# Connectivity of the set of irrational points in the plane

I'm working on a problem that asks the following:

Let $$Y$$ denote the subset of all points in the plane $$\mathbb{R}^2$$ both of whose coordinates are rational numbers and let $$X$$ denote the compliment of $$Y$$ in the plane. $$S$$ inherits the subspace topology from the standard topology on the plane; is $$X$$ connected?

They provide a hint: Choose any $$x \in X$$, and let $$C$$ denote the union of all half rays in $$\mathbb{R}^2$$ which begin a $$x$$ and are completely contained in $$X$$. Show that the closure of $$C$$ in $$X$$ is equal to $$X$$.

1. I have two constradictory arguements I can't distinguish between:

I could use the arguement presented here: Connectedness of points with both rational or irrational coordinates in the plane? to show $$\mathbb{Q} \times \mathbb{Q}$$ is connected. Then so is $$\overline{\mathbb{Q} \times \mathbb{Q}} = \mathbb{R}^2$$. $$\mathbb{R}^2 = (\mathbb{Q} \times \mathbb{Q}) \bigcup (\mathbb{R-Q} \times \mathbb{R-Q})$$ is a separation. But no separation can exist because $$\mathbb{R}^2$$ is connected so $$(\mathbb{R-Q} \times \mathbb{R-Q})$$ will be connected.

Alternatively I could produce a separation between any two elements $$(p_1,p_2),(p_3,p_4) \in \mathbb{R-Q} \times \mathbb{R-Q}$$: between $$p_1$$ and $$p_3$$, choose a rational $$q_1$$ and between $$p_2$$ and $$p_4$$, choose a rational $$q_2$$. The vertical and horizontal lines passing through $$(q_1,q_2)$$ will contain all rational pts (so it won't hit $$(\mathbb{R-Q} \times \mathbb{R-Q})$$) and the half plane $$A=\{(m,n)|m,n \in \mathbb{Q}, m > q_1, n>q_2\}$$ separates our points from the rest of $$\mathbb{R-Q} \times \mathbb{R-Q}$$, making it completely disconnected.

2. I want to understand the hint. Part of the confusion is that I don't understand what a "ray" in $$\mathbb{R}^2$$ is. Is this a ray in each basis element (so an open half plane)?

• o wait, the second argument won't work, there are "holes" in the boundary of $A$. nevertheless, how do I use the hint then to argue directly? Sep 30 '18 at 13:50
• o wait, will the arguement for the irrationals be the same for the one in the linked post. rays are just the half lines in which a path between two points of $X$ is contained Sep 30 '18 at 14:01

Consider two points $$x,y$$ in $$S$$. So both points have at least one irrational coordinate.
Note that all lines of the form $$\{p\} \times \mathbb{R}$$ and $$\mathbb{R} \times \{p\}$$ are connected (copies of the reals) and when $$p$$ is irrational, lie completely inside $$S$$. Can you connect a few of those lines together to form a angular path from $$x$$ to $$y$$ inside $$S$$? There are a few cases depending on whether the first or second coordinate is irrational in either point.
Your first argument fails because it is not true that we have$$\mathbb{R}^2=(\mathbb{Q}\times\mathbb{Q})\bigcup\bigl((\mathbb{R}\setminus\mathbb{Q})\times(\mathbb{R}\setminus\mathbb{Q})\bigr)).$$Think about $$\left(0,\sqrt2\right)$$, for instance.
On the other hand, if $$l$$ is a straight line and $$P\in l$$, then $$P$$ divides $$l$$ into two halves, each of wich is a ray. If $$x\in X$$ and if you consider a ray starting at $$x$$ with rational slope, then evexh element of the ray will be an element of $$X$$ too. But if you take $$x^\star\in X$$, there will be elements from such rays as close as you wish from $$x^\star$$. And the union $$C$$ of these rays is connected (each ray is connected and $$x$$ belongs to each of them). So, since $$C\subset X\subset\overline C$$, $$X$$ is connected too.
• in your post, what is $P$? Sep 30 '18 at 14:05
• @yoshi A point of the straight line $l$. Sep 30 '18 at 14:06
• Why doesn't $\overline{C}$ include $\mathbb{Q} \times \mathbb{Q}$? Can't you get arbitrarily close to these points too? Sep 30 '18 at 14:10