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.

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

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    $\begingroup$ It has some issues, which may very well be typos, but they are issues anyway. First, note the typo in your second condition when defining the family. Second, is it obvious that $Y$ is a set? Third, if $F^*$ is infinite, it is not necessarily countable, so what you did there needs to be modified. Also, it is maximality rather than minimality that is contradicted. Fourth, if $F^*$ is finite but nonempty, you must modify what you did, since your third condition in the definition of $Y$ asks all pieces of the partition to be infinite. $\endgroup$ – Andrés E. Caicedo Sep 14 '18 at 11:44
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    $\begingroup$ (If you don't insist that the pieces of the partition should be infinite, as your final paragraph suggests, then the problem is trivial, as you can partition any set into singletons. So the last paragraph should be modified as well.) $\endgroup$ – Andrés E. Caicedo Sep 14 '18 at 11:48
  • $\begingroup$ Hi @AndrésE.Caicedo I have fixed the all issues. I only require my partition to contain countable sets ^^ Please have a check! $\endgroup$ – Navier_Stokes Sep 14 '18 at 11:59
  • $\begingroup$ Again, unless you require all pieces of the partition to be infinite, the problem is trivial, and it would be ridiculous to use this approach. You must fix what you do when $F'$ is finite. $\endgroup$ – Andrés E. Caicedo Sep 14 '18 at 12:33
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    $\begingroup$ Well, write that down on the body of the question, not in comments. $\endgroup$ – Andrés E. Caicedo Sep 14 '18 at 12:40

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

  • $\begingroup$ Just one minor suggestion, I would be better if you use Mathjax to format your answer. $\endgroup$ – Navier_Stokes Sep 15 '18 at 3:48

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.


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