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Possible Duplicate:
Arcwise connected part of $\mathbb R^2$

As the topic,if $A\subset\mathbb{R^2}$ is countable, does $\mathbb{R^2}\setminus A$ path connected??? I know the answer is it is path connected but not sure how to prove it.

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1 Answers1

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Here is a marginal variation:

Choose $x_0,x_1 \in A^c$. Let $d \neq 0$ be orthogonal to $x_1-x_0$. Define the collection of paths $\gamma_\alpha(t) = x_0 + t(x_1-x_0) + \alpha t(1-t) d$. Since $\{\gamma_\alpha\}_{\alpha \in [0,1]}$ is uncountable, there exists $\alpha_0 \in [0,1]$ such that $\gamma_{\alpha_0}[0,1] \cap A = \emptyset$. Hence $A^c$ is path connected.

copper.hat
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  • How do these paths look like? – abeliangrape Jan 06 '20 at 12:55
  • They look like segments of a parabola. It is not hard to generate a little example in the plane. – copper.hat Jan 06 '20 at 14:42
  • Why the downvote after 8 years? – copper.hat Jan 08 '20 at 04:44
  • @abeliangrape: Did you downvote this? – copper.hat Jan 08 '20 at 04:44
  • ?? No, your answer was helpful. Why would I? – abeliangrape Jan 08 '20 at 07:02
  • @abeliangrape: Thanks for reply. Someone downvoted after 8 years, I was curious why. – copper.hat Jan 08 '20 at 15:42
  • @copper.hat +1.. can you give me some reference that how you derive the formula $\gamma_\alpha(t) = x_0 + t(x_1-x_0) + \alpha t(1-t) d$ or tell me some book name where this concept are explain – jasmine Feb 01 '21 at 18:17
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    @jasmine There is no reference, I just made it up. Try an example in $\mathbb{R}^2$, take $x_0 =(0,0), x_1 = (1,0)$ and $d = (0,1)$ and plot a few curves for various values of $\alpha$. – copper.hat Feb 01 '21 at 18:19
  • oks now im thinking@copper.hat $\gamma_\alpha(t) = x_0 + t(x_1-x_0) + \alpha t(1-t) d$

    $=(0,0) +t\left((1,0)-(0,0)\right) + \alpha t(1-t)(0,1)=(t,0) +(0,\alpha t(1-t))=(t,\alpha t(1-t))\implies \gamma_\alpha(t)=(t,\alpha t(1-t))$

    $\gamma_\alpha(t)= (x(t),y(t))$ where $x(t)=t$ and $y(t)= \alpha t-\alpha t^2$

    This mean that in $\gamma_\alpha(t)$ there are uncountably many disjoint curve .Is it correct or not ?

     – jasmine Feb 01 '21 at 18:38
  • Well, the argument is a little subtle. For each $\alpha$ there is a range of points $R_\alpha={ \gamma_\alpha(t) | 0 < t < 1 }$ (the end points are all the same which is why I am ignoring them). If $\alpha \neq \alpha'$ then $R_\alpha $ and $R_{\alpha'}$ are disjoint. Since $\alpha$ ranges through $[0,1]$ there are uncountably many of them. Since $A$ is countable, at most a countable number of $R_\alpha$ contain an element of $A$. Hence there are lots of paths that do not pass through $A$. – copper.hat Feb 01 '21 at 18:54
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    @jasmine Sry, forgot to add the @ thing to my response. – copper.hat Feb 01 '21 at 21:34
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    @copper.hat it oks no problem ,btw thanks u – jasmine Feb 02 '21 at 03:13