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In the book 'Introduction to Set Theory' by Hrbacek and Jech, cardinal addition is defined as

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

where $|A_i|=k_i$ for all $i \in I$ and $\langle A_i\mid i \in I\rangle$ is a system of mutually disjoint sets.

The author stated that: Without the Axiom of Choice, there may exists two system $\langle A_n\mid n \in \mathbb{N}\rangle$, $\langle A^{\prime}_n\mid n \in \mathbb{N}\rangle$ of mutually disjoint sets such that each $A_n$ and each $A^{\prime}_n$ has two elements, but $\bigcup_{n=0}^\infty{A_n}$ is not equipotent to $\bigcup_{n=0}^\infty{A^{\prime}_n}$.

I try to construct such example but I fail to do so. I tried the set of even natural numbers and natural numbers but they have the same cardinality after infinite union.

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  • $\begingroup$ You can't construct such an example without using some extra set-theoretic principles. $\endgroup$ – Zhen Lin Oct 19 '14 at 9:09
  • $\begingroup$ May I know what are the needed extra set-theoretic principle? $\endgroup$ – Idonknow Oct 19 '14 at 9:11
  • $\begingroup$ It is consistent with ZF that $\Bbb{R}$ is countable union of countable subsets of $\Bbb{R}$. See the related thread in MO $\endgroup$ – Hanul Jeon Oct 19 '14 at 9:13
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You can't construct such example. The fact that it is only consistent and not provable is exactly the issue here. Not only you can't construct one, you can't prove that one exists when the axiom of choice fails.

These remarks, at the introductory level, are meant to guide the reader and suggest that a choice-free proof cannot be found. This is used to both (1) reassure the reader when they can't think of a choice free argument; and (2) to bring to the attention of the reader that even if they think that they evaded the axiom of choice, they have not.

Cardinality of union of pairwise disjoint elements needs choice? and axiom of choice: cardinality of general disjoint union have more information about situations where this happens.

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