Suppose we have a set of ordinals such that their supremum is a cardinal (not in the original set).
$\sup\{\alpha\}=\kappa$
I'm interested in the supremum of the cardinalities of those ordinals: $\sup\{|\alpha|\}$. I've seen many proofs that seem to assume that $\sup\{|\alpha|\}=\kappa$ also. I can see how this would be the case if the set $\{\alpha\}$ contained an infinite subset of cardinals whose supremum was also $\kappa$, but I can think of a simple counterexample that would seem to indicate that's not always the case.
Take our set to be $\{\alpha\mid\omega\le\alpha\lt\omega_1\}$. The supremum of this set is clearly $\omega_1$ (with cardinality $\aleph_1$ if you prefer). But each ordinal $\alpha$ is countably infinite, so $\forall\alpha,|\alpha|=\omega$ (or $\aleph_0$ if you insist). But then $$\sup\{|\alpha|\}=\sup\{\omega\}=\omega\ne\omega_1=\sup\{\alpha\}.$$
What am I missing?
Motivation: I'm trying to show that if $\kappa$ is an infinite cardinal, these definitions for cofinality are equivalent:
$$cof(\kappa)=\inf\{\beta\mid\{\alpha_\xi\}_{\xi\lt\beta}\text{ is cofinal in }\kappa\}\equiv\inf\{\delta\mid\sum_{\xi<\delta}\kappa_\xi=\kappa\}.$$
To establish the RHS, (after taking $\{\kappa_\xi\}=\{|\alpha_\xi|\}$) it's easy to show the summation is bounded above by $\kappa$. To show that it is also bounded below by $\kappa$, an example proof I found claims that since each $$\kappa_\xi\le\sum_{\xi<\delta}\kappa_\xi$$ then the summation, as an upper bound of all of the $\kappa_\xi$'s, must be larger than the smallest upper bound (supremum) of the $\kappa_\xi$'s, so $$\kappa\le\sup\{\alpha_\xi\}=\sup\{|\alpha_\xi|\}=\sup\{\kappa_\xi\}\le\sum_{\xi<\delta}\kappa_\xi$$ but I'm not so sure about the first equality. Any tips would be appreciated.