Infinite Cardinal Addition Without the Axiom of Choice

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.

• You can't construct such an example without using some extra set-theoretic principles. – Zhen Lin Oct 19 '14 at 9:09
• May I know what are the needed extra set-theoretic principle? – Idonknow Oct 19 '14 at 9:11
• It is consistent with ZF that $\Bbb{R}$ is countable union of countable subsets of $\Bbb{R}$. See the related thread in MO – Hanul Jeon Oct 19 '14 at 9:13