The following is a lemma in Just/Weese on page 179:
Lemma 17: Let $\kappa$ be an infinite cardinal. Then $\kappa$ is singular iff there exist an $\alpha < \kappa$ and a set of cardinals $\{\kappa_\xi : \xi < \alpha \}$ such that $\kappa_\xi < \kappa$ for every $\xi < \alpha$ and $\kappa = \sum_{\xi < \alpha}\kappa_\xi$.
I think the following constitutes a proof of the $\implies$ direction, can you please tell me where it's wrong? Thank you!
Let $\alpha = \mathrm{cf}(\kappa) < \kappa$. Then by a previous exercise (23 (f)) there exists a strictly increasing function $f: \alpha \to \kappa$ such that the range of $f$ is cofinal in $\kappa$. Now for $\xi < \alpha$ define $\kappa_\xi = |f(\xi)|$. Now we want to show that $\kappa = \sum_{\xi < \alpha}\kappa_\xi$. To see this observe that "$\ge$" immediately follows from theorem 14 on the same page.
The part I need you to check starts here:
To show $\le$, assume that we have strict inequality $ \sum_{\xi < \alpha}\kappa_\xi < \kappa $. We distinguish two cases: $\kappa$ is either an infinite successor cardinal or a limit cardinal. If $\kappa = \left ( \sum_{\xi < \alpha}\kappa_\xi \right )^+$ is an infinite successor cardinal then by corollary 15 (on the same page) $\kappa$ cannot be the union of fewer than $\kappa$ sets each of which has cardinality less than $\kappa$. But $\bigcup_{\xi < \alpha}f(\xi) = f[\alpha] = \kappa $ where $|f(\xi)| = \kappa_\xi < \kappa$ and $\alpha < \kappa$ which would be a contradiction. Hence $\kappa$ must be a limit cardinal so that there exists a cardinal $\eta$ with $\sum_{\xi < \alpha}\kappa_\xi < \eta < \kappa$. But $f$ is cofinal in $\kappa$ hence there is $\xi$ such that $f(\xi) \ge \eta$. Then $\kappa_\xi = |f(\xi)| \ge \eta$. Which is a contradiction.
We have:
And also, the reason why I am asking this question: in the proof in the book, $\kappa_\xi$ are defined as $\kappa_\xi = \left | f(\xi) \setminus \sum_{\eta < \xi} f(\eta) \right |$.