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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)?

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  • $\begingroup$ 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? $\endgroup$ – yoshi Sep 30 '18 at 13:50
  • $\begingroup$ 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 $\endgroup$ – yoshi Sep 30 '18 at 14:01
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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.

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  • $\begingroup$ in your post, what is $P$? $\endgroup$ – yoshi Sep 30 '18 at 14:05
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    $\begingroup$ @yoshi A point of the straight line $l$. $\endgroup$ – José Carlos Santos Sep 30 '18 at 14:06
  • $\begingroup$ Why doesn't $\overline{C}$ include $\mathbb{Q} \times \mathbb{Q}$? Can't you get arbitrarily close to these points too? $\endgroup$ – yoshi Sep 30 '18 at 14:10
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    $\begingroup$ @yoshi I made an error in my final sentence. I suppose that it is correct now. $\endgroup$ – José Carlos Santos Sep 30 '18 at 14:15
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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.

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