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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.

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up vote 2 down vote accepted

Suppose that $x\in U_\beta$ with $q>\sup x$. Let $p$ be any rational number satisfying $\sup x<p<q$. Then by the induction hypothesis there is $y\in U_\alpha$ such that $x\subseteq y$ and $p\ge\sup y$. Now choose your $r\in(p,q)$ and concatenate it with $y$: $\sup(y^{\frown}r)=r<y$.

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Just beat me to it. ;) –  Arthur Fischer Mar 29 '12 at 17:24
    
@Arthur: Figured it would be you. :-) –  Brian M. Scott Mar 29 '12 at 17:25
    
been a long day. Thanks guys. –  Paul Slevin Mar 29 '12 at 17:29
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