Partition an infinite set into countable sets

Question

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:

1. $F_i\subseteq X$ for all $i\in I$

2. $F_{i_1}\cap F_{i_2}=\emptyset$ for all $i_1,i_2\in I$ and $i_1\neq i_2$.

3. $|F_i|=\aleph_0$ 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.

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Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

Answer 1

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 $F_j = \{ x_i \mid j \le i < j' \}$.

Answer 2

Assuming Axiom of Choice, $|X|=|X\times\Bbb N|$ if $|X|\ge\aleph_0$.

Thus there is a bijection $f:X\times\Bbb N\to X$. We define a family $(X_i\mid i\in X)$ by $X_i=f[\{i\}\times\Bbb N]$ for all $i\in X$.

Since $f$ is bijective, $X_{i_1}\cap X_{i_2}=\emptyset$ for all $i_1,i_2\in X$ and $i_1\neq i_2$, and $\bigcup\limits_{i\in X}X_i=\bigcup\limits_{i\in X}f[\{i\}\times\Bbb N]=f[\bigcup\limits_{i\in X}\{i\}\times\Bbb N]=f[X\times\Bbb N]=X$.

Furthermore, $|X_i|=|f[\{i\}\times\Bbb N]|=|\{i\}\times\Bbb N|=|\Bbb N|=\aleph_0$ since $f$ is bijective.

Hence $\{X_i\mid i\in X\}$ is a partition of $X$ in which each element is countably infinite.