1. **Problem statement:** We want to show that the union of a countable sequence of countable sets $A_n$ is itself countable. 2. **Recall definitions:** - A set is **countable** if it is either finite or has the same cardinality as the natural numbers $\mathbb{N}$. - A sequence of sets $A_n$ means we have sets indexed by natural numbers $n=1,2,3,\dots$. 3. **Key idea:** The union $\bigcup_{n=1}^\infty A_n$ can be thought of as all elements appearing in any of the $A_n$. 4. **Strategy:** Since each $A_n$ is countable, we can list its elements as $A_n = \{a_{n1}, a_{n2}, a_{n3}, \dots\}$ (possibly finite). 5. **Construct a listing of the union:** Consider the elements arranged in a two-dimensional array: $$ \begin{matrix} a_{11} & a_{12} & a_{13} & \cdots \\ a_{21} & a_{22} & a_{23} & \cdots \\ a_{31} & a_{32} & a_{33} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{matrix} $$ 6. **Enumerate elements diagonally:** We can list elements by moving along diagonals: - First $a_{11}$, - Then $a_{12}, a_{21}$, - Then $a_{13}, a_{22}, a_{31}$, - And so on. 7. **Remove duplicates:** Since sets may overlap, we skip repeated elements to get a sequence without repetition. 8. **Conclusion:** This process produces a sequence listing all elements of $\bigcup_{n=1}^\infty A_n$, showing it is countable. **Final answer:** The countable union of countable sets is countable.