If $A\subseteq\Bbb{R}^2$ is countable then $\Bbb{R}^2\setminus A$ is polygonally connected

Question

Show that if $A\subseteq \Bbb{R}^2$ is countable then $\Bbb{R}^2\setminus A$ is polygonally connected.

I have only come by similar questions here, but that set no guidelines from which I can yield and understand. I thought I had some ideas buy nothing seems correct at the moment. I just can't think of anything other than taking a point and trying to connect it with another point. It seems like I could always intersect some point in $A$, but on the other hand, from that point I can draw a random line in a random direction, and even the slightest continuous rotation will cover uncountably many line segments. Suppose I have two points in $\Bbb{R}^2\setminus A$, following my intuitive, how can I make sure the proccess never ends? Is this intuitive direction correct, and if so, how can it be formalized? I have been struggling with that quite a bit.

For a slight variation, see http://math.stackexchange.com/a/240468/27978. — copper.hat, Dec 23 '16 at 04:46

score 5 · Accepted Answer · answered Dec 23 '16 at 04:26

5

Let's try to link two points $a,b \in \mathbb R^2$.

Consider all the lines in $a$. Since they are parametrized by their slope, they are not countable. You can pick two of them without points in $A$.

As for $a$, we can pick a line in $b$ with no points in $A$. This line can't be parallel to both of the previous lines, thus we found a polygonal path from $a$ to $b$.

answered Dec 23 '16 at 04:26

Maffred

4,166

This was quite creative. It fit well my need to have every little path formally confirmed. – Meitar Dec 23 '16 at 04:33

score 3 · Answer 2 · answered Dec 23 '16 at 04:25

Hint Pick two points $x,y \notin A$.

There are uncountably many lines through $x$. Only countably many of them can contain points in $A$.

Pick one such line $l_1$.

There are uncountably many lines through $y$. Only countably many of them can contain points in $A$. One more is parallel to $l_1$.

Pick a line $l_2$ which passes through $y$, and is not parralel to $l_1$ nor contains points in $A$.

score 1 · Answer 3 · answered Dec 23 '16 at 04:24

If I understand you correctly, it seems like you're going in the right direction: It is true that, for any point, there are uncountably many lines passing through it and only countably many of these passing through any point in $A$, hence uncountably many miss every point in $A$. To show that $\mathbb R^2\setminus A$ is polygonally connected, just take two points $x$ and $y$ and consider some line $\ell$ not passing through $x$ or $y$. Note that, aside from the endpoints, the set of paths consisting of a segment from $x$ to $z$ then $z$ to $y$ for $z\in \ell$ is pairwise disjoint. Thus, only countably many such paths may intersect $A$. However, there are uncountably many such paths, so there is some $z$ such that the segment from $x$ to $z$ then $z$ to $y$ does not intersect $A$, thus we are done.

If $A\subseteq\Bbb{R}^2$ is countable then $\Bbb{R}^2\setminus A$ is polygonally connected

3 Answers3