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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?

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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 $\Bbb R^2\setminus C$.

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    $\begingroup$ +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 $b$. $\endgroup$ Oct 4 '20 at 23:03
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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.

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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.

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