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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

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

2 Answers 2

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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 $n \geq 0$.

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

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

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