# What does it mean for a sequence of self-homeomorphism of $\mathbb{R}^n$ to converge to a point?

Let $\{f_j\}$ be a family of self-homeomorphisms of $\overline{\mathbb{R}}^n$ and $x,y\in\overline{\mathbb{R}}^n$, where $\overline{\mathbb{R}}^n$ is the one-point compactification of $\mathbb{R}^n$. What does it mean for $\{f_j\}$ to converge c-uniformly to y in $\overline{\mathbb{R}}^n\setminus \{x\}$? That is, what does "$\lim_{j\rightarrow\infty} f_j =y$, c-uniformly in $\overline{\mathbb{R}}^n\setminus {x}$" mean? I'm only confused about what function the point $y$ represents. I would have assumed it meant the function that sends everything to the point $y$, but this is not a bijection on $\overline{\mathbb{R}}^n$.

The source for this question is the paper "Discrete Quasiconformal Groups I" by Gehring and Martin, Proc. London Math. Soc. (3) 55, 1987. Definition 3.3 and Theorem 3.7. My interest here is in understanding convergence groups, in particular their use in defining relatively hyperbolic groups.

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I'm pretty sure it is meant that for every compact subset $K \subset \overline{\mathbb{R}}^{n} \smallsetminus \{x\}$ and every neighborhood $U$ of $y$ there is $N$ such that $f_{j}(K) \subset U$ for all $j \geq N$. Think of a hyperbolic automorphism of the (closed) Poincaré disk. It has an attractive point $y$ and a repelling point $x$ and precisely this behaviour. – t.b. Feb 11 '11 at 23:35
That makes perfect sense, and answers my question. Thanks! – Jeremy Macdonald Feb 12 '11 at 0:54
Very good. I'll post this as an answer, please accept it, so that the question gets closed. – t.b. Feb 12 '11 at 0:58

I'm pretty sure it is meant that for every compact subset $K \subset \overline{\mathbb{R}}^{n} \smallsetminus \{x\}$ and every neighborhood $U$ of $y$ there is $N$ such that $f_{j}(K) \subset U$ for all $j \geq N$. Think of a hyperbolic automorphism of the (closed) Poincaré disk. It has an attractive point $y$ and a repelling point $x$ and precisely this behaviour.