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I was using Van Kampen's Theorem and induction to show that $\mathbb{R}^n$ with finitely many points removed is simply connected for $n\geq 3$. However, this made me wonder whether we can remove countably many points and get the same conclusion. I'm not sure how I would go about proving it, nor can I think of a counterexample. I'm not even sure whether $\mathbb{R}^n\setminus\mathbb{Q}^n$ is simply connected.

Anonymous
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  • It's pretty easy to see that $ \mathbb{R}^n \backslash \mathbb{Q}^n$ is simply connected for $n \geq 2$. Think of them as lots of (hyper)planes that are connected together. – Calvin Lin Oct 15 '19 at 20:55
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    @CalvinLin Do you mean $n\geq 3$? – Zubin Mukerjee Oct 15 '19 at 20:56
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    @CalvinLin I assume you mean $n\geq 3$. I get the idea would be to homotope a loop into one of the hyperplanes with an irrational coordinate, but I don't find it obvious how to do that. – Anonymous Oct 15 '19 at 20:57
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    $\mathbb{R}^n \setminus \mathbb{Q}^n$ is path-connected via lines with at least one irrational coordinate. However, path-connectedness is just the "easy" part of proving something is simply connected – Zubin Mukerjee Oct 15 '19 at 20:59
  • Ah, I missed out the 2nd part. – Calvin Lin Oct 15 '19 at 21:00
  • @ZubinMukerjee Yea, I'm familiar with that fact. It's partially what inspired me to ask this question. – Anonymous Oct 15 '19 at 21:00
  • Possibly you can invoke the proof in https://math.stackexchange.com/a/1186988 replacing the word "finite" with "countable". – Chris Culter Oct 15 '19 at 21:05
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    @ChrisCulter I'm not sure that proof works. It appears he uses the fact that $\mathbb{R}^n\setminus X$ is open in $\mathbb{R}^n$ when $X$ is finite, but this fails if $X$ is countable. – Anonymous Oct 15 '19 at 21:08

1 Answers1

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Yes, it is simply connected. See the proof given here: It will apply to any countable subset of any simply-connected topological manifold of dimension $\ge 3$. A related MSE question is here.

Moishe Kohan
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