Answer
See explanation
Work Step by Step
Let \(A\) and \(B\) be any two countable sets. We wish to show that \(A \cup B\) is countable. There are two cases to consider:
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**Case 1. One of the sets is finite.**
- Suppose, without loss of generality, that \(A\) is finite and \(B\) is countable (which might be countably infinite).
- From the result of one exercise, we know that the disjoint union of a finite set and a countably infinite set is countably infinite.
- (And if \(B\) were finite as well, then \(A \cup B\) would clearly be finite and hence countable.)
- Therefore, \(A \cup B\) is countable.
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**Case 2. Both \(A\) and \(B\) are countably infinite.**
- Since each of \(A\) and \(B\) is countably infinite, there exist bijections \(f:\mathbb{N} \to A\) and \(g:\mathbb{N} \to B\).
- The result of the previous exercise shows that we can interleave these two listings to form an enumeration of \(A \cup B\). For instance, define a function \(h: \mathbb{N} \to A \cup B\) by
\[
h(n) =
\begin{cases}
f\left(\tfrac{n+1}{2}\right), & \text{if } n \text{ is odd}, \\
g\left(\tfrac{n}{2}\right), & \text{if } n \text{ is even}.
\end{cases}
\]
- Every element of \(A\) appears as some \(f(k)\) and every element of \(B\) appears as some \(g(k)\), so every element of \(A \cup B\) appears in the sequence \(\{h(n)\}_{n \in \mathbb{N}}\).
- Thus, \(A \cup B\) is countably infinite.
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**Conclusion:**
In either case, whether one set is finite or both are countably infinite, the union \(A \cup B\) is countable. This proves that the union of any two countable sets is countable.
