Question about defining sequence via transfinite induction.

Question

I have sequence $A_i$ where $i\in\mathbb{N}$ and every $A_i$ is countable.

Let $A=\bigcup_{i\in\mathbb{N}}A_i$.

I want to find new sequence of countable subsets of $A$ such that $C_i\subseteq A_i$ and $\forall n\neq m:C_n\cap C_m=\emptyset$.

I do it that way. First, by Zelmero theorem, I take $(A,\leq)$ to be well-ordered set.

Now I define a injective sequence ($B_i$) of elements of $A$ such that:
the smallest $b_1\in A$ such that $b_1\in A_1$
the smallest $b_2\in A\setminus\{b_1\}$ such that $b_2\in A_1$
the smallest $b_3\in A\setminus\{b_1,b_2\}$ such that $b_3\in A_2$
the smallest $b_4\in A\setminus\{b_1,b_2,b_3\}$ such that $b_4\in A_1$
the smallest $b_5\in A\setminus\{b_1,b_2,b_3,b_4\}$ such that $b_5\in A_2$
the smallest $b_6\in A\setminus\{b_1,b_2,b_3,b_4,b_5\}$ such that $b_6\in A_3$
$\dots$

Although not formally defined such sequence exist. Now let $d_i$ be fractal sequence. Then we take order isomorphism $f$, such that $f(d_i)=b_i$ and take $C_n=\{f(d_i):d_i=n\}$. Now I believe we have found sequence we were looking for.

Is such reasoning correct?

---

Possibly improved answer

Let $A=\bigcup_{i\in\mathbb{N}}A_i$. And let $(A,\leq)$ be well ordered set.

$B=\Sigma_{n=0}^\infty n$ where addition is meant as defined for ordinal numbers.

Now $B\subset\omega\times\omega$
Let's define order isomorphism $f:\omega\to B$ and projection $h:B\to\omega$ such that $h(x,y)=y$. Finally let's take $b_n=g\circ f(n)$

If everything is correct, then $b_n$ is what I called fractal sequence earlier (but now starting with 0). So $b_0=0,b_1=0,b_2=1,b_3=0,b_4=1,b_5=2,b_6=0,b_7=1\dots$

Now let's define another injective sequence:
$d_1=\min_{(A,\leq)}\{x:x\in A_{b_1}\}$
$\dots$
$d_n=\min_{(A,\leq)}\{x:x\in A_{b_n}$ and $\forall m<b_n: x\neq b_m\}$

Finally, let's define sequence of subsets of elements of sequence $A_i$

$$ C_n=\{d_k:k\in h^{-1}(n)\} $$

Answer 1

As written, I neither see how this would yield a well-defined sequence (the crux lies in "...") nor do I understand what $f$ (and thus - by extension - $C_n$) is supposed to be.

It seems that you are looking for a sequence $(C_n \mid n \in \mathbb N)$ of pairwise disjoint sets such that $C_n \subseteq A_n$ and such that $A = \bigcup_{n \in \mathbb N} C_n$. This is easily arranged:

Recursively define

• $C_0 := A_0$ (or begin with $C_1 := A_1$ in case you don't consider $0$ to be a natural number) and
• $C_{n+1} := A_{n+1} \setminus \bigcup_{i \le n} A_i$.

Clearly $C_n \subseteq A_n$ and thus $\bigcup_{n \in \mathbb N} C_n \subseteq A$. It's also easy to see that the sequence $(C_n \mid n \in \mathbb N)$ is pairwise disjoint. So let me show that $A \subseteq \bigcup_{n \in \mathbb N} C_n$:

Fix $x \in A$ and let $n \in \mathbb N$ be minimal such that $x \in A_n$. Then $x \not \in \bigcup_{i \le n - 1} A_i$ and thus $x \in C_n \subseteq \bigcup_{m \in \mathbb N} C_m$.