Cardinality of disjoint infinite union

Question

Let be $M:=\bigcup\limits_{n=1}^{\infty}M_n$ a disjoint union. Show that $|M|=\sum\limits_{n=1}^{\infty}|M_n|$. (Hint: $|\cdot|$ denotes the cardinality)

The sample solution lists three cases:

1.) If there exists a $n_0\in\mathbb{N}$ such that $|M_{n_0}|=\infty$ then $\sum\limits_{n=1}^{\infty}|M_n|=\infty$ and also $|M|=\infty$.

2.) If there exist infinitely many $n$ such that $|M_n|\geq 1$ then $\sum\limits_{n=1}^{\infty}|M_n|=\infty$ and $|M|=\infty$.

3.) If there exists a $N\in\mathbb{N}$ such that for all $n>N$ we have $|M_n|=0$, then $|M|=\sum\limits_{n=1}^N|M_n|=\sum\limits_{n=1}^{\infty}|M_n|$.

---

Is it possible to merge all cases into one? We could simply write $$ \left|\bigcup\limits_{n=1}^mM_n\right|=\sum\limits_{n=1}^m|M_n|\implies \lim\limits_{m\to\infty}\left|\bigcup\limits_{n=1}^mM_n\right|=|M|=\sum\limits_{n=1}^{\infty}|M_n| $$ because of the rules of cardinality of finite disjoint unions and of the properties of limits of sequences.

Answer 1

Actually, this is exactly the definition of the sum of infinitely many cardinal numbers, so there's nothing to prove.

(Hrbacek & Jech-Introduction to set theory, page 155) 1.1 Definition Let $\langle A_i\mid i \in I\rangle$ be a system of mutually disjoint sets, and let $|A_i|=\kappa_i$ for all $i \in I$. We define the sum of $\langle \kappa_i\mid i \in I\rangle$ by $$\sum_{i \in I}\kappa_i=\left|\bigcup_{i \in I}A _i\right|.$$

I think by $\infty$ you mean $\aleph_0$ or $2^{\aleph_0}$. Your proof is invalid since if one $\infty_1=\aleph_0$ and the other $\infty_2=2^{\aleph_0}$, then $\infty_1<\infty_2$.

However, the definition above holds for arbitrary sums and arbitrary disjoint sets with arbitrary cardinality (could be greater than $\aleph_0$).