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Here is an exercise I'm trying to solve:

Prove that for $\kappa$ an infinite cardinal, cof($\kappa$) is the least $\lambda$ such that there is $\langle A_\alpha \subseteq \kappa\ | \ \alpha < \lambda \rangle $ such that $\bigcup_{\alpha < \lambda} A_\alpha = \kappa$ and $|A_\alpha| < \kappa$ for all $\alpha < \lambda$

So, the sketch of what I'd do is the following:

  • Given $f:\lambda\rightarrow \kappa$ cofinal, I'd define $\langle A_\alpha \ | \ \alpha < \lambda \rangle $ by $A_\alpha = f(\alpha)$, which satisfies all conditions required.
  • Given $\langle A_\alpha \ | \ \alpha < \lambda \rangle $ as specified, I'd define $f:\lambda \rightarrow \kappa$ as follows: $$\begin{align}f:\lambda &\longrightarrow \kappa\\\alpha &\longmapsto \text{ot}(\bigcup_{\beta \le \alpha} A_\beta)\end{align}$$ By ot$()$ I mean the order type, i.e. the (unique) ordinal isomorphic to a well-ordered set.

Now $f$, as defined in the second step, is well-defined, since by hypothesis $|f(\alpha)|<\kappa$ and therefore $f(\alpha)<\kappa$. I have some problems though in showing that it is cofinal, which would complete basically the proof. Is this approach correct? If this is the case, some hints for proving $f$ cofinal? Thanks

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Here is an another approach: assume that $\kappa$ is partitioned by $\mu<\lambda$ sets $\langle B_\xi\mid \xi<\mu\rangle$, whose cardinality is strictly less than $\kappa$. Especially, we have $$\sup_{\xi<\mu}|B_\xi| = \kappa$$ (since $|\bigcup_{\xi<\mu} B_\xi|=\sup_{\xi<\mu}|B_\xi|$.) It contradicts with the assumption that $\lambda$ is the cofinality of $\kappa$.

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  • $\begingroup$ Thank you. Taking the cardinality of the sets is certainly the way to go. The problem with Kamsma's answer was that taking the sup of the sets is not well defined, since the sets $A_\alpha$ may be unbounded in $\kappa$. $\endgroup$
    – Lorenzo
    Feb 8 '20 at 16:15
  • $\begingroup$ @Lorenzo Yes, you are right. I have to remove this comment. $\endgroup$
    – Hanul Jeon
    Feb 8 '20 at 16:17
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    $\begingroup$ This was indeed the trick to fix my answer, instead of taking downwards closure (which of course doesn't work, thanks for pointing that out), we take the cardinality. $\endgroup$ Feb 8 '20 at 16:35

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