Axiom of Choice: General associative and commutative law for infinite sum/product of cardinals

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In his book Basic Set Theory, Levy wrote:

The full associative law and associative-commutative law can be proved only by means of axiom of choice.

(page 104 in the Dover edition)

Here Levy is referring to infinite sum of well-ordered cardinals (a.k.a. alephs), which is defined as follows:

$$\sum_{i\in I} \kappa_i:=\left|\bigcup_{i\in I}\{i\}\times\kappa_i\right|$$

I don't know what version of the associative-commutative law Levy has in mind, but one is as follows:

$$\sum_{i\in I} \kappa_i=\sum_{j\in J}\sum_{i\in X_j}\kappa_i$$

where $\{X_j :j\in J\}$ is a partition of $I$.

I can't see how this uses the axiom of choice. One could say $\sum_{i\in X_j} \kappa_i$ might not be a well-ordered cardinal and thus the sum on the right hand side is not defined, but I could view this as a statement purely about existence of bijections between two sets, and the proof is choice-free. Am I missing something obvious or Levy has other versions of the general associative law whose requires the axiom of choice?

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The problem is not that the sum is by definition the cardinality of the union. It is whether or not this definition really holds up to cardinality.

The cardinality of $\sum_{n<\omega}2$ should be the cardinality of any countable union of pairwise disjoint of size $2$. Not just the cardinality of $\omega\times2$. But without choosing bijections, we cannot guarantee that this statement holds. For example, there could be a proper class of pairwise incomparable cardinals, all of which are "countable union of pairs".

And even if you insist on staying within the limited range of ordinals. Remember that a countable union of countable sets could have size $\aleph_1$.


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