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I am thinking about whether the two spaces $\Bbb{R}^3 \setminus \{0 \}$ and $\Bbb{R}^3 \setminus \{0,1 \}$ are homeomorphic or not?

I guess they are not homeomorphic but cannot find out the proper reason. Till now I have come to the following :

$S^2$ is a deformation retract of $\Bbb{R}^3 \setminus \{0\}$ where as I think one can deform the space $\Bbb{R}^3 \setminus \{0,1 \}$ on to two spheres with a single common point, i.e. Wedge of two Spheres ( For this I try to see the deformation visually). But this means both of the space has trivial First fundamental group. So I think this idea didn't work...!!

So how can I distinguish these to space topologically. Any suggestion is appreciated. Thank you.

P.S: Let me clear that I am very new to Algebraic topology. I recently started the first fundamental groups and its properties and try to use it to distinguish two spaces. The spaces in the question is a very random that I thought that it could be solved using fundamental groups. So if this two spaces cannot be distinguished using General Topology and tools in First Fundamental group then let me know. Thank you..

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    $\begingroup$ "trivial First fundamental group" then look at the second. $\endgroup$
    – Arthur
    May 15, 2019 at 11:01
  • $\begingroup$ If you know about homology or Euler characteristic, you could check those too $\endgroup$ May 15, 2019 at 11:05
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    $\begingroup$ what do you mean by $\Bbb R^3\setminus \{0,1\}$? Because $\Bbb R^3$ consists of ordered tuples. $\endgroup$ May 15, 2019 at 11:05
  • $\begingroup$ @RyleeLyman I don't know homology or Euler characteristics....I am studying Homotopy nowdays. $\endgroup$
    – sigma
    May 15, 2019 at 11:06
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    $\begingroup$ When I said "then look at the second", I didn't mean the second space. I meant the second fundamental group (which is usually called "homotopy group" rather than "fundamental group", but still). $\endgroup$
    – Arthur
    May 15, 2019 at 11:13

3 Answers 3

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If you don't want to use algebraic invariants you can use ends. I showed in this post that $\Bbb R^3-\{0\}$ (or $S^2\times \Bbb R$) has two ends. You can convince yourself (or prove) that $\Bbb R^3-\{0,(1,0,0)\}$ has three ends.

Edit: For a proof you could take the sequence of compact subsets $$K_n=B_c(0,n)-B_o(0,1/n)\cup B_o(1,1/n).$$ $K_n$ is just ball getting larger as $n$ increases, with two holes in it around $0$ and $1$ getting smaller. By construction $(\Bbb R^3-\{0,1\})-K_n$ has three components: $B_o(0,1/n)$, $B_o(1,1/n)$ and $\Bbb R^3-B_c(0,n)$. Also the collection $\{\stackrel{\circ}{K_n}\}_n$ covers $\Bbb R^3-\{0,1\}$. This tells you how to create the three ends.

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    $\begingroup$ How $K_n$ is compact? Its a open ball minus a union of two open balls inside it...its not closed...!! $\endgroup$
    – sigma
    May 15, 2019 at 11:43
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$\Bbb R^3 - \{0\}$ deformation retracts to $S^2$ , whereas $\Bbb R^3 - \{0,1\}$ is homotopic (actually deformation retracts) to $S^2 \lor S^2$ but $S^2$ is clearly not homotopic to $S^2 \lor S^2$ since, $H_2 (S^2) \cong \Bbb Z$ but $H_2 (S^2 \lor S^2) \cong \Bbb Z^2$ , thus $\Bbb R^3 - \{0\}$ is not homotopic to $\Bbb R^3 - \{0,1\}$ , and hence in particular not homeomorphic!

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    $\begingroup$ I guess the OP said he is new. So using homology is a big tool for this problem. $\endgroup$ May 15, 2019 at 15:33
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I'm assuming you'd like a hint but also be given the opportunity to work out the details yourself. If the two spaces were homeomorphic so would be their 1-point compactifications. It's fun to work out what these are. Using the van Kampen theorem one can show that the fundamental groups of their 1-point compactifications are groups that are probably well-known to you and which are not isomorphic groups.

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  • $\begingroup$ That's a cool way to do it with just $\pi_1$. Nice. $\endgroup$ May 16, 2019 at 4:45

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