13
$\begingroup$

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).

Thanks in advance

Improve this question
$\endgroup$
11
  • 6
    $\begingroup$ 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, $d\ge 2$. $\endgroup$
    – YCor
    Jan 6, 2021 at 18:01
  • 2
    $\begingroup$ en.wikipedia.org/wiki/End_(topology) I also posted an answer. $\endgroup$
    – YCor
    Jan 6, 2021 at 18:07
  • 1
    $\begingroup$ Why is the fundamental group OK, but not higher homotopy groups? $\endgroup$
    – LSpice
    Jan 6, 2021 at 18:08
  • 4
    $\begingroup$ @LSpice because it is meant for a basic course in Algebraic Topology, which covers only up to the fundamental group $\endgroup$
    – gigi
    Jan 6, 2021 at 18:10
  • 2
    $\begingroup$ 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. $\endgroup$
    – Yaakov Baruch
    Jan 6, 2021 at 18:58

2 Answers 2

Reset to default
41
$\begingroup$

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 $n \geq 0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Really interesting, where can I find this result? $\endgroup$
    – gigi
    Jan 6, 2021 at 18:51
  • 20
    $\begingroup$ @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. $\endgroup$
    – mme
    Jan 6, 2021 at 18:55
  • $\begingroup$ Wow, what a cool fact! $\endgroup$
    – Nik Weaver
    Jan 7, 2021 at 12:20
24
$\begingroup$

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).

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ I'm sure it really is clear, but why does $X$ clearly satisfy $P_n$ but not $P_{n - 1}$? $\endgroup$
    – LSpice
    Jan 6, 2021 at 18:10
  • 4
    $\begingroup$ I wouldn't say clear, but it's elementary and I think it's more interesting to figure it out than reading a proof. $\endgroup$
    – YCor
    Jan 6, 2021 at 18:12
  • $\begingroup$ +1. It certainly exercises the brain... How did you come up with this? $\endgroup$
    – Yaakov Baruch
    Jan 6, 2021 at 18:32
  • 8
    $\begingroup$ @YaakovBaruch: This is just a translation of Yves' earlier comment on the number of ends. $\endgroup$
    – Moishe Kohan
    Jan 6, 2021 at 18:43
  • 1
    $\begingroup$ @‍YCor's earlier comment referenced by @MoisheKohan. $\endgroup$
    – LSpice
    Jan 7, 2021 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.