Answer
See explanation
Work Step by Step
A common way to show that the disjoint union of a **finite** set and a **countably infinite** set is again **countably infinite** goes like this:
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## 1. Terminology
- A set \(F\) is **finite** if it has only finitely many elements.
- A set \(C\) is **countably infinite** if there is a bijection between \(C\) and the positive integers \(\mathbb{Z}^+\) (or equivalently \(\mathbb{N}\)).
We want to prove that if \(A\) is finite and \(B\) is countably infinite, then their **disjoint union** \(A \cup B\) is also **countably infinite**.
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## 2. Outline of the Proof
1. **Finite sets are countable.**
By definition, a finite set is “countable” (just not *infinitely* countable). In other words, a finite set is a special case of a countable set.
2. **Union of two countable sets is countable.**
A standard theorem says: if \(X\) and \(Y\) are both countable sets (finite or infinite), then \(X \cup Y\) is countable.
- Intuition: you can list out elements of \(X\) (in some finite or infinite list) and then list out elements of \(Y\). If both can be enumerated by natural numbers, their union can also be enumerated.
3. **At least one of them is infinite**
Since \(B\) is countably infinite, \(A \cup B\) must contain infinitely many elements (because it contains all of \(B\)). So \(A \cup B\) is not just countable but also **infinite**.
Putting these observations together:
- \(A\) is countable (since it is finite).
- \(B\) is countably infinite (given).
- Therefore \(A \cup B\) is the union of two countable sets, hence **countable**.
- Because \(B\subseteq A\cup B\) and \(B\) is infinite, \(A\cup B\) must also be infinite.
Thus \(A \cup B\) is **countably infinite**.
