Existence of certain $\left\langle{\alpha_n | n \in \omega}\right\rangle$

Question

Let $\beta$ be a countable limit ordinal. Prove $\exists$ sequence $\left\langle{\alpha_n | n \in \omega}\right\rangle$ with the following properties:

$(1): \alpha_0 = 0\;;$
$ (2): \forall n \in \omega [\alpha_0 \in \beta, \alpha_n <\alpha_{n+1}]\;;$
$(3): \forall \epsilon < \beta\,\exists n \in \omega [\epsilon \leq \alpha_n]\;.$

Could anyone advise me on this problem? I'm totally clueless. The hint is to fix bijection $f: \omega \to \beta$ and define $ \alpha_n$ by induction. Could anyone explain why? Thank you.

Answer 1

Having such a bijection $f$, define a sequence $\langle i_n \rangle_{i \in \omega}$ as follows:

• $i_0 = f^{-1} ( 0 )$;
• $i_{n+1} = \min \{ i \in \omega : i > i_n , f ( i ) > f ( j )\text{ for all }j \leq i_n \}$.

Since $\beta$ is a limit ordinal it is possible to find such a sequence. Note that it is strictly increasing.

Given $\epsilon < \beta$, since $f^{-1} ( \epsilon ) = k < \omega$, it follows that $i_{k} \geq k$, and therefore $f ( i_{k+1} ) > f ( k ) = \epsilon$.

We then define $\alpha_n = f ( i_n )$.

Answer 2

You’re being asked to construct a strictly increasing sequence whose limit (supremum) is $\beta$. Let $f:\omega\to\beta$ be a bijection; there is one, because $\beta$ is a countable limit ordinal. Moreover, you may assume that $f(0)=0$: if $f(0)=n>0$ and $f(k)=0$, just switch the values of $f(0)$ and $f(k)$. Now let $\alpha_0=f(0)$. Given $\alpha_n$, let $$\alpha_{n+1}=f\big(\min\{k\in\omega:f(k)>\alpha_n\}\big)\;.$$ This certainly ensures that $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence of ordinals less than $\beta$. You still have to show that $\sup_n\alpha_n=\beta$, and I’ll leave that for you to try.