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

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

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

1 Answer 1


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.


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