11

Is the set of points in the plane whose coordinates are either both irrational, or both rational connected?

Martin Sleziak
  • 56,060
Minghao Liu
  • 1,088

1 Answers1

16

Let $S$ be your set, and suppose $S = U \cup V$ where $U$ and $V$ are disjoint, and are both closed and open in $S$. Note that if $x$ and $y$ are both rational, then the diagonal lines $\{(x+t, y+t): t \in {\mathbb R}\}$ and $\{(x+t,y-t): t \in {\mathbb R}\}$ are subsets of $S$. Using a path of diagonal line segments, it is possible to get from any point of ${\mathbb Q} \times {\mathbb Q}$ to any other while staying in $S$. Therefore one of $U$ and $V$, let's say $U$, contains all of ${\mathbb Q} \times {\mathbb Q}$. But ${\mathbb Q} \times {\mathbb Q}$ is dense in $S$, and $U$ is closed in $S$ so $U = S$.

For extra credit, show that $S$ is path-connected. In fact, if $a < b$ and $c < d$ with $(a,b)$ and $(c,d)$ in $S$, there is a continuous increasing function $f: [a,b] \to [c,d]$ such that $f(x)$ is rational if and only if $x$ is rational.

Robert Israel
  • 470,583
  • @MattN. This is not necessary (and one can argue it is better not to). – Did Apr 23 '12 at 16:13
  • 1
    Can anyone prove the part about path-connectedness? Why does such a function $f(x)$ exist? – Antonio AN Jan 07 '18 at 11:46
  • 5
    Take two-sided increasing sequences of rationals $x_n$ and $y_n$ with $x_n \to a$ and $y_n \to b$ as $n \to -\infty$, $x_n \to c$ and $y_n \to d$ as $n \to +\infty$, and join $(x_n, y_n)$ to $(x_{n+1},y_{n+1})$ with straight line segments. – Robert Israel Jan 08 '18 at 02:21
  • I happened to see this just now, being directed here by the recently posted question Subset of both rational or both irrational coordinates is connected. Regarding the path connectedness result, see these sci.math posts: 15 May 2002 AND 16 May 2002 AND 19 October 2005. – Dave L. Renfro Jul 05 '22 at 21:28
  • could you elaborate more? How does one get from an arbitrary point in $\Bbb{Q} \times \Bbb{Q}$ to a point in $\Bbb{I} \times \Bbb{I}$ where $\Bbb{I}=\Bbb{R} \setminus \Bbb{Q}$ – MyMathYourMath Jul 05 '22 at 23:14
  • @Hossien Sahebjame: How does one get from an arbitrary point in --- See the 3rd comment by Lukas Geyer (begins with "Tell me if I'm wrong, but isn't there ...") under Partitioning $\mathbb{R}^2$ into disjoint path-connected dense subsets (BTW, for the reference I gave in the 4th comment, see here) and the paragraph beginning with "Here's the easy way. Given two points in the space, ..." in this 15 May 2002 sci.math post. – Dave L. Renfro Jul 06 '22 at 13:31
  • For the book by Sieradski, I found a copy here -- see Example 2 on page 187 (= .pdf page 195). Incidentally, the date mismatch between the 15 May 2002 sci.math post I just linked to and the reference to the 14 May 2002 sci.math post I mentioned in the question by Lukas Geyer is presumably due to how the google-groups archived version (the 15 May version) and the Math Forum archived version (the 14 May version, no longer exists, or maybe behind a paywall) assigned the posting time/day. – Dave L. Renfro Jul 06 '22 at 13:47
  • @DaveL.Renfro Are you sure the proof you cite in Sieradski's book is actually correct? How do you know that infinitely many straight line paths assemble into a single path $f: [0,1] \to X$? It seems like you'd have to be doing those paths infinitely fast. – David White Nov 12 '23 at 02:01
  • @RobertIsrael Are you sure the proof you sketch is actually correct? How do you know that infinitely many straight line paths assemble into a single path $f: [0,1] \to X$? It seems like you'd have to be doing those paths infinitely fast. – David White Nov 12 '23 at 02:02
  • @DavidWhite Yes, it's correct. If $x_n$ and $y_n$ are ($2$-sided) increasing sequences with finite limits as $n \to -\infty$ and $n \to +\infty$, the union of the line segments joining $(x_n, y_n)$ to $(x_{n+1}, y_{n+1})$ is the graph of a continuous increasing function. I don't understand what you mean by "doing those paths infinitely fast". – Robert Israel Nov 12 '23 at 06:26
  • What I mean is, given paths $f_1$ from p to q, and $f_2$ from q to r, then, to define a path $f$ from p to r, we have to do each of the two pieces twice as fast, so we go from p to q on [0,1/2] and from q to r on [1/2,1]. For three line segments, f has to go three times as fast on each one. With countably infinitely many line segments, you can define a sequence of continuous functions whose limit is f (e.g., $g_n$ is 0 outside of [1/n,1/(n+1)] and does the path $f_n$ on that interval at n times speed (replacing t by nt)) but this sequence doesn't converge uniformly, so why is $f$ continuous? – David White Nov 12 '23 at 12:18
  • But you don't have to do it that way. Just use the $x$ coordinate as the parameter. – Robert Israel Nov 12 '23 at 16:53
  • @RobertIsrael Ok, now I'm satisfied, thanks. First, it's intuitively clear that the image of f is homeomorphic to a closed interval, yielding the path needed for path connectedness (e.g., proving that any two points are in the same path component). As for constructing f from all the $f_n$, and proving it's continuous, this can be accomplished by the infinite pasting lemma, Exercise 9 on page 112 of Munkres, since the collection of intervals $[x_n,x_{n+1}]$ is locally finite. – David White Nov 12 '23 at 18:16