we need to prove that the set $\{[q,+\infty )|q\in \mathbb Q\}$ is a producer of a borel sigma algebra at R. I had the following solution but i think there are a lot of mistakes in it.
we will take a set $[d,+inf)$ and we want to take inclusions, complements ... We take the complement and we get $[0,d)$. If we do $[o,d) \bigcap [e,+inf)$ we get $[e,d)$. if we take $\bigcup$ from all of $[(q-1)/n,b)$ we get $(g,b)$, this is an open set. We know that every open set in $Q$ is a set in $R$.
i'm stuck here and sure if the above is correct.
For any family $F$ of subsets of $\Bbb R$ let $A(F)$ be the set of $\sigma$-algebras on $\Bbb R$ that each have $F$ as a subset. The $\sigma$-algebra on $\Bbb R$ generated by $F$ is $B(F)=\cap A(F)$.
From this definition it is easy to show that if $F_1,F_2$ are families of subsets of $\Bbb R$ with $F_1\subset B(F_2)$ and $F_2\subset B(F_1)$ then $B(F_1)=B(F_2).$ (See $Footnote$.)
Let $S=\{[q,\infty):q\in\Bbb Q\}.$ Let $O$ be the set of open subsets of $\Bbb R.$
The set of Borel sets of $\Bbb R$ is $B(O)$. We have $[q,\infty)=\Bbb R\setminus (-\infty,q)\in B(O).$ So $S\subset B(O).$ Now we show $O\subset B(S)$ to conclude that $B(S)=B(O).$
(i). For every $x\in \Bbb R$ we have $(x,\infty)=\cup T(x),$ where $T(x)=\{[q,\infty):x<q\in \Bbb Q\}$ is a countable subset of $S.$ So $(x,\infty)\in B(S).$
(ii). For every $y\in \Bbb R$ we have $(-\infty,y)=\cup U(y),$ where $U(y)=\{(-\infty,q): y>q\in \Bbb Q\}=\cup \{\Bbb R\setminus [q,\infty): y>q\in \Bbb R\}$ is a countable subset of $B(S).$ So $(-\infty,y)\in B(S).$
(iii). By (i) and (ii), for every $x,y \in \Bbb R$ we have $(x,y)=(x,\infty)\cap (-\infty,y)\in B(S).$
(iv). For any $W\in O$ (i.e. any open $W\subset \Bbb R$) we have $W=\cup V(W)$ where $V(W)=\{(q_1,q_2): q_1,q_2\in \Bbb Q \land (q_1,q_2)\subset W\}$ is countable, and by (iii), $V(W)\subset B(S).$ So $W\in B(S).$
So $O\subset B(S).$
$Footnote$.Suppose $F_1\subset B(F_2)$ and $F_2\subset B(F_1).$ Then $B(F_2)\in A(F_1)$ so $B(F_2)\supset\cap A(F_1)=B(F_1).$ Interchanging the subscripts $1,2$ we also have $B(F_1)\supset B(F_2).$