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Let $A$ be a countable subset of $\mathbb{R}^2$. How to show that $\mathbb{R}^2 - A$ is path-connected?

My effort:

Let $A = \{ a_1, a_2, a_3, \ldots \}$, where $a_n = (\alpha_n, \beta_n) \in \mathbb{R}^2$ for each $n \in \mathbb{N}$.

Let $x_1 = (\xi_1, \eta_1)$ and $x_2 = (\xi_2, \eta_2)$ be any two given points in $\mathbb{R}^2 - A$.

How to find (explicitly) a continuous function $f \colon [0,1] \to \mathbb{R}^2 - A$ such that $f(0) = x_1$ and $f(1) = x_2$?

Or, how to find (explicitly) a polygonal path (i.e. a path consisting of a finite number of line segments ) from $x_1$ to $x_2$ none of whose edges intersects $A$?

Saaqib Mahmood
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  • But the number of points may be infinite. – Nishant May 05 '15 at 13:25
  • Oh yeah, misread the problem, I'll delete the comment. – Gregory Grant May 05 '15 at 13:25
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    In $\mathbb{R}^2$, you can connect any two points by circular arcs. How many distinct circular arcs connecting two given points are there? – Daniel Fischer May 05 '15 at 13:27
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    @DanielFischer Or just two straight line segments, one from each point to a common point on the perpendicular bisector between them. – Arthur May 05 '15 at 13:29
  • @Arthur, do I get it right? Given two points $x, y \in \mathbb{R}^2 - A$, the perpendicular bisector of the segment joining them necessarily has a point not in $A$. So we can join $x$ to that point which in turn gets joined to $y$. But what is the gaurantee that neither of these intermediate paths will intersect $A$? – Saaqib Mahmood May 05 '15 at 13:42
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    Nothing in your construction guarantees that. But how many different such paths is there? And how many points of $A$ can possibly be "in the way"? Can you guarantee that there exists enough paths that at least one of them does not contain any points of $A$? – Arthur May 05 '15 at 13:46
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    Yes, @Arthur, since there are uncountably many such paths and since there are at most countably many points of $A$ to be lying on these uncountably many paths, there must be at least one path not containing points of $A$. Is it correct logic? But I wonder how to formulate a "continuous function" argument! – Saaqib Mahmood May 05 '15 at 13:58
  • This has already been asked here: http://math.stackexchange.com/questions/1193545/discrete-subset-of-mathbb-r2-such-that-mathbb-r2-setminus-s-is-path-conn/1193610#1193610 – Jesus RS May 05 '15 at 17:27
  • @Brian M. Scott, can you please answer my question using the continuous-function argument? – Saaqib Mahmood May 06 '15 at 05:05
  • @Saaqib: You can’t really find an explicit function, since you don’t have an explicit description of $A$. You can observe that there must be a point $p$ on the perpendicular bisector of $\overline{xy}$ such that the segments $\overline{xp}$ and $\overline{py}$ are disjoint from $A$, and you can then let $f(t)=2tp+(1-2t)x$ for $0\le t\le\frac12$ and $f(t)=(2t-1)y+(2-2t)p$ for $\frac12\le t\le 1$, but that’s about as explicit as you can get. – Brian M. Scott May 06 '15 at 20:04

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