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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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    $\begingroup$ Is $B_\epsilon$ the open ball or the closed ball ? (though no matter what, your heuristic is not right : take any two non homeomorphic spaces $X,Y$ and consider $\coprod_{n\in \mathbb N} X \sqcup \coprod_{n\in \mathbb N}Y$, then you can remove one copy of $X$ and get the same thing, or remove one copy of $Y$ and get the same thing too : you can take the complements of nonhomeomorphic things and get homeomorphic things - and in fact if you take the closed ball, you'll see an even more natural example) $\endgroup$ – Maxime Ramzi Feb 6 at 11:22
  • $\begingroup$ @Max $B_{\epsilon}$ is the closed ball $\endgroup$ – vidyarthi Feb 6 at 11:32
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    $\begingroup$ You might find it more intuitive that these sets are homeomorphic (even though the complements are not) if you consider the version of this problem over $\Bbb R^1$. $\endgroup$ – Omnomnomnom Feb 6 at 11:44
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    $\begingroup$ Interestingly, the fact that the complements of homeomorphic sets can be non-homeomorphic makes an appearance in knot-theory. Two knots are considered the same when there complements are homotopy-equivalent, but each "knot itself" is a loop in $\Bbb R^3$ which is homeomorphic to the unit circle. $\endgroup$ – Omnomnomnom Feb 6 at 11:47
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    $\begingroup$ Interestingly, the fact that the complements of homeomorphic sets can be non-homeomorphic makes an appearance in General Relativity in questions like "what is the topology of a black hole's singularity?". There is no meaningful answer to that question. $\endgroup$ – emacs drives me nuts Feb 6 at 12:00
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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 $\mathbb R^n$.

I dont think so. This is because the origin and the ball are not homeomorphic

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 $(r,\varphi,\theta)\mapsto (r+\epsilon,\varphi,\theta)$.

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  • $\begingroup$ The function is required to have a continuous inverse too (which this happens to have) $\endgroup$ – skyking Feb 6 at 11:37
  • $\begingroup$ This is not a homeomorphism and it's not well defined. $r+ \epsilon$ is not a defined operation. $\endgroup$ – HelloDarkness Feb 6 at 11:38
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    $\begingroup$ @HelloDarkness the map $f(r,\theta) = (r + \epsilon, \theta)$ is a well-defined map on polar coordinates, and polar coordinates with $r>0$ and $0 \leq \theta < 2 \pi$ correspond uniquely to points in the punctured plane. $\endgroup$ – Omnomnomnom Feb 6 at 11:41
  • $\begingroup$ Oh, I see! Thanks! $\endgroup$ – HelloDarkness Feb 6 at 11:43
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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 $n\geq 1$.

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