4

I'm trying to show that if $A$ is a countable subset of $\mathbb{R}^{2}$ then $\mathbb{R}^{2}\backslash A$ is path connected. I already have the idea of the proof. It is clear to me that for every two points in the plane there is an uncountable collection of paths between them that intersect only at the edges. Thus given two points not in $A$ since $A$ is countable it is impossible that all these paths intersect $A$. Thus for every two points in $\mathbb{R}^{2}\backslash A$ there is path between them that doesn't intersect $A$

What I want is an explicit construction of such a collection of paths given two points $\left(x_{1},y_{1}\right),\left(x_{2},y_{2}\right)\in\mathbb{R}^{2}$. Help would be appreciated.

Serpahimz
  • 3,971
  • I think this was answered before. – Asaf Karagila Jun 25 '13 at 09:58
  • 2
    See http://math.stackexchange.com/questions/240453/if-a-subset-mathbbr2-is-countable-does-mathbbr2-setminus-a-path-conn and http://math.stackexchange.com/questions/155952/arcwise-connected-part-of-mathbb-r2 – Asaf Karagila Jun 25 '13 at 09:59
  • Also http://math.stackexchange.com/questions/145912/formal-proof-that-mathbbr2-setminus-mathbbq-times-mathbbq-subset and http://math.stackexchange.com/questions/65705/is-mathbbr2-minus-a-countable-number-of-points-skew-manhattan-connected and http://math.stackexchange.com/questions/380918/prove-that-the-complement-of-mathbbq-times-mathbbq-in-the-plane-math – Asaf Karagila Jun 25 '13 at 10:05
  • Thanks Asaf. I did do a search for "countable path connected" but I didn't look into that first post since it wasn't marked as answered :( From all the posts you referred to only the first one (which isn't marked answered) offers an explicit construction by the way. – Serpahimz Jun 25 '13 at 10:07
  • @Serpahimz: It’s no more constructive than the argument suggested in my hint for the second one: in one you have to pick a point on a specific line in such a way that two line segments miss $A$, and in the other you have to pick two angles to accomplish the same end. Neither can be made constructive unless you have a precise description of $A$. – Brian M. Scott Jun 25 '13 at 19:00
  • Is there a way to prove this using contradiction. Namely, suppose that $\mathbb{R}^{2}\setminus A$ is disconnected. Then construct a continuous surjective map $j: \mathbb{R}^{2}\setminus A \to {0,1}$? – Rene Cabrera Jan 18 '17 at 02:15

0 Answers0