Infinite Series of infinite cardinals in ZFC

$$\sum_{n=0}^\omega 2^{\aleph_n}=2^{\aleph_\omega}$$

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

• What do you even mean by the first sum? Are you defining it as the union of representatives for those cardinalities where $n<\omega$? Feb 12 at 20:08

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.