so $\omega+1=\{\{0,1,2,...\},\{\omega\}\}$, it seems to be the case that $cf(\omega+1)=1$, since $1=|\{\omega\}|$ is the least cardinality of cofinal subsets of $\omega+1$, but $1$ is also the cofinality of the ordinal $2$, which is much less than $\omega+1$. How can I overcome this contradiction?

  • 2
    $\begingroup$ Rather than $\{\{0,1,2,...\},\{\omega\}\}$, you mean the union of this set. $\endgroup$ – Andrés E. Caicedo Aug 27 '17 at 2:19

There's no contradiction; cofinality is not a strictly increasing function. Indeed, the cofinality of every non-zero successor ordinal (i.e., ordinals $\alpha$ such that $\alpha = \beta + 1$ for another ordinal $\beta$) is $1$.

One way to think about cofinality of an ordinal $\alpha$ is that it measures how long a sequence needs to be in order to "reach" $\alpha$, not how big $\alpha$ is.

  • $\begingroup$ But then $cf(\omega)=\omega$, and $\omega$ is less than $\omega+1$ and has a bigger cofinality? $\endgroup$ – Sid Caroline Aug 26 '17 at 21:26
  • $\begingroup$ Yes. In fact, $cf(\omega) > cf(\aleph_{\aleph_\omega} + 1)$, or indeed any other suitably monstrously huge successor ordinal. $\endgroup$ – Duncan Ramage Aug 26 '17 at 21:28
  • 1
    $\begingroup$ @SidCaroline Why would you expect $cf(\alpha)$ to be strictly increasing? $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 21:29
  • $\begingroup$ I guess my intuition was wrong. I thought that a bigger ordinal must have a bigger cardinality, so it must have a bigger cofinality. $\endgroup$ – Sid Caroline Aug 26 '17 at 21:34
  • 1
    $\begingroup$ Duncan, "every non-zero successor ordinal", since when is zero a successor ordinal? :D $\endgroup$ – Asaf Karagila Aug 26 '17 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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