Let$Q$ be the linearly ordered set of all rational numbers con-sidered as a category, while $R^+$ is the set of reals with a symbol $\infty$ adjoined. In $Sets^Q$, prove that the subobject classifier $\Omega$ has $\Omega (q) = {r: r\in R^+, r\geq q}$.
I coud just write this: $Hom_{Sets} (Hom_Q (q,-),\Omega) \cong sub_{Sets} (Hom_Q(q,-))=\Omega (q)$
For each $q\in Q , r\geq q$ define a subfunctor of $Hom_Q(q,-)$
$\infty$ defines a subfunctor for each $Hom(q,-)$
Then $\Omega (q) = \{r: r\in R^+, r\geq q\}$.
Please help...
You're on the right track here by applying the Yoneda lemma. You've correctly observed that at $q$ the subobject classifier is precisely determined by the subobjects of $y(q) = \hom(-, q)$. However, as written you're considering $\mathbf{Set}^\mathbb{Q}$, so we are not considering presheaves on $\mathbb{Q}$, but rather on $\mathbb{Q}^{\mathsf{op}}$. In other words, $\Omega(q)$ is the set of "upwards closed subsets of rational numbers larger than $q$".
Thinking of real number in terms of Dedekind cuts $(A, B)$, we can see that real numbers larger than $q$ precisely determine upwards closed nonempty subsets of rational numbers. In order to account for the empty subset, we throw in $\infty$ to obtain your desired characterization.