$\Bbb R^n-\{k$ points$\}$ are all homeomorphic If $p$ and $q$ are any two points in $\Bbb R^n$, then $\Bbb R^n-\{p\}$ and $\Bbb R^n-\{q\}$ are homeomorphic, by a translation. Is this true for $k$ points? I.e., if $p_1,\dots,p_k,q_1,\dots,q_k \in \Bbb R^n$ then do we necessarily have $\Bbb R^n -\{p_1,\dots,p_k\}\cong \Bbb R^n-\{q_1,\dots,q_k\}$?
 A: Yes, this is true, and as AlexL says it's kind of annoying to write down such a homeomorphism explicitly, although for $k = 2$ a translation, a rotation, and a scale suffices, and for $n = 1$ we just get a union of open intervals (two of which are infinite) and we can scale each interval separately.
I think this works: they are in fact diffeomorphic, and we can write down a diffeomorphism from one to the other by writing down a vector field on $\mathbb{R}^n$ whose flow $\phi_t$ at some time $t$ satisfies $\phi_t(p_i) = q_{\pi(i)}$ for some permutation of the indices $\pi$ (although we don't really need the permutation when $n \ge 2$). We can do this by writing some smooth curves connecting each pair of points, parameterized so that they're all traversed in the same time $t$, then writing down a smooth vector field which restricts to the velocity vector of each curve. There's a huge amount of slack available in how to do this (one of the reasons it's kind of annoying to write down); we can do it using bump functions, for example.
