# $\mathbb{R}$ \ $\mathbb{Q}$ and $\mathbb{R}^2\setminus\mathbb{Q}^2$ disconnected?

If I want to prove that $\mathbb{R} \setminus \mathbb{Q}$ is disconnected, does it suffice to say that there are two open disjoint sets that cover $\mathbb{R}\setminus\mathbb{Q}$, namely:

$$(- \infty, 0), (0, \infty)\text{ ?}$$

Along the same lines, I want to prove or disprove that $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$ (points $(x, y) \in S$ that have at least one irrationais connected. I feel that it is also disconnected; does it suffice to say that there are two open disjoint sets that cover $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$, namely $((- \infty, - \infty), (0, 0))$ and $((0, 0), (\infty, \infty))$? (Very iffy on my assertion and my notation, sorry.)

Thanks!

• What do your mean by $((−∞,−∞),(0,0))$ ? Anyway, the set $S$ is connected. Oct 6, 2013 at 18:52
• I meant the set where both coordinates are < 0. And oh wait... is $\mathbb{R}$ \ $\mathbb{Q}$ connected as well? Oct 6, 2013 at 18:57
• – JDH
Oct 6, 2013 at 20:05

## 4 Answers

The answer to your first question is yes: $$\{(\leftarrow,0),(0,\to)\}$$ is a clopen partition of the irrationals, and its existence shows that they are not connected.

$$\Bbb R^2\setminus\Bbb Q^2$$, on the other hand, is connected, and even path connected: you can get from any point of it to any other point along a path lying entirely within $$\Bbb R^2\setminus\Bbb Q^2$$. In fact, you can do it alone straight line segments, using at most three of them; just make sure that each horizontal segment lies on a line $$y=a$$ with irrational $$a$$, and each vertical segment lies on a line $$x=a$$ with irrational $$a$$. See if you can work out the details for yourself; I’ve written them up below and left them spoiler-protected.

Suppose that $$\langle a_,b\rangle$$ is a point with at least one irrational coordinate. Without loss of generality let $$a$$ be irrational. Let $$\langle c,d\rangle$$ be any other point with at least one coordinate irrational. If $$d$$ is irrational, you can travel along the line $$x=a$$ to the point $$\langle a,d\rangle$$, and then travel along the line $$y=d$$ to $$\langle c,d\rangle$$; this path lies entirely in $$\Bbb R^2\setminus\Bbb Q^2$$. If $$c$$ is irrational, travel along the line $$x=a$$ to any $$\langle a,u\rangle$$ with irrational $$u$$, then along $$y=u$$ to the point $$\langle c,u\rangle$$, and finally along the line $$x=c$$ to the point $$\langle c,d\rangle$$; again the path lies entirely in $$\Bbb R^2\setminus\Bbb Q^2$$.

• Is this use of ← and → for $-\infty$ and $\infty$ something that you started doing recently? I don't think I'd seen it before a week or two ago.
– MJD
Oct 6, 2013 at 19:21
• @MJD: I’ve used it for about $40$ years now, and I’ve almost always used it here, so you just haven’t noticed. Oct 6, 2013 at 19:22
• Did you make it up, or does anyone else do the same thing?
– MJD
Oct 6, 2013 at 19:24
• @MJD: It’s not my invention. It was in pretty common use among people working with linearly ordered spaces back when I was doing so, and I much prefer it to the $\pm\infty$ notation: it doesn’t give the impression that there’s an object out there on the end of the order. Oct 6, 2013 at 19:32
• @BrianM.Scott Very Elegant Sir Apr 9, 2020 at 10:20

As the others have noted, the second statement is wrong. One can arrive at a less direct, but maybe more elegant proof of its converse by proving the stronger fact that $\mathbb R^2\setminus A$ is path-connected for every countable set $A$. As far as I remember this simple proof is due to Cantor. I only do not give it here to not spoil anything, I hope that having said the word “countable” is a sufficient hint.

• This should rather be a comment that an answer. Oct 6, 2013 at 19:33
• @StefanH, I have made my comment more answer-like. Oct 6, 2013 at 20:01
• In fact, $\mathbb{R}^2\setminus A$ is path connected for any $A$ of size less than continuum, whether or not $A$ is countable. See math.stackexchange.com/a/145958/413
– JDH
Oct 6, 2013 at 20:04
• @JDH good point. Of course not many people have actually encountered subsets of this kind that are not countable ;) Oct 6, 2013 at 20:07
• Well, the idea is that there are uncountably many disjoint paths between the two given points, so choosing a point in $A$ from each path would give an uncountable subset of $A$. Oct 6, 2013 at 20:52

You are right about the disconnectedness of $\Bbb R-\Bbb Q$.

The set $S$, on the other hand, is connected, even path connected. Choose irrational $\alpha,\beta$. Each point $(x,y)$ can be joined to the point $(\alpha,\beta)$ by the path $p:I\to\Bbb R^2-\Bbb Q^2$$p(t)=\begin{cases} 2t(\alpha,y)+(1-2t)(x,y) &\text{ if }t\le\frac12\\ (2t-1)(\alpha,\beta)+(2-2t)(\alpha,y) &\text{ if }t\ge\frac12 \end{cases}$$ if$y$is irrational, and $$p(t)=\begin{cases} 2t(x,\beta)+(1-2t)(x,y) &\text{ if }t\le\frac12\\ (2t-1)(\alpha,\beta)+(2-2t)(x,\beta) &\text{ if }t\ge\frac12 \end{cases}$$ if$x\$ is irrational

Actually, it is rather an obvious fact that the set of all rational numbers are disconnected under the subspace topology induced from $$\mathbb R$$, and even if $$\mathbb Q$$ is given under the order topology, the order being induced from $$\mathbb R$$, then also it is disconnected. As because, $$\mathbb Q$$ does not has the lub property. And any ordered set $$X$$ with the $$order$$ topology is connected $$iff$$ it is a linear continuum .