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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)

QED
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  • 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 $(e,e)$. – user284331 Dec 09 '17 at 06:35
  • Exactly, but this is a problem from the book on topology by Munkres – QED Dec 09 '17 at 06:36
  • Hint: For $P,Q\in \mathbb{R}^2\backslash A$, consider the straight lines through $P$ and lines through $Q$. – Phil. Z Dec 09 '17 at 06:37
  • https://math.stackexchange.com/questions/155952/arcwise-connected-part-of-mathbb-r2 – user284331 Dec 09 '17 at 06:38
  • @Phil.Z Ok, so there must be a line through $P$ and another line through $Q$ which intersect. Right? – QED Dec 09 '17 at 06:41
  • As @Brian M. Scott has indicated in the link, yes. – user284331 Dec 09 '17 at 06:42
  • Thanks a lot. But the case of $\Bbb{R}$ was easy to visualise through the concept of lines passing through a point. What about a general case of an uncountable connected space $X$. – QED Dec 09 '17 at 06:45
  • @AbishankaSaha A general uncountable connected space could be disconnected by removing a countable set, indeed even a finite set, indeed even a one-point set. – Angina Seng Dec 09 '17 at 06:58
  • Indeed, $\mathbb{R} \setminus {0}$ is not path-connected. – mechanodroid Dec 09 '17 at 07:43
  • Does the context that $X$ is connected, and we are talking about the space $X^2$ make any difference? @LordSharkTheUnknown – QED Dec 09 '17 at 07:48
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    IMO this is just "a soft duplicate" since it asks the second, more general question. Here is a counterexample for this second question: Consider $ℝ^ℝ$ (with the product topology). It is path-connected as a product of path-connected spaces. Moreover $ℝ^ℝ \times ℝ^ℝ \cong ℝ^ℝ$. But singleton disconnects $ℝ^ℝ$, hence $ℝ^ℝ-x$ is not even connected (details). – Kamil Jul 03 '24 at 20:12

2 Answers2

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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.

William Elliot
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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 $L_P \cap L_0 \ne \emptyset$.

CopyPasteIt
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