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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!

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    $\begingroup$ What do your mean by $((−∞,−∞),(0,0))$ ? Anyway, the set $S$ is connected. $\endgroup$ Oct 6, 2013 at 18:52
  • $\begingroup$ I meant the set where both coordinates are < 0. And oh wait... is $\mathbb{R}$ \ $\mathbb{Q}$ connected as well? $\endgroup$
    – r123454321
    Oct 6, 2013 at 18:57
  • $\begingroup$ See math.stackexchange.com/a/145958/413 $\endgroup$
    – JDH
    Oct 6, 2013 at 20:05

4 Answers 4

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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 $\Bbb R^2\setminus\Bbb Q^2$.

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  • $\begingroup$ 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. $\endgroup$
    – MJD
    Oct 6, 2013 at 19:21
  • $\begingroup$ @MJD: I’ve used it for about $40$ years now, and I’ve almost always used it here, so you just haven’t noticed. $\endgroup$ Oct 6, 2013 at 19:22
  • $\begingroup$ Did you make it up, or does anyone else do the same thing? $\endgroup$
    – MJD
    Oct 6, 2013 at 19:24
  • $\begingroup$ @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. $\endgroup$ Oct 6, 2013 at 19:32
  • $\begingroup$ @BrianM.Scott Very Elegant Sir $\endgroup$
    – MathMan
    Apr 9, 2020 at 10:20
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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.

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  • $\begingroup$ This should rather be a comment that an answer. $\endgroup$ Oct 6, 2013 at 19:33
  • $\begingroup$ @StefanH, I have made my comment more answer-like. $\endgroup$
    – Carsten S
    Oct 6, 2013 at 20:01
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    $\begingroup$ 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 $\endgroup$
    – JDH
    Oct 6, 2013 at 20:04
  • $\begingroup$ @JDH good point. Of course not many people have actually encountered subsets of this kind that are not countable ;) $\endgroup$
    – Carsten S
    Oct 6, 2013 at 20:07
  • $\begingroup$ 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 $A$. $\endgroup$ Oct 6, 2013 at 20:52
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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

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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 .

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