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Is there a countable family of open subsets of ${\bf R}$ or $[0,1]$ such that each rational belongs to only finitely many of the open sets and each irrational belongs to infinitely many of the sets?

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  • $\begingroup$ What do you mean by "each rational belongs to only finitely many of the open sets". Do you mean each rational belongs to finitely many of them and does not belong to any of the other open sets? $\endgroup$ – 0XLR Jul 15 at 9:41
  • $\begingroup$ @ZeroXLR I think so $\endgroup$ – miraunpajaro Jul 15 at 9:42
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Yes. We can take the countable many open intervals with rational center $\frac ab$ and radius $\frac1{b^2}$, $$U_{a,b}:=\left]\frac ab-\frac1{b^2},\frac ab-\frac1{b^2}\right[,$$ with $a\in\Bbb N_0$, $b\in \Bbb N$, $a\le b$, $\gcd(a,b)=1$.

If $\frac ab\in[0,1]$ is rational, then the distance to any other rational $\frac cd$ is at least $\frac1{bd}$ and this is $>\frac1{d^2}$ for almost all $\frac cd$, hence $\frac ab\in U_{c,d}$ at most for the finitely many cases when $0\le c\le d\le b$.

On the other hand, if $\alpha\in[0,1]$ is irraional, then we find infinitely many $a,b,c,d$ with $$\tag1\frac ab<\alpha<\frac cd\quad\text{and}\quad ad-bc=-1, \text{ i.e., }\frac cd-\frac ab=\frac1{bd}-$$ More concretely, we can take $(a_0,b_0,c_0,d_0)=(0,1,1,1)$ and if we have $(a_n,b_n,c_n,d_n)$ such that $(1)$ holds, then either $(a_{n+1},b_{n+1},c_{n+1},d_{n+1}):=(a_n,b_n,a_n+c_n,b_n+d_n)$ or $(a_{n+1},b_{n+1},c_{n+1},d_{n+1}):=(a_n+c_n,b_n+d_n,c_n,d_n)$ works as next choice, depending on whether $\alpha<\frac{a_n+c_n}{b_n+d_n}$ or $\alpha>\frac{a_n+c_n}{b_n+d_n}$ (equality cannot occur) As $$\frac cd-\frac ab=\frac 1{bd}<\frac1{b^2}+\frac 1{d^2},$$ we conclude that $$\tag2\alpha\in U_{a_n,b_n}\cup U_{c_n,d_n}$$ for each of our tuples. And as $\frac1{b_nd_n}\to 0$, we have both $\frac{a_n}{b_n}\to \alpha$ and $\frac{c_n}{d_n}\to \alpha$, hence no $U_{a_n,b_n}$ or $U_{c_n,d_n}$ occurs infinityly often in $(2)$.


Alternative use of the same idea, but perhaps with nicer proof:

For every quadruple $(a,b,c,d)\in\Bbb N_0^4$ with $b,d\ge1$ and $ad-bc=-1$ let $$U_{a,b,c,d} =\left]\frac ab,\frac cd\right[.$$ Assume $\frac xy$ is a rational number in $U_{a,b,c,d}$. Then from $\frac ab<\frac xy$, we have $\frac xy-\frac ab=\frac{bx-ay}{by}>0$, hence for the numerator $bx-ay\ge 1 $ and simimlarly we find $yc-xd\ge1. $ Hence $$y=(bc-ad)y=d(bx-ay)+b(yc-xd)\ge b+d.$$ It follows that a rational number $\frac xy\in[0,1]$ can only be in the finitely many $U_{a,b,c,d}$ with $b+d\le y$, $0\le a<b$, $1\le c\le d$.

On the other hand, each irrational $\alpha\in[0,1]$ is certainly $\in U_{0,1,1,1}$. If there were only finitely many of such open sets containing $\alpha$, then there'd be one with maximal $b+d$. But one verifies that $$U_{a,b,c,d}=U_{a,b,a+c,b+d}\cup U_{a+c,b+d,c,d}\cup\left\{\frac {a+c}{b+d}\right\} $$ and hence $\alpha$ must be in one of the intervals on the right, contradicting maximality.

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  • $\begingroup$ What do you do about 0/b? $\endgroup$ – William Elliot Jul 15 at 21:36
  • $\begingroup$ @William Elliot. What is "o/b"? $\endgroup$ – user558840 Jul 15 at 21:44
  • $\begingroup$ @Hagen von Eitzen. Many thanks indeed. $\endgroup$ – user558840 Jul 15 at 22:21
  • $\begingroup$ @user558840. It is the rational number 0/1 = 0/2 = 0/3 = 0/4 = ... $\endgroup$ – William Elliot Jul 16 at 2:30
  • $\begingroup$ @WilliamElliotIn my first variant, $0\in U_{0,1}$ (and none else), in the second variant, $0$ is in none of the $U_{a,b,c,d}$ - which is arguably a finite number. If we additionally demand that our open sets cover $[0,1]$, we can simply take $\Bbb R$ itself as a single additional open set, and that does not change anything about finiteness $\endgroup$ – Hagen von Eitzen Jul 16 at 6:22
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It's a fact (which I'm not going to use except as motivation) that any two countable dense subsets of $\mathbb R$ are "equivalent" in the sense that there is a homeomorphism of $\mathbb R$ with itself that maps one to the other. Therefore, if the statement you want to prove is true, it remains true if the set $\mathbb Q$ of all rational numbers is replaced with any other countable dense subset of $\mathbb R$. Hence it seems natural to seek a proof which is set-theoretical rather than arithmetical.

In fact, the set doesn't even have to be dense; if $A$ is any countable subset of $\mathbb R$, there is a countable family $\mathcal U$ of open sets such that no element of $A$, but every element of $\mathbb R\setminus A$, belongs to infinitely many members of $\mathcal U$.

Case 1. $A$ is finite.

Let $U$ be the collection of all rational open intervals which are disjoint from $A$.

Case 2. $A$ is countably infinite.

Let $A=\{a_n:n\in\mathbb N\}$ (enumerated without repetition) and let $\mathcal U=\{U_n:n\in\mathbb N\}$ where $U_n=\mathbb R\setminus\{a_1,\dots,a_n\}$.


The statement is still true, but slightly less trivial, if you want the members of $\mathcal U$ to be bounded open intervals. Suppose $A=\{a_n:n\in\mathbb N\}$. Let $$\mathcal U=\mathcal W\cup\bigcup_{n\in\mathbb N}\mathcal V_n$$ where $\mathcal W$ is the set of all rational open intervals disjoint from $A$, and $\mathcal V_n$ is the set consisting of the bounded components of $\mathbb R\setminus\{a_1,\dots,a_n\}$ together with the intervals $(L_n-1,L_n)$ and $(R_n,R_n+1)$ where $L_n=\min\{a_1,\dots,a_n\}$ and $R_n=\max\{a_1,\dots,a_n\}$. It is easy to see that $\{U\in\mathcal U:x\in U\}$ is finite if $x\in A$, infinite if $x\notin A$.

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  • $\begingroup$ Many thanks for this too. This is easier to extend to more general situations. $\endgroup$ – user558840 Jul 16 at 9:10

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