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?
Countable family of open sets
general-topology
general-topology
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?
Best Answer
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.