Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $y$: $\sup(y^{\frown}r)=r<y$.

share|cite|improve this answer
Just beat me to it. ;) – arjafi 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.