I aim to construct by induction an uncountable collection of sets $U_{\alpha}$ which contain increasing transfinite sequences of rational numbers. I want $|U_\alpha | \le \aleph_0$ for each $\alpha < \omega_1$, and I want to satisfy the condition that for each $\beta< \alpha$, each $x \in U_\beta$, and each $q > \sup x$, there exists $y \in U_\alpha$ such that $x \subseteq y$ and $q \ge \sup y$. I'm having some trouble with the successor step however:
Let $U_0 = \{ \emptyset \}$. Given $U_\alpha$, define $U_{\alpha + 1}=\{x::r \mid x \in U_\alpha,\ r > \sup x \}$ (i.e. we just add an extra element on to the end of each sequence $x$. Now this $U_{\alpha+1}$ is supposed to satisfy the conditions above, but I am not sure why it satisfies the second one.
Let $\beta \le \alpha, x \in U_\beta$ and $q > \sup x$. Then by inductive hypothesis I know that there is a $y \in U_{\alpha}$ such that $x \subseteq y$ and $q \ge \sup y$. If $\sup y < q$, choose some $r \in (\sup y, q)$ and concatenate it to y. Then $\sup (y::r) = r \le q$, as desired. However, how do I deal with the situation where $\sup y = q$? Can I make a different choice of $y$ somehow, to ensure $\sup y < q$? I can't see how to deal with this without getting involved in recursive definitions.
Any help would be appreciated.