# Cardinal sums over successor ordinals are equal?

Can you tell me if the following claim and subsequent proof are correct? Thanks.

Claim: If $\alpha = \delta + 1$ is an infinite successor ordinal then $\sum_{\xi < \alpha } \kappa_\xi = \sum_{\xi < \delta} \kappa_\xi$.

Proof: Let $f: \delta + 1 \to \delta$ be a bijection. Then $f$ induces a bijection $F: \bigcup_{\xi < \alpha} \kappa_\xi \times \{\xi \} \to \bigcup_{\xi < \delta} \kappa_\xi \times \{\xi \}$ so that $\sum_{\xi < \alpha} \kappa_\xi = \sum_{\xi < \delta} \kappa_\xi$.

Thanks for your help.

-

I think your claim is false, just let all $\kappa_\xi=\omega$ for $\xi\leq\omega$ and let $\kappa_{\omega+1}=\omega_1$. Then one sum is $\omega$, the other is $\omega_1$. In fact, you need to assume in your proof that you have bijections between different $\kappa_\xi$'s (which does not work in my example obviously).

It is quite easily shown that $\sum\limits_{\xi<\alpha}\kappa_\xi=\alpha\cdot\sup\limits_{\xi\leq\alpha}\kappa_\xi$. This makes clear that your claim is true if $\kappa_\delta$ is not bigger than $\sup\limits_{\xi<\delta}\kappa_\xi$.

-
No, this claim is false. But it is false because you don't put restrictions on $\kappa_\alpha$. It might be that $\kappa_\alpha$ is much much larger than $\alpha$ and any of the previous cardinals, and in which case there is no way that the sums are equal.
If, on the other hand, you put some limitation on $\kappa_\alpha$ (e.g. $\kappa_\alpha\leq\sup\{\kappa_\xi\mid\xi<\alpha\}$ then the equality holds.
Your proof, however, is flawed regardless. To say that $f$ induces a bijection between the unions is to assume that the sums are already equal.