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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.

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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.

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    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?2012-02-23
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    @Paul: I don't understand your comment.2012-02-23
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    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.2012-02-23
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    To me it seems you are saying the sequence $\langle \aleph_{\beta} \mid \beta < \gamma \rangle$ is such a sequence.2012-02-23
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    @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.2012-02-23
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    But how do I know it is approaching $\kappa$? And surely that sequence is automatically continuous? Do you mean $\langle \alpha_{\xi} \mid \xi < \lambda \rangle$, the original sequence?2012-02-23
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    @Paul: Let $\kappa=\aleph_\gamma$. To say that the cofinality of $\kappa$ is $\lambda$ means that there is a strictly increasing sequence $\alpha_\xi$ for $\xi<\lambda$ such that $\sup\alpha_\xi=\gamma$. This means that $\sup\aleph_{\alpha_\xi}=\aleph_\gamma=\kappa$. The sequence $\alpha_\xi$ has only $\lambda$ many limit points, let $\beta_\tau$ for $\tau<\lambda$ be the sequence composed by $\alpha_\xi$'s and their limit points. This is a continuous sequence and thus $\aleph_{\beta_\tau}$ is a continuous sequence of cardinals whose limit is still $\kappa$.2012-02-23
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    Don't you mean $\sup \alpha_{\xi} = \kappa$?2012-02-23
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    @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.2012-02-23
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    OK, I will try my best to understand this. Thanks for all the help so far.2012-02-23
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2595/discussion-between-paul-slevin-and-asaf-karagila)2012-02-23