Answer
\(\mathbb{Z}\times \mathbb{Z}\) Is Countably Infinite
Being an infinite set that is at most countable implies that \(\mathbb{Z}\times \mathbb{Z}\) is **countably infinite**.
Work Step by Step
A common way to prove \(\mathbb{Z} \times \mathbb{Z}\) is **countably infinite** is to use the fact that \(\mathbb{Z}\) itself is countably infinite and that a countable union of countable sets is countable.
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## 1. Express \(\mathbb{Z}\times \mathbb{Z}\) as a Countable Union of Countable Sets
Observe that
\[
\mathbb{Z}\times \mathbb{Z}
\;=\;
\bigcup_{m\in \mathbb{Z}}\bigl(\{m\}\times \mathbb{Z}\bigr).
\]
- For each fixed \(m\in \mathbb{Z}\), the set \(\{m\}\times \mathbb{Z}\) is in bijection with \(\mathbb{Z}\) itself (simply “ignore” the first coordinate). Since \(\mathbb{Z}\) is countably infinite, each slice \(\{m\}\times \mathbb{Z}\) is countably infinite.
- There are countably many choices for \(m\) (because \(\mathbb{Z}\) is countably infinite), so we have a **countable union** of countably infinite sets.
By a standard theorem, a countable union of countable sets is countable. Hence \(\mathbb{Z}\times \mathbb{Z}\) is **at most** countable.
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## 2. \(\mathbb{Z}\times \mathbb{Z}\) Is Infinite
Clearly, \(\mathbb{Z}\times \mathbb{Z}\) has infinitely many elements (for example, \((0,n)\) is distinct for each integer \(n\)). So it is not finite.
