How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? 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.
 A: I'll first make a comment about the concern in the title of the question, and it seems to break down into successor and limit cases.


*

*If $\kappa = \lambda^+$ is a successor cardinal, then $| \alpha | \leq \lambda$ for all ordinals $\alpha < \kappa$, and it easily follows that $\sup \{ | \alpha | : \alpha < \kappa \} = \lambda$.  (This is what you noticed for $\kappa = \aleph_1 = \aleph_0^+$.)

*If $\kappa$ is a limit cardinal, then $\mu^+ < \kappa$ for each $\mu < \kappa$.  From here it easily follows that $\sup \{ | \alpha | : \alpha < \kappa \} = \kappa$.


The above dichotomy also holds whenever $\{ \alpha_\xi \}_{\xi < \nu}$ is an arbitrary family of ordinals cofinal in $\kappa$.
Of course, it is relatively easy to show that successor cardinals are regular:  If $\delta < \kappa = \lambda^+$ and $\{ \alpha_\xi \}_{\xi < \delta}$ is an (increasing) sequence of ordinals $< \kappa$, then let $\beta = \sup_{\xi < \delta} \alpha_\xi = \bigcup_{\xi < \delta} \alpha_\xi$.  Note that $$| \beta | = \left| \bigcup_{\xi < \delta} \alpha_\xi \right| \leq \sum_{\xi < \delta} | \alpha_\xi | \leq \sum_{\xi < \delta} \lambda = | \delta | \cdot \lambda = \lambda < \kappa.$$  Thus $\{ \alpha_\xi \}_{\xi < \delta}$ is not cofinal in $\kappa$.  Note that this also contains the idea for proving the summation characterisation of cofinality in this case: if $\{ \kappa_\xi \}_{\xi < \delta}$ is a family of cardinals $< \kappa = \lambda^+$, then $\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \lambda = | \delta | \cdot \lambda = \max \{ |\delta| , \lambda \}$, so if this sum equals $\kappa = \lambda^+$, it must be that $| \delta | = \kappa$.)
For a limit cardinal $\kappa$, letting $\mu = \mathrm{cof} ( \kappa )$, note that there is an (increasing) sequence $\{ \kappa_\xi \}_{\xi < \mu}$ of cardinals $< \kappa$ which is cofinal in $\kappa$.  It is relatively easy to show that $\sum_{\xi < \mu} \kappa_\xi = \kappa$.


*

*Clearly $\kappa_\eta \leq \sum_{\xi < \mu} \kappa_\xi$ for all $\eta < \mu$, and therefore $\kappa \leq \sum_{\xi < \mu} \kappa_\xi$.

*Given $\delta < \mu$ note that $\kappa_\xi < \kappa_\delta$ for all $\xi < \delta$ and so $\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \kappa_\delta = | \delta | \cdot \kappa_\delta = \max \{ | \delta | , \kappa_\delta \}$.  As $\kappa_\delta < \kappa$, and $| \delta | < \kappa$, it follows that $\sum_{\xi < \delta} \kappa_\xi < \kappa$.  Thus, $\sum_{\xi < \mu} \kappa_\xi \leq \kappa$.


But suppose $\delta < \mu$ and $\{ \kappa_\xi \}_{\xi < \delta}$ is any sequence of cardinals $< \kappa$.  Then $\{ \kappa_\xi \}_{\xi < \delta}$ is not cofinal in $\kappa$, and so there is a cardinal $\nu < \kappa$ such that $\kappa_\xi \leq \nu$ for all $\xi < \delta$.  Then $$\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \nu = | \delta | \cdot \nu = \max \{ | \delta | , \nu \}.$$
As $\nu , | \delta | < \kappa$, we have that $\sum_{\xi < \delta} \kappa_\xi < \kappa$.
A: Well, as you noted the supremum of the cardinalities is never larger than the supremum of ordinals.
To have a better understanding of this equivalence I suggest to divide into two cases:


*

*If $\kappa$ is regular then the least $\beta$ is $\kappa$ itself, and since we have: 
$$\inf\left\{\beta\mid\exists \{\alpha_\xi\}_{\xi<\beta}\text{ cofinal in }\kappa\right\}=\kappa\geq\inf\left\{\delta\mid\exists\{\kappa_\xi\}_{\xi<\delta}: \sum_{\xi<\delta}\kappa_\xi=\kappa\right\}$$
Note that $\kappa\geq\inf$ because we can always write $\kappa$ as sum of $\kappa$ singletons, so trivially we have the above.
On the other hand if $\{\kappa_\xi\}_{\xi<\delta}$ is a sequence of cardinals of minimal order type then the ordinals $\alpha_\xi=\left(\sum_{\zeta<\xi}\kappa_\zeta\right)+\kappa_\xi$ (this sum is an ordinal sum!) form a cofinal sequence in $\kappa$, and therefore the equality follows.

*If $\kappa$ is singular then we know it can be written as a limit of cardinals, then we indeed have $\sup\{\alpha_\xi\mid\xi<\delta\}=\sup\{|\alpha_\xi|\mid\xi<\delta\}$ as wanted.
It is therefore sufficient to show that there is a minimal cofinal sequence made of cardinals, but this is simple.
