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The Assignment:

Let $K_1(0) := \{x \in \mathbb{R}^2: \|x\|_2 < 1\}$ and $S := K_1(0) \setminus \mathbb{Q}^2$. Is S path connected? Explain your answer.

I don't think S is path-connected since we're removing an infinite amount of points from the path-connected $K_1(0)$. I think I should assume that it is path-connected and thus for every $x,y \in S$ there's a continuous function $\gamma$ with $\gamma(x)=0$ and $\gamma(y)=1$, but I cannot find the contradiction.

I'd appreciate any help.

Nhat
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  • So you are saying if i remove infinitely many points from a path connected space i will be left with a space which is not path connected –  Jun 21 '14 at 14:11
  • @PraphullaKoushik That's the way I'm picturing it. I don't have a theorem proving my assertion, but I think it does make a difference if you remove infinitely or finitely many points. If only finitely many points of a connected subset of $\mathbb{R}^2$ are removed, the subset will still easily be path-connected. – Nhat Jun 21 '14 at 14:16
  • and why is it so? –  Jun 21 '14 at 14:16
  • @PraphullaKoushik What now? – Nhat Jun 21 '14 at 14:18
  • I have asked a question.. why do you think if you remove finitely many points you will be left with path connected space? –  Jun 21 '14 at 14:18
  • Well, if I take two arbitrary points in a subset of $\mathbb{R}^2$ which was formerly path connected and draw the line between them, I might get into trouble if one of the points between them was removed. But since every arbitrary ball around my removed point without my removed point is path connected, there is a continuous function "around" the removed point for the two arbitrary points. – Nhat Jun 21 '14 at 14:27
  • now... do the same for countably many points... see that i can have uncountably many lines passing through a point... –  Jun 21 '14 at 14:28

1 Answers1

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The set $S$ is path-connected.

See How is $\mathbb R^2\setminus \mathbb Q^2$ path connected?

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Ewan Delanoy
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