# 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/… Commented Jun 9, 2012 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 $?$ Commented May 22, 2013 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. Commented May 22, 2013 at 20:21
• @TeresaLisbon I first asked it as new question but it was closed and repititive requests to open it were discarded in CURED chatroom and they suggested that new questions shouldn't be asked but bounty should be put on old question. Which I did now.
– user775699
Commented Sep 23, 2021 at 6:02
• @No-One Got it, thanks. I think what I'll do, if I plan to answer this, is place your bounty text in my answer, to make sure that my answer targets your query, along with the question in general. Commented Sep 23, 2021 at 9:20

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 $X$ Commented Jun 9, 2012 at 9:32
• @Mohamed: Exactly. (Or if you’re really lucky, one of them is the line through $p$ and $q$.) Commented Jun 9, 2012 at 9:37
• I have not thought about this one!! Thank's!! (Edit [p,q] not [a,b] ) Commented Jun 9, 2012 at 9:51
• From here. In order to show path $f:[0,1]\to \Bbb{R}^2-A$ is continuous.we can construct a function: $f(t)=f_1(t)$ ,if $t\in [0,1/2]$ and $f(t)=f_2(t)$ ,if $t\in [1/2,1]$. $f$ is continuous by pasting lemma. And we're done. Am I right?@Briam M Scott Commented Jul 16, 2020 at 11:57
• @AmanPandey: Yes, that’s right. Commented Jul 16, 2020 at 16:35

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.)

I am unable to understand the answer of Brian M Scott. I didn't asked user: mohammed or user: Brian M Scott because Both are inactive for a pretty long time. My question: I am looking for a rigorious argument. The answer of brian is not rigorious and also I was unable to complete the argument, understand the comment of user: mohammmed. Pathwise connectedness is not my strong point. > E

Let $$p, q$$ be two distinct points in $$\mathbb{R}^2 \setminus D$$. If the straight line segment joining $$p$$ to $$q$$ lies in $$\mathbb{R}^2 \setminus D$$, then you are done: one possible parametrization of the path is $$\gamma(t) = (1 - t)p + tq$$, $$0 \leq t \leq 1$$.

So, suppose that the above does not work. Now, note that there are uncountably many straight lines passing through the point $$p$$. Since $$D$$ is only countable, in particular there are uncountably many straight lines passing through $$p$$ and not intersecting $$D$$. Fix any one such line $$L_1$$. Once more, there are uncountably many straight lines passing through $$q$$ that also intersect $$L_1$$. In particular, there are uncountably many such lines that also do not intersect $$D$$. Fix any one such line $$L_2$$.

Suppose that $$L_1$$ and $$L_2$$ intersect at the point $$r$$. Then, you can take the path joining $$p$$ and $$q$$ to be the one that starts at $$p$$, traverses along the line $$L_1$$ to the point $$r$$, then traverses along the line $$L_2$$ to the point $$q$$. One possible parametrization of this path is $$\gamma(t) = \begin{cases} (1 - 2t)p + 2tr, & 0 \leq t \leq \tfrac{1}{2};\\ (2 - 2t)r + (2t - 1)q, & \tfrac{1}{2} \leq t \leq 1. \end{cases}$$ Note that the pasting lemma shows that $$\gamma$$ is indeed continuous.

Lastly, for the sake of completeness, note that this path is injective, so it defines an embedding $$[0,1] \to X$$, as required by the definition of an arcwise connected space.

As for how one arrives at these specific parametrizations, one first needs to know the parametric equation of a straight line. For instance, see Parametric form of a line for a quick refresher. Secondly, one needs to adjust the parameter $$t$$ by scaling and translating in order to get it to lie in the "correct" range for one's purposes. This is another linear change, so it should not be too difficult.

But, just to belabor the point further, here's one way you can mentally figure out the parametrization $$\gamma$$ in the second case. To join the points $$p$$ and $$r$$, we want to vary $$t$$ from $$0$$ to $$1/2$$ such that at $$t = 0$$ we are at $$p$$ and at $$t = 1/2$$ we are at $$r$$. So, if the parametrization for this part looks like $$\gamma(t) = f(t)p + g(t)r$$, then we want $$f(0) = 1$$ and $$f(1/2) = 0$$, and we also want $$g(0) = 0$$ and $$g(1/2) = 1$$. Can we find linear functions $$f(t) = \alpha_1 t + \beta_1$$ and $$g(t) = \alpha_2 t + \beta_2$$ that satisfy these conditions? Sure, just substitute those values of $$(t, f(t))$$ and $$(t, g(t))$$ that we wrote down for $$t = 0, 1/2$$ into the respective equations to solve for $$\alpha_i, \beta_i$$. Repeat the process for the line joining $$r$$ to $$q$$, and you are done.

• For completeness, you should at least state that your path is also injective, which implies that it defines an embedding $[0,1]\to X$, as required by the definition of an arcwise connected space. Commented Sep 28, 2021 at 17:57

Ok I will take a stab at clarifying Brian M. Scott's Answer instead of writing a fully different answer.

Let $$p,q\in \mathbb{R}-D$$. Then Define $$A:=\{\text{ Lines Through }p\}$$, $$B:=\{ \{\text{ Lines through }q\}$$.

One could write the definitions of these sets more formally if desired, but the meaning is clear. I claim the following: $$\exists l\in A, \text{ such that } l\cap D=\emptyset.$$ Similarly $$\exists n\in B, \text{ such that } l\cap D=\emptyset.$$ Lets us understand why this should be clear. Now define a mapping $$m:\mathbb{R} \to A$$ by $$m(x):=$$ the line in $$A$$ with slope $$x$$. This mapping is clearly $$1-1$$ and onto (almost see below*), as any line through a point $$p$$ is determined by its slope, and any slope uniquely determines a line through $$p$$. You could formalize this more by writing any line through $$p$$ in point slope form, but I leave that to the reader.

A similar argument follows for $$B$$, thus $$||A||=||B||=||\mathbb{R}||$$.

Now let us show that there are lines in $$A$$ and $$B$$ which do not contain points in $$D$$. define a map $$\phi:D\to A\times B$$, by $$\phi(d):=\{(l,m)\in A\times B| d=l\cap m\}$$. That is each point $$d\in D$$ gets mapped to the pair of lines in $$A$$ and $$B$$ which meet at $$d$$. This mapping is $$1-1$$, as any two points determine a unique line. By assumption $$D$$ is countable, therefor $$\phi$$ cannot be onto thus there exists a pair of lines, $$(l^*,m^*)$$, which do not contain any points in $$D$$. This follows as $$A\times B- \phi(D)\neq \emptyset$$.

we have a piecewise linear path $$p \to l^*\cap m^* \to q$$. QED.

The Crux of the argument, here is where we assert that the mapping $$\phi$$ cannot be onto, given the assumption that $$D$$ was countable.

One small note, the map $$m$$ is not quite onto, as the vertical line through $$p$$ has undefined slope. However, this doesn't disturb the argument as the main idea is that $$A$$ and $$B$$ are both uncountable. We might also be able to circumvent this by taking a mapping from the extended reals.

• Great stuff, really nice answer, thanks for adding the clarification in the first line. Commented Sep 25, 2021 at 12:32