I am reading Jech's & Hrbacek's book "Introduction to set theory" and this is question 2.2 from chaper 9:

Prove $cf(\aleph_{\omega_1})=\omega_1$

Isn't it clear from the fact that $\aleph_{\omega_1} = \lim\limits_{\alpha \rightarrow \omega_1}\aleph_\alpha$?

I am not sure what is there to prove here..

Thanks for any help..


1 Answer 1


It's true that it is immediately clear that $\operatorname{cf}(\aleph_{\omega_1})\leq\omega_1$, but you still have to argue that it is not strictly smaller.

More generally, however, you can show that if $\delta$ is a limit ordinal then $\operatorname{cf}(\aleph_\delta)=\operatorname{cf}(\delta)$.

  • 2
    $\begingroup$ Additionally, note that not every question in a textbook needs to be very difficult. Sometimes a question is given to make you have confidence in your ability to understand the definitions. $\endgroup$
    – Asaf Karagila
    Commented Feb 24, 2015 at 17:21
  • 1
    $\begingroup$ I see thank you. I will try your other suggestion. $\endgroup$
    – user135172
    Commented Feb 24, 2015 at 17:23

You must log in to answer this question.

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