# Elementary proof that $\mathbb{R}^3 \setminus \{p_1,\dots,p_n\}$ is not homeomorphic to $\mathbb{R}^3$

I was wondering if there were a proof of the fact that $$\mathbb{R}^3 \setminus \{p_1,\dots,p_n\} \: \text{is not homeomorphic to} \: \mathbb{R}^3$$ for every $$n \geq 1$$ that does not use cohomology or higher homotopy groups techniques (of course the fundamental group is allowed).

• The set of ends of a locally compact space (say locally path connected and path connected) is easy to compute. For $S^d$ minus $n$ points, $d\ge 2$, the number of ends is $n$. Idem for a closed $d$-dimensional manifold minus $n$ points, $d\ge 2$.
– YCor
Jan 6, 2021 at 18:01
• en.wikipedia.org/wiki/End_(topology) I also posted an answer.
– YCor
Jan 6, 2021 at 18:07
• Why is the fundamental group OK, but not higher homotopy groups? Jan 6, 2021 at 18:08
• @LSpice because it is meant for a basic course in Algebraic Topology, which covers only up to the fundamental group
– gigi
Jan 6, 2021 at 18:10
• Closing seems a bit drastic. What if the OP had asked instead for a list of different "elementary" proofs? There are many such questions on MO. Jan 6, 2021 at 18:58

The fundamental group of the one-point compactification of $$\mathbf R^3 \setminus \{p_1,\ldots,p_n\}$$ is a free group on $$n$$ generators, for any $$n \geq 0$$.

• Really interesting, where can I find this result?
– gigi
Jan 6, 2021 at 18:51
• @gigi Prove it! It is a good exercise. Equivalently, this is $S^3$ with $n+1$ distinct points identified. That may help, but the description Dan gives is enough.
– mme
Jan 6, 2021 at 18:55
• Wow, what a cool fact! Jan 7, 2021 at 12:20

Let $$P_n$$ be the property for a Hausdorff topological space $$X$$: for every compact subset $$K$$ of $$X$$, there exists a compact subset $$L$$ of $$X$$ such that $$K\subset L$$ and $$X-L$$ has exactly $$n$$ components. Obviously $$P_n$$ is a homeomorphism invariant.

Let $$X$$ be a compact connected $$d$$-dimensional manifold, possibly with boundary, minus $$n$$ points, with $$d\ge 2$$. One can check elementarily that $$X$$ satisfies $$P_n$$ but not $$P_{n-1}$$. This applies to $$\mathbf{R}^{d\ge 2}$$ minus $$n-1$$ points ($$=\mathbf{S}^d$$ minus $$n$$ points).

• I'm sure it really is clear, but why does $X$ clearly satisfy $P_n$ but not $P_{n - 1}$? Jan 6, 2021 at 18:10
• I wouldn't say clear, but it's elementary and I think it's more interesting to figure it out than reading a proof.
– YCor
Jan 6, 2021 at 18:12
• +1. It certainly exercises the brain... How did you come up with this? Jan 6, 2021 at 18:32
• @YaakovBaruch: This is just a translation of Yves' earlier comment on the number of ends. Jan 6, 2021 at 18:43
• @‍YCor's earlier comment referenced by @MoisheKohan. Jan 7, 2021 at 0:20