Suppose $$\kappa= \sup_{\alpha < \lambda} \kappa_\alpha$$ where $\kappa$ is an infinite cardinal and $\kappa_\alpha$ are cardinals, $\lambda$ is a non-zero limit ordinal, $\lambda < \kappa$ and $\kappa_\alpha < \kappa$ for every $\alpha < \lambda$.
Does there exist a sequence $\{\kappa_\alpha': \alpha < \theta\}$ with $\kappa_\alpha' < \kappa, \theta < \kappa$, $\theta$ is a limit ordinal and $$\kappa = \sup_{\alpha < \theta} \kappa'_\alpha; \quad (\alpha < \beta \implies\kappa'_{\alpha}< \kappa'_\beta)$$
I saw the answers here: Question on singular cardinals but I was not able to understand them. If someone could give an explicit construction, I'd be glad:
My attempt: Using unions, one can make the sequence increasing, and then one wants to try to assume a subset that is strictly increasing but I'm not sure how to completely make this formal.
Yes.
Simply consider $\{\kappa_\alpha\mid\alpha<\lambda\}$ as a set of ordinals, it is a naturally well-ordered set. So we may simply consider its order type as $\theta$. Because $\kappa$ is an infinite cardinal, this $\theta$ has to be a limit ordinal as well.
But we can do better. The set is isomorphic to an ordinal (by considering the order type of the set), so it has a cofinal sequence which is indexed by a cardinal which in turn has to be the cofinality of $\kappa$.