A minimal generating set for the order relation of rationals and reals

Question

Does there exist a binary relation $R$ on $\mathbb{Q}$ such that the transitive closure of $R$ is the order relation $<$, and that the transitive closure of any proper subset of $R$ is not the order relation? My second question is the same thing, but with $\mathbb{R}$ instead of $\mathbb{Q}$.

Answer 1

I'll just spell out bof's argument from the comments.

Suppose $(L,<)$ is a dense linear order with more than one element. I claim that there is no minimal binary relation on $L$ whose transitive closure is $<$.

Suppose $R$ is a binary relation on $L$ whose transitive closure is $<$. Since $L$ has more than one element, $R\neq \varnothing$. It suffices to show that for any $(a,b)\in R$, if we let $R' = R\setminus \{(a,b)\}$, then the transitive closure of $R'$ is also equal to $<$, since then $R$ is not minimal.

Since $(a,b)\in R$, we have $a<b$, and since $L$ is dense, there is some $c$ with $a<c$ and $c<b$. Since $<$ is the transitive closure of $R$, there exists $m\geq 0$ and $d_1,\dots,d_m$ with $a R d_1 R d_2\dots R d_m R c$ and there exist $n\geq 0$ and $e_1,\dots,e_n$ with $cR e_1 R e_2 R \dots R e_n R b$. Putting these together, and noting that no instance of $R$ in either chain is $(a,b)$, we have $$a R' d_1 R' d_2\dots R' d_m R' cR' e_1 R' e_2 R' \dots R' e_n R' b$$ so $(a,b)$ is in the transitive closure of $R'$. It follows that the transitive closure of $R'$ is equal to the transitive closure of $R$, which is $<$.