Let $X$ be an infinite set. Find a partition of $X$ where each element in the partition is countable.
Let $Y$ be the set of all families $\{F_i\mid i\in I\}$ in which each family satisfies 3 below conditions:
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$F_i\subseteq X$ for all $i\in I$
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$F_{i_1}\cap F_{i_2}=\emptyset$ for all $i_1,i_2\in I$ and $i_1\neq i_2$.
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$|F_i|=\aleph_0$ for all $i\in I$
We define a partial order $<$ on $Y$ by $$\{F_i\mid i\in I\}<\{F_i\mid i\in J\}\iff\{F_i\mid i\in I\}\subseteq\{F_i\mid i\in J\}$$
For any chain $Z$ in $Y$, let $T=\bigcup_{F\in Z}F$, then $T\in Y$ and $T$ is an upper bound of chain $Z$. Thus the requirement of Zorn's Lemma is satisfied. Hence $Y$ has a maximal family $\bar{F}$.
Let $F'=X\setminus\bigcup_{F\in\bar{F}}F$, then $F'$ is finite. If not, $F'$ is infinite. Then there exists $F^\ast\subseteq F'$ such that $|F^\ast|=\aleph_0$. Thus $\bar{F}\cup\{F^\ast\} \in Y$ and $\bar{F} \subsetneq \bar{F}\cup\{F^\ast\}$, which clearly contradicts the maximality of $\bar{F}$. Hence $F^\ast$ is finite.
To sum up $\bar{F}\cup\{F'\}$ is the required partition of $X$ where each element is either finite or countable infinite.
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
Best Answer
Alternate proof:
Well order the set so that it is order isomorphic to an initial ordinal $k$.
$$X = \{ x_i \mid 0 \le i < k \}$$
For each limit ordinal $j$, let $j'$ be the next limit ordinal. Let $F_0 = \{ x_i \mid i \text{ is non-negative integer}\}$.
For each limit ordinal $j < k$, let $F_j = \{ x_i \mid j \le i < j' \}$.