# Arcwise connected part of $\mathbb R^2$

Here's a question that I share: Show that if $D$ is a countable subset of $\mathbb R^2$ (provided with its usual topology) then $X=\mathbb R^2 \backslash D$ is arcwise connected.

• This is even true if you only require $D$ to have lebesgue-measure 0: math.stackexchange.com/questions/77791/… – Listing Jun 9 '12 at 9:20
• I have two further questions: 1) Is this result holds in higher dimension i.e., on $\mathbb{R}^n$, where $n \in \mathbb{N}$. 2) Can one characterize the topological space for which this kind of properties hold $?$ – Tapan May 22 '13 at 18:47
• @Tapan: It would be much preferred for you to post the questions from your answer/comment as a new separate question. You can always add a link to this question to make it clear that it is connected to a previous question. – user642796 May 22 '13 at 20:21

HINT: Not only is $\Bbb R^2\setminus D$ arcwise connected, but you can connect any two points with an arc consisting of at most two straight line segments.
Suppose that $p,q\in\Bbb R^2\setminus D$. There are uncountably many straight lines through $p$, and only countably many of those lines intersect $D$, so there are uncountably many straight lines through $p$ that don’t hit $D$. Similarly, there are uncountably many straight lines through $q$ that don’t hit $D$. Can you finish it from here?
• Yes, I can : two of them at lest intersect. One auther way is to consider the bissector $\Delta$ of $[ab]$ and all straight lines $[p,\delta]\cup[\delta,q]$ where $\delta \in \Delta$, On of them at lest is a subset of $X$ – Mohamed Jun 9 '12 at 9:32
• @Mohamed: Exactly. (Or if you’re really lucky, one of them is the line through $p$ and $q$.) – Brian M. Scott Jun 9 '12 at 9:37
Any two points in $\mathbb R^2\setminus D$ are connected by uncountably many disjoint arcs of circles, of which only countably many may intersect $D$.