$\lambda_i=\kappa_i+1$, $I=\omega$, and $\sum\kappa_i=\aleph_1$, $\prod\lambda_i=2^{\aleph_0}$

Question

The following is part of the answer of this question,

Note that if you simply require that $\kappa_i$ are ordinals and that $\kappa_i<\lambda_i$, we can get an odd situation, with $\lambda_i=\kappa_i+1$, $I=\omega$, and $\sum\kappa_i=\aleph_1$, $\prod\lambda_i=2^{\aleph_0}$, but the two cardinals are incomparable in that case!

I cannot think of what $\kappa_i$ should be for these equalities to hold. I this is worth a separate question.

Thanks in advance!

Answer 1

This is a consistency result. It is not an actual example, since under the axiom of choice the countable sum of countable ordinals is countable and not $\aleph_1$. But your question was about the failure of König's theorem without choice.

In short, if we take a singular cardinal $\lambda$, we can make it into $\aleph_1$ of a universe of $\sf ZF$, in which case its cofinality will be countable. That means that there will be some sequence like that, although it is not necessary that this sequence is somehow "easy to define".