Why can I write a singular cardinal as the limit of an increasing, continuous sequence of cardinals.

Question

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.

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.