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
Best Answer
Here is another approach: assume that $\kappa$ is partitioned by $\mu<\lambda$ many 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 $\kappa$.