Answer
See explanation
Work Step by Step
A standard proof uses the idea of **listing** (enumerating) each set \(A_i\) and then **interleaving** those lists so that every element of every \(A_i\) appears somewhere in a single infinite list. Here’s a common approach:
---
## 1. Each \(A_i\) is Countable
Because each \(A_i\) is countable, there is a way to list its elements. Concretely, for each \(i \in \mathbb{N}\),
\[
A_i = \{\, a_{i,1},\; a_{i,2},\; a_{i,3},\;\dots \}.
\]
(If any \(A_i\) is finite, we can still list its elements in a finite list and pad with a dummy “no element” beyond that—this won’t affect the argument.)
---
## 2. Arrange the Union in a Grid
Imagine writing these countably many sets in a two‐dimensional grid:
\[
\begin{array}{cccccc}
A_1 : & a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & \dots \\
A_2 : & a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} & \dots \\
A_3 : & a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} & \dots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\]
Each row corresponds to one set \(A_i\), and each row is countably infinite in length.
---
## 3. Enumerate via a Diagonal (or Zigzag) Argument
We now produce a **single** infinite list that covers **every** entry in the grid:
1. Start with \(a_{1,1}\).
2. Then list \(a_{1,2}, a_{2,1}\).
3. Then list \(a_{1,3}, a_{2,2}, a_{3,1}\).
4. Then \(a_{1,4}, a_{2,3}, a_{3,2}, a_{4,1}\).
5. And so on…
Concretely, you can think of summing the indices:
\[
\text{Step }k \text{ lists all }a_{i,j}\text{ where }i+j=k+1.
\]
This procedure visits **every** pair \((i,j)\) of natural numbers exactly once, so eventually it will list any element \(a_{i,j}\) in some finite step.
---
## 4. Conclude the Union is Countable
Because we can place all elements of
\[
\bigcup_{i=1}^{\infty} A_i
\]
into a **single** infinite sequence (by the diagonal listing above), we have effectively defined an **injection** from
\(\bigcup_{i=1}^{\infty} A_i\) into \(\mathbb{N}\). This shows that
\(\bigcup_{i=1}^{\infty} A_i\) is **at most countable**.
Since it is clearly infinite if any \(A_i\) is infinite (or if infinitely many \(A_i\) are nonempty), we deduce it is **countably infinite** in that case. More formally:
1. A countable union of countable sets is at most countable.
2. If the union is infinite, it must be **countably infinite**.
Hence,
\[
\boxed{\text{A countably infinite union of countable sets is itself countable.}}
\]
