Let $\kappa$ be a singular cardinal, such that $\operatorname{cf}\kappa = \lambda < \kappa$. Now because $\operatorname{cf}\kappa = \lambda$, then I can write down an increasing sequence of ordinals $\langle \alpha_{\xi} \mid \xi < \lambda \rangle$ such that $\displaystyle\lim_{\xi \rightarrow \lambda}\alpha_{\xi}=\kappa$.

But why is it possible to construct an increasing, continuous sequence of cardinals $\langle \beta_{\xi} \mid \xi < \lambda \rangle$?

I think that replacing each $\alpha_{\xi}$ with $| \alpha_{\xi}|^+$ yields an increasing sequence of cardinals (since $\kappa$ is not a successor cardinal), but how do I guarantee continuity? Do we require that $\lambda > \omega ?$

Any help would be appreciated.

  • $\begingroup$ Note that the result (proof by Asaf below) works equally well for any limit cardinal $\kappa$, even regular ones (i.e. inaccessible cardinals). It does not depend in any way on having a singular cardinal. $\endgroup$
    – PatrickR
    Mar 2 at 10:33

1 Answer 1


Suppose $\kappa$ is a limit cardinal, whose cofinality is $\lambda<\kappa$. This means that there exists a strictly increasing sequence $\langle\kappa_\xi\mid\xi<\lambda\rangle$ of cardinals such that $\sup\kappa_\xi=\kappa$.

When is a sequence like that is continuous? Exactly if the following condition holds:

If $\delta<\lambda$ is a limit ordinal, then $\kappa_\delta=\sup\{\kappa_\beta\mid\beta<\delta\}$.

Define a new sequence $\kappa^\prime_\xi$ as: $$\kappa^\prime_\xi=\begin{cases}\kappa_\xi & \xi=\alpha+1\\\sup\{\kappa_\beta\mid\beta<\xi\} &\xi\text{ is a limit ordinal}\end{cases}$$ To see that this sequence is continuous note that whenever $\delta$ is a limit ordinal then $\kappa^\prime_\delta$ is defined to be the correct cardinal (recall that the limit of cardinals is a cardinal).

We need to see that $\sup\kappa^\prime_\xi=\kappa$, but since $\kappa^\prime_{\xi+1}=\kappa_{\xi+1}$ their $\sup$ is also the same.

  • $\begingroup$ What I mean is, how can I construct the sequence of cardinals with limit $\kappa$. How can I be sure that $\displaystyle\lim_{\alpha \rightarrow \lambda} \aleph_{\alpha} = \kappa$? Is $\kappa$ the $\gamma$-th cardinal? $\endgroup$ Feb 23, 2012 at 13:02
  • $\begingroup$ @Paul: I don't understand your comment. $\endgroup$
    – Asaf Karagila
    Feb 23, 2012 at 13:18
  • $\begingroup$ My goal is to construct a continuous, increasing sequence of cardinals of length $\lambda$ with supremum $\kappa$, given that $\kappa$ is singular and $\operatorname{cf}\kappa = \lambda$. I can only construct an increasing sequence of ordinals which might not be continuous. $\endgroup$ Feb 23, 2012 at 13:28
  • 1
    $\begingroup$ @Paul: Since you already have $\langle\aleph_\beta\mid\beta<\lambda\rangle$ which is an increasing sequence approaching $\kappa$, I claim that it has at most $\lambda$ many limit points. Add those and you will have a sequence of length $\lambda$ which is continuous. $\endgroup$
    – Asaf Karagila
    Feb 23, 2012 at 14:03
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    $\begingroup$ @Paul: No, these are ordinals approaching $\gamma$. Read my comments (and the answer) again more closely. Recall that the cofinality of $\aleph_\alpha$ is $\aleph_\alpha$ if $\alpha$ is not a limit ordinal and it is the cofinality of $\alpha$ if it is a limit ordinal. $\endgroup$
    – Asaf Karagila
    Feb 23, 2012 at 14:55

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