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This question already has an answer here:

Let $A\subset\Bbb{R}^2$ be countable. I need to prove that $\Bbb{R}^2\setminus A$ is path connected.

I know that through each of $\Bbb{R}^2\setminus A$, there pare uncountably many straight lines, and as there are only countably many points in $A$, uncountably many of these lines will not contain any point of $A$. But why do I construct a path between any two points.

Also can this result be generalised, so that:

If $X$ is uncountable and $A$ is a countable subset of $X^2$, then should $X^2\setminus A$ be path connected.? (where $X$ and $X^2$ are path connected of course)

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marked as duplicate by GNUSupporter 8964民主女神 地下教會, Brahadeesh, Arnaud D., supinf, José Carlos Santos general-topology Mar 1 at 13:59

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  • $\begingroup$ I just can't imagine if $A$ is the set of all rational coordinates, then how one does the continuous path between, say, $(\pi,\pi)$ and $(e,e)$. $\endgroup$ – user284331 Dec 9 '17 at 6:35
  • $\begingroup$ Exactly, but this is a problem from the book on topology by Munkres $\endgroup$ – Abishanka Saha Dec 9 '17 at 6:36
  • $\begingroup$ Hint: For $P,Q\in \mathbb{R}^2\backslash A$, consider the straight lines through $P$ and lines through $Q$. $\endgroup$ – Phil. Z Dec 9 '17 at 6:37
  • $\begingroup$ math.stackexchange.com/questions/155952/… $\endgroup$ – user284331 Dec 9 '17 at 6:38
  • $\begingroup$ @Phil.Z Ok, so there must be a line through $P$ and another line through $Q$ which intersect. Right? $\endgroup$ – Abishanka Saha Dec 9 '17 at 6:41
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Let a,b be two points on the plain and not in A.
Draw an arc of a circle of radius r with ab as a cord.
There are uncountable many such arcs and as they are pairwise disjoint execept at the endpoints, almost all of them will miss A. Thus a circlular arc from a to b missing A.

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  • $\begingroup$ There are only two semicircles with a given diameter. Now, if you were talking about circular arcs.... $\endgroup$ – Lord Shark the Unknown Dec 9 '17 at 6:56
  • $\begingroup$ @LordSharktheUnknown. r is not fixed. $\endgroup$ – William Elliot Dec 9 '17 at 11:28
  • $\begingroup$ The radius of a circle is always half its diameter. $\endgroup$ – Lord Shark the Unknown Dec 9 '17 at 11:34
  • $\begingroup$ I was going to post a complicated construction but after seeing this, I won't...............+1 $\endgroup$ – DanielWainfleet May 12 '18 at 14:47
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Recall that a countable set can only be partitioned into a countable family of blocks. This does not require the Axiom of Choice (just saying).

Exercise 1: Show that there is a line of $\Bbb{R}^2$ wholly contained in $\Bbb{R}^2\setminus A$.

Exercise 2: Choose a line $L_0 \subset \Bbb{R}^2\setminus A$. Show that any point $P \in \Bbb{R}^2\setminus A$ with $P \notin L_0$ belongs to a line $L_P$ in $\Bbb{R}^2$ such $L_P \subset \Bbb{R}^2\setminus A$ and $L_P \cap L_0 \ne \emptyset$.

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