Disjoint refinement of a countable set of sets

Question

Say I have a set $A=\{X_{n}|n\in\mathbb{N}\}$ where for each $n\in\mathbb{N}$, $X_{n}$ is countable. I need to prove that there exists a disjoint refinement $D=\{D_{n}|n\in\mathbb{N}\}$ of $A$ such that: For each $n\in\mathbb{N}, D_{n}\subseteq X_{n}$ and $D_{n}$ is infinite and $m < n \Rightarrow D_{m}\cap D_{n}=\emptyset$.

I had two ideas: say that for every $m < n$ we managed to find such $D_{m}$. So, defining $D_{n}=X_{n}\backslash\bigcup_{m<n} D_{m}$, but I can't prove that such $D_{n}$ is infinite.

Another idea I had is defining a well - ordering on each of the $X_{n}$ (possible because of AC) and defining $a_1=min\{x|x\in X_{n}\land x\notin\bigcup_{m<n} D_{m}\}$, but it seems to give the same result.

Is one of my ideas the right way or at least close?

Answer 1

Since $\bigcup_{n\in\Bbb N}X_n$ is countable, we may without loss of generality assume that each $X_n$ is a subset of $\Bbb N$.¹ Let $P=\{\langle \ell_k,m_k\rangle:k\in\Bbb N\}$ be an enumeration of $\Bbb N\times\Bbb N$. I’ll construct the sets $D_n$ by recursion.

Let $D_n(0)=\varnothing$ for each $n\in\Bbb N$. At stage $k$ let

$$D_n(k+1)=\begin{cases} D_n(k)\cup\left\{\min\left(X_n\setminus\bigcup_\limits{i\in\Bbb N}D_i(k)\right)\right\},&\text{if }n=\ell_k\\ D_n(k),&\text{otherwise.} \end{cases}$$

In other words, if $n=k_\ell$, we add to $D_n$ the first member of $X_n$ that has not yet been assigned to any $D_i$. This is always possible, since $X_{\ell_k}$ is infinite and $\bigcup_{i\in\Bbb N}D_i(k)$ is finite, so the recursive construction goes through to construct $D_n(k)$ for each $n,k\in\Bbb N$.

Now let $D_n=\bigcup_{k\in\Bbb N}D_n(k)$ for each $n\in\Bbb N$; clearly $D_n\subseteq X_n$. For each $n\in\Bbb N$ the set of $k\in\Bbb N$ such that $\ell_k=n$ is infinite, since $P$ includes a pair $\langle n,m\rangle$ for each $m\in\Bbb N$, so $D_n$ is infinite. As it is also clear that the construction produces pairwise disjoint sets $D_n$, we’re done.

¹ To see this, let $X=\bigcup_{n\in\Bbb N}X_n$. $X$ is countably infinite, so there is a bijection $\varphi:X\to\Bbb N$. We can replace each $X_n$ by its image $\varphi[X_n]\subseteq\Bbb N$, find infinite pairwise disjoint subsets $D_n$ of theses images, and then use $\varphi^{-1}$ to carry them back to infinite pairwise disjoint subsets $\varphi^{-1}[D_n]$ of the sets $X_n$.