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The question is in the title. In particular, I am interested in the case when $n≥2$.

Maybe if $S_1$ is dense and $S_2$ is not, then $\mathbb{R}^n - S_1$ is not homeomorphic to $\mathbb{R}^n - S_2$, but I don't know if this is true, and if so, how to show this rigorously?

For example, is $\mathbb{R}^n-\mathbb{Q}^n$ homeomorphic to $\mathbb{R}^n-\mathbb{Z}×\{0\}^{n-1}$?

We cannot use a connectedness argument here (for $n≥2$), because removing countably many points from $\mathbb{R}^n$ still leaves it path connected (take any two points $x$ and $y$ which aren't removed, then look at the plane through these two points. since there are uncountably many lines through a point but only countably many points removed, we can find two intersecting lines, one through $x$ and one through $y$ such that these lines don't contain any removed point, and this gives a path between the points).

Further, the answer in Are $\Bbb R^2\setminus \Bbb Q^2$ and $\Bbb R^2\setminus \Bbb Q^2\cup \{(0,0)\}$ homeomorphic? tells us that $\mathbb{R}^n - S_1$ is homeomorphic to $\mathbb{R}^n - S_2$ if both $S_1$ and $S_2$ are countable and dense. So any counter example would require atleast one of them to be not dense.

What if $S_1$ and $S_2$ are required to be countably infinite?

(If we are allowed to remove uncountably many points, then it's much easier to get examples. For example, $(a,b)^n$ is homeomorphic to $(c,d)^n$ for any $a<b$ and $c<d$, and in both cases we have removed uncountably many points.)

Avyaktha Achar
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    The two examples you mentioned are not homeomorphic since the second is locally compact and the first is not. – Moishe Kohan Mar 09 '25 at 14:35
  • If you want to get something less trivial, one can prove that $\mathbb C\setminus \mathbb Z$ is not even homotopy-equivalent to $\mathbb C\setminus {\frac{1}{n}: n\in \mathbb N}$. – Moishe Kohan Mar 09 '25 at 15:23

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For any $n\geq 1$, the spaces $\mathbb{R}^n - \{\text{one point}\}$ and $\mathbb{R}^n - \{\text{two points}\}$ are not homeomorphic. This can be shown using homotopy groups or homology. For example in $\mathbb{R}^2$ these two spaces contain different numbers of "essentially different" simple incontractible loops, but this number is a homeomorphism invariant.

Alex B.
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  • Nice, upvoted since it's an answer to the question asked, since finite sets are countable. I wonder if the OP meant countably infinite. – Ethan Bolker Mar 09 '25 at 14:21
  • Yes, this answers my question, but I am more interested in the case when we remove countably infinitely many points. I have added that to my question... – Avyaktha Achar Mar 09 '25 at 14:32
  • @AvyakthaAchar Then the question is becoming a moving target, and the answers become obsolete almost as soon as they are posted. Instead of repeatedly changing the question, you should ask a new one (after thinking carefully what it is that you actually want to ask). – Alex B. Mar 09 '25 at 14:34
  • I am sorry, I had $n≥2$ and countably infinite in mind in the beginning, but I forgot to write them in the original question. The answers so far are perfectly valid and I've upvotes them, but I am also looking for examples in the other cases that I mentioned. – Avyaktha Achar Mar 09 '25 at 14:37
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For $n=1$, one can consider $S_1=\Bbb Z$ and $S_2=\{1/n:n\in\Bbb N\}$. The first has locally connected complement, but the second does not, since any $0$ neighbourhood of zero contains a component of the form $(1/(n+1),1/n)$.

This works for all $n\ge2$, too, (embed $S_1,S_2$ into the first coordinate axis) but the method of proof we use to distinguish the complements is different. Local compactness is the trick (thanks to @Christophe Boilley). One space is locally compact but the second is not: any neighbourhood of zero contains some punctured open $B(1/n,\delta)\setminus\{1/n\}$ and we can take a sequence tending to $1/n$, then; then all neighbourhoods of zero have a sequence not convergent in this space, so it cannot be locally compact by the compact iff. sequentially compact criterion for metrisable spaces.

FShrike
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  • This is a nice example for the case when $n=1$. I forgot to mention in the question, but in particular I want examples for $n≥2$. I'll edit the question now... – Avyaktha Achar Mar 09 '25 at 13:32
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    Your example can be embedded in $\mathbb R^n$ for $n>1$. The null vector in $(\mathbb R\setminus S_2)\times \mathbb R^{n-1}$ is the only one without compact neighborhood. – Christophe Boilley Mar 09 '25 at 15:08
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    @ChristopheBoilley Ah that’s nice. I saw of course to embed my example in higher dimension but did not think of a way to distinguish the spaces. – FShrike Mar 09 '25 at 15:25
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    @AvyakthaAchar It works now for all $n$. Thanks to Christophe – FShrike Mar 09 '25 at 15:30