# Solution verification: Proving that $\mathbb R^2 \setminus C$ is connected if $C$ is countable.

In an exercise I'm asked to prove the following:

Let $$C$$ be any countable subset of $$\mathbb R^2$$. Prove that the space $$\mathbb R^2 \setminus C$$ is path-connected.

I came up with a proof for this but I'm a little skeptical about whether it's valid or not. This is what I came up with:

So, first let's do this in the complex plain as it will simplify some notation along the way. Let $$C^* \subset\mathbb C$$ such that $$C^* = \{x + iy:(x,y ) \in C\}.$$ Now, we have that $$\mathbb R^2 \setminus C \cong \mathbb C \setminus C^*$$, with $$f(x,y) = x+iy$$ being an homeomorphism.

If $$\mathbb C \setminus C^*$$ is path-connected then $$\mathbb R^2 \setminus C$$ is also path-connected.

Let $$a,b \in \mathbb C \setminus C^*$$. Let's construct a path between $$a$$ and $$b$$.

Let's define the following family of lines: $$R_\varphi \subset \mathbb C$$, such that $$R_\varphi = \{a + re^{i\varphi}, r\in \mathbb R^+\}$$, for $$\varphi \in (-\pi,\pi]$$.

Lemma 1: $$\exists \varphi' \in (-\pi, \pi]: R_{\varphi'} \subset \mathbb C\setminus C^*$$.

Proof for lemma 1:

Let's assume that $$\forall \varphi \in (-\pi, \pi], \exists c \in C^*: c \in R_\varphi$$.

This would mean that $$\text{card } C^* \geq \text{card } (-\pi, \pi]$$. This is false, so we have that $$\exists \varphi' \in (-\pi, \pi]: R_{\varphi'} \subset \mathbb C\setminus C^*$$.

So, let $$\varphi '$$ be that value for $$\varphi$$ such that $$R_\varphi \subset \mathbb C \setminus C^*$$.

Let $$c \in R_{\varphi'}$$ and let $$R_c$$ be the line whose endpoints are $$c$$ and $$b$$.

Lemma 2: $$\exists c' \in R_{\varphi '}: R_{c'} \subset \mathbb C \setminus C^*$$.

Proof for lemma 2: Let's assume that $$\forall c \in R_{\varphi'}, \exists k \in C^*: c \in R_c$$.

This would mean that $$\text{card } C^* \geq \text{card } R_{\varphi'}$$. This is false, so we have that $$\exists c' \in R_{\varphi '}: R_{c'} \subset \mathbb C \setminus C^*$$.

So now we are ready to construct our path: Let $$\gamma : [0,1] \to \mathbb C \setminus C^*$$ be the straight like that connected $$a$$ and $$c'$$, and let $$\delta : [0,1] \to \mathbb C \setminus C^*$$ be the straight line connecting $$c'$$ and $$b$$.

Let $$f : [0,1]\to \mathbb C \setminus C^*$$, such that:

$$f(x) = \begin{cases} \gamma(2x) & 0 \leq x < \frac{1}{2} \\ c' & x = \frac{1}{2} \\ \delta(2x) & \frac{1}{2} < x \leq 1 \end{cases}$$

Then $$f$$ is a continuous path connecting $$a$$ and $$b$$ and therefore $$\mathbb C \setminus C*$$ is path connected, meaning that $$\mathbb R^2 \setminus C$$ is also path connected.

This is my proof. It seams a little bit convoluted and over-complicated, but I don't know what other ways there are to solve this. Is my proof correct? If not, where did I made a mistake? What other ways are there to prove this?

The idea looks correct, but you can simplify it a lot. Let $$a$$ and $$b$$ be two distinct points of $$\Bbb R^2\setminus C$$. Fix some line $$l$$ in $$\Bbb R^2$$ which is not the line defined by $$a$$ and $$b$$. For each $$c\in l$$, consider the path $$p_c$$ in $$\Bbb R^2$$ which goes in a straight line from $$a$$ to $$c$$ and then goes in a straight line from $$c$$ to $$b$$. Then, if $$c\ne d$$, the paths $$p_c$$ and $$p_d$$ have no points in common other than $$a$$ and $$b$$. Since there are uncountably many such paths and since $$C$$ is countable, some $$p_c$$ maus be a path in $$\Bbb R^2\setminus C$$.

• +1. In other words there is an uncountable family of triangles, each with two of its vertices at $a$ and $b$, that do not intersect each other except at $a$ and $b$. Oct 4 '20 at 23:03

The essence of the idea is clear and you seem to understand the principle: For every $$a,b \in \Bbb C$$ with $$a \neq b$$ there is a set of (continuous) paths $$P(a,b)=\{f_i: [0,1] \to \Bbb C, i \in J \}$$ from $$a$$ to $$b$$ such that $$f_i[[0,1]] \cap f_j[[0,1]]= \{a,b\}$$ for all $$i \neq j,i,j \in J$$ and $$J$$ is uncountable.

Then if $$C$$ is countable, and $$a,b \notin C$$ we then have that only countably many $$f_i$$ can intersect $$C$$ (all the possible intersection points are different!) so there is at least (in fact uncountably many), $$f_i \in P(a,b)$$ with $$f_i[[0,1]] \subseteq \Bbb C\setminus C$$ and this $$f_i$$ shows that $$\Bbb C\setminus C$$ is path-connected and hence connected.

Either you can claim the existence of $$P(a,b)$$ as "geometrically obvious" (just combine two straight lines) or go detailed and prove formulas for such paths (say for $$0$$ and $$1$$ (as points) first and then apply translation and stretching to do it for any pair of points... Or whatever detail is required by the teacher/referee etc. The other answers provide some extra help in that as well.

The main thing to prove is that there is too many path according to $$C$$ points, i.e construct an uncountable set of strictly different path. On one hand the @José Carlos Santos one, which is the closest to your answer works, properly.

On the other hand, remind that a circle is entirely defined by three distinct points on it. Since $$a$$ et $$b$$ are two points, the set of all arcs of a circle which pass by $$a$$ and $$b$$ is an uncountable set of strictly different paths.