setminus \bigcup_n \operatorname{int} R'_n$ fit together to a homeomorphism $h'$ on $\mathbb R^2$ such that $h'(0) = 0$.
Step 3.
Using the above $h'$, we can assume w.l.o.g. that $x_n = 1/n$ for all $n$. We can now construct inductively a sequence of pairwise disjoint rectangles $R''_n$ such that $0 \notin R''_n$, $p_n =(1/n,y_n)$ and $(1/n,0)$  are contained in $\operatorname{int} R''_n$ and $\lvert R''_n \rvert \to 0$. The last condition can be achieved because $y_n \to 0$. Choose boundary-fixing homeomorphisms $h''_n$ on $R''_n$ such that the $h''_n(p_n) = (1/n,0)$. As in step 1 the $h''_n$ and the identity on  $E'' = \mathbb R^2 \setminus \bigcup_n \operatorname{int} R''_n$ fit together to a homeomorphism $h''$ on $\mathbb R^2$ such that $h''(0) = 0$.
