Is the punctured $3d$ space, $\mathbb{R}^3\setminus\{(0,0,0)\}$ homeomorphic to $\mathbb{R}^3\setminus B_{\epsilon}$, where $B_{\epsilon}$ is a closed ball of radius $\epsilon$?
I don't think so. This is because the origin and the ball are not homeomorphic, so I think their complements must also not be. Any hints? Thanks beforehand.
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Another way to see that they are homeomorphic (and perhaps another way to understand it) is to use $x\mapsto \frac{x}{||x||^2}$. This is a homeomorphism $\mathbb R^3\setminus B_\epsilon \to \mathrm{Int}(B_{1/\epsilon})\setminus \{0\}$ (the inverse is given by the same formula).
Now $\mathrm{Int}(B_{1/\epsilon})$ is an open ball, so it's homeomorphic to $\mathbb R^3$, so we get the desired result.
Essentially, with $\epsilon = 1$, you're turning $\mathbb R^3$ around itself along the radius $1$ sphere, $0$ becomes infinity and infinity becomes $0$. This allows me to understand better what's happening : since the "large sphere at infinity" can become $0$, but it can also become a small sphere, the two work the same way.
Note that this proof works for $\mathbb R^n$ for any $n\geq 1$.
Yes. You can map $\mathbb R^3\setminus\{(0,0,0)\}$ continuously to $\mathbb R^3\setminus B_\epsilon$ by means of $r\mapsto r+\epsilon$ provided $B_\epsilon$ is the closed ball and $r$ denotes the radial component of polar coordinates in $\mathbb R^3$. The inverse mapping is obviously $r\mapsto r-\epsilon$.
Moreover, similar applies to any dimension, i.e. to $\mathbb R^n$.
The origin and the ball are not in your (open) sets; hence it does not matter. Would be different if their borders were part of the sets, because the first one would be just $\mathbb R^3$ in that case which is a open set whereas the second one would be neither open nor close.
Note: $r\mapsto r+\epsilon$ is short for $(r,\varphi,\theta)\mapsto (r+\epsilon,\varphi,\theta)$.