1
$\begingroup$

I have a question related to cofinality:

Let $\alpha$ is a limit ordinal and is not a cardinal. How can I construct an increasing sequence $\alpha_{\xi}$ in $\alpha$ with the length $|\alpha|$ from a bijection $f: |\alpha| \rightarrow \alpha$ such that $\lim_{\xi \rightarrow |\alpha|} \alpha_{\xi} = \alpha$.

I have read one construction from here where the sequence was built based on a set $S$:

$S = \{ \beta \in |\alpha| \mid \forall \gamma \prec \beta: f(\gamma) \prec f(\beta) \}$

However, let take an example for $\alpha = \omega +1$ where the bijection $f$ is defined with $f(0) = \omega$ and $f(n+1) = n$ for all $n \in \omega$. We can easily get $S = \emptyset$.

I cannot comment to that question so my post (as an answer) has been deleted by moderators. I hope your helps in this case and so sorry if I made a mistake.

Edited: $\alpha$ is a limit ordinal.

$\endgroup$
3
  • 1
    $\begingroup$ Just to clarify, your answer was not deleted by the moderators. It was deleted by users who decided (correctly) that it is not an answer. Asking new questions should be done by asking new questions, not by writing an answer on a different question. This is the platform, use it appropriately. $\endgroup$
    – Asaf Karagila ♦
    Dec 7, 2020 at 23:57
  • $\begingroup$ Thank you @AsafKaragila. I noted it. $\endgroup$
    – Chau Long
    Dec 8, 2020 at 3:55
  • $\begingroup$ In the question you cite, the result is the existence of a sequence of length $\leq |\alpha|$ and not of length exactly $|\alpha|$. In general for ordinals $\beta,\alpha$ where $\alpha$ is a limit, there is a non-decreasing cofinal function $\beta \rightarrow \alpha$ if and only if $\operatorname{cf}(\beta)=\operatorname{cf}(\alpha)$. $\endgroup$
    – nombre
    Dec 9, 2020 at 11:59

1 Answer 1

Reset to default
2
$\begingroup$

In general this is not possible: the best that can be hoped for is a cofinal increasing sequence of length $\operatorname{cf}\alpha$. It certainly isn’t possible if $\alpha$ is not a limit ordinal, but it isn’t possible for all limit ordinals, either. For instance, let $\alpha=\omega_1+\omega$; then $|\alpha|=\omega_1$, but $\operatorname{cf}\alpha=\omega$. There are certainly increasing $\omega$-sequences cofinal in $\alpha$, the simplest being $\langle \omega_1+n:n\in\omega\rangle$, but no increasing $\omega_1$-sequence can be cofinal in $\alpha$.

$\endgroup$
5
  • $\begingroup$ Indeed, the most we can do is construct a cofinal sequence of minimal length, but not necessarily of type $|\alpha|$. $\endgroup$
    – Asaf Karagila ♦
    Dec 7, 2020 at 23:57
  • $\begingroup$ Why are all your answers always so clear, really amazing $\endgroup$
    – user655132
    Dec 8, 2020 at 3:05
  • 1
    $\begingroup$ @mugiwara: Thank you. I’ve had a lot of practice explaining things. $\endgroup$
    – Brian M. Scott
    Dec 8, 2020 at 3:07
  • $\begingroup$ Thank you very much for your answer @BrianM.Scott. Firstly, $\alpha$ is a limit ordinal. Secondly, could you please tell me if it's possible to construct the sequence with a length $\leq |\alpha|$ instead of $|\alpha|$? $\endgroup$
    – Chau Long
    Dec 8, 2020 at 3:52
  • $\begingroup$ @ChauLong: If $\alpha$ is a limit ordinal, the shortest sequences cofinal in $\alpha$ will all have length $\operatorname{cf}\alpha$, and the longest is of course $\alpha$ itself. If $\beta$ is the length of a sequence cofinal in $\alpha$, then $\operatorname{cf}\beta=\operatorname{cf}\alpha$. The only cardinal that is the length of a sequence cofinal in $\alpha$ is $|\alpha|$. $\endgroup$
    – Brian M. Scott
    Dec 9, 2020 at 17:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .