Answer
See explanation
Work Step by Step
A standard argument proceeds by noting that each of \(A\) and \(B\) can be “listed” (put into bijection with \(\mathbb{N}\)), and then using that listing to enumerate all ordered pairs in \(A\times B\). Here’s a concise version:
---
## 1. Enumerate \(A\) and \(B\)
Since \(A\) and \(B\) are **countably infinite**, there exist bijections
\[
f \colon \mathbb{N} \,\to\, A
\quad\text{and}\quad
g \colon \mathbb{N} \,\to\, B.
\]
In other words, we can label the elements of \(A\) as \(\{a_1, a_2, a_3, \dots\}\) and the elements of \(B\) as \(\{b_1, b_2, b_3, \dots\}\).
---
## 2. Set Up a Bijection \(\mathbb{N}\times \mathbb{N} \to A\times B\)
Consider the function
\[
h \colon \mathbb{N}\times \mathbb{N} \;\to\; A\times B,
\quad
h(m,n) \;=\; \bigl(f(m),\,g(n)\bigr).
\]
- Because \(f\) and \(g\) are bijections, every pair \(\bigl(a_i,b_j\bigr)\) in \(A\times B\) is hit by exactly one pair \((m,n)\in \mathbb{N}\times \mathbb{N}\).
- Hence \(h\) is itself a **bijection**.
Since \(\mathbb{N}\times \mathbb{N}\) is **countably infinite**, any set in bijection with it (namely \(A\times B\)) must also be countably infinite.
---
## 3. Alternate “Union of Slices” Argument
Another common proof uses the fact that a **countable union of countable sets is countable**:
1. For each \(a \in A\), the “slice” \(\{a\}\times B\) is in bijection with \(B\), hence countable.
2. Because \(A\) is countably infinite, there are countably many such slices.
3. Taking the union of all slices \(\bigcup_{a\in A}(\{a\}\times B)\) recovers \(A\times B\).
4. A countable union of countably many countable sets is again countable.
Since \(A\times B\) is clearly infinite, it follows it is **countably infinite**.
---
**Conclusion**: In either approach, we conclude that the Cartesian product of two countably infinite sets, \(A\times B\), is itself **countably infinite**.
