Is $\mathbb{R}^3\setminus\{0\}$ homeomorphic to $\mathbb{R}^3\setminus B_{\epsilon}$

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.

• 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) – Maxime Ramzi Feb 6 at 11:22
• @Max $B_{\epsilon}$ is the closed ball – vidyarthi Feb 6 at 11:32
• 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 $\Bbb R^1$. – Omnomnomnom Feb 6 at 11:44
• 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. – Omnomnomnom Feb 6 at 11:47
• 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. – emacs drives me nuts Feb 6 at 12:00

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

• The function is required to have a continuous inverse too (which this happens to have) – skyking Feb 6 at 11:37
• This is not a homeomorphism and it's not well defined. $r+ \epsilon$ is not a defined operation. – HelloDarkness Feb 6 at 11:38
• @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. – Omnomnomnom Feb 6 at 11:41
• Oh, I see! Thanks! – HelloDarkness Feb 6 at 11:43

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 $$n\geq 1$$.