Whether $\Bbb{R}^3\setminus \{0\}$ and $\Bbb{R}^3 \setminus \{0,1 \}$ are homeomorphic or not?

Question

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..

Answer 1

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.

Answer 2

$\Bbb R^3 - \{0\}$ 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!

Answer 3

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.