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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.

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This is even true if you only require $D$ to have lebesgue-measure 0:… – 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. – arjafi May 22 '13 at 20:21
up vote 19 down vote accepted

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?

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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
I have not thought about this one!! Thank's!! (Edit [p,q] not [a,b] ) – Mohamed Jun 9 '12 at 9:51

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 $D$.

(This is adapted from one of my student's solutions to a related exam problem.)

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Of course, this notion can be generalized for higher dimension. Because, you can use the same argument (as Brian/Bob) for $\mathbb{R^n}$. Note that, in $\mathbb{R^n}$ we have straight lines and sphere too.

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