$\sum_{n=0}^\omega 2^{\aleph_n}=2^{\aleph_\omega}$
Is this true?
And is there a way in ZFC to let $\infty$ range over ALL infinite ordinals (not a concrete one as in the example above) ?
$\sum_{n=0}^\infty 2^{\aleph_n}=?$
If not -why not ? And is it possible to state this in other set theoretical axiomatic systems or in the surreal number system?
Best, Michael
The first question is independent of $\mathsf{ZFC}$. It is possible that $2^{\aleph_n}=2^{\aleph_\omega}$ for all $n\in\omega$ (see here) and then $$\sum_{n\in\omega}2^{\aleph_n} = \aleph_0 2^{\aleph_\omega}=2^{\aleph_\omega}$$
On the other hand, if $2^{\aleph_n}<2^{\aleph_\omega}$ for all $n\in\omega$ then by König's Theorem (Theorem 5.10 in Jech Set Theory 3rd Edition) $$\sum_{n\in\omega}2^{\aleph_n}<\prod_{n\in\omega}2^{\aleph_\omega}=\left(2^{\aleph_\omega}\right)^{\aleph_0}=2^{\aleph_\omega\aleph_0}=2^{\aleph_\omega}$$
For the second question, you can define $\bigcup_{\lambda\in\text{Ord}} \mathcal{P}(\omega_\lambda)$ where $\mathcal{P}$ denotes powerset. This is a proper class so it doesn't formally have a cardinality.
I don't know what the answer is for the surreal number system.