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To prove that the union of two countable sets \( X \) and \( Y \) is countable, we will proceed as follows:

### Step-by-Step Proof

#### Definitions:
1. **Countable Set**: A set \( A \) is countable if there exists a bijection between \( A \) and the set of natural numbers \( \mathbb{N} \), meaning \( A \) can either be finite or countably infinite.
2. **Union of Two Sets**: The union \( X \cup Y \) consists of all elements in \( X \), \( Y \), or both.

#### Goal:
We need to prove that if \( X \) and \( Y \) are countable sets, then their union \( X \cup Y \) is also countable.

### Strategy:

We construct functions from \( X \) and \( Y \) to the set \( \mathbb{N} \) and use these to define a function on the union \( X \cup Y \).

#### Step 1: Countable Functions for \( X \) and \( Y \)
Since \( X \) and \( Y \) are countable, there exist bijections:
- \( f: X \to \mathbb{N} \)
- \( g: Y \to \mathbb{N} \)

Thus, each element in \( X \) and \( Y \) can be uniquely assigned a natural number.

#### Step 2: Define a Function on the Union \( X \cup Y \)
To construct a bijection for \( X \cup Y \), we must map the elements of \( X \cup Y \) to \( \mathbb{N} \).

There are two cases to consider:
- \( X \cap Y \neq \emptyset \), meaning \( X \) and \( Y \) may have some common elements.
- \( X \cap Y = \emptyset \), meaning \( X \) and \( Y \) are disjoint.

To handle both cases, we can proceed as follows:
- Consider the function \( h: X \cup Y \to \mathbb{N} \times \{0, 1\} \) defined by:
  \[
  h(z) =
  \begin{cases}
  (f(z), 0) & \text{if } z \in X \setminus Y \\
  (g(z), 1) & \text{if } z \in Y \setminus X \\
  (f(z), 0) & \text{if } z \in X \cap Y
  \end{cases}
  \]
This function creates a distinct pair for elements of \( X \) and \( Y \) by pairing the outputs of \( f \) and \( g \) with an indicator value (0 or 1), ensuring that even if an element is shared by both sets, it is assigned consistently.

#### Step 3: Map \( \mathbb{N} \times \{0, 1\} \) to \( \mathbb{N} \)
Next, we need to show that \( \mathbb{N} \times \{0, 1\} \) is countable. This can be done by interleaving the pairs to produce a single natural number using a known bijection between \( \mathbb{N} \times \mathbb{N} \) and \( \mathbb{N} \). A simple function that works is:
\[
\varphi(n, 0) = 2n, \quad \varphi(n, 1) = 2n + 1
\]
This function maps all pairs in \( \mathbb{N} \times \{0, 1\} \) to distinct natural numbers, preserving the countability of the union.

#### Step 4: Conclude that \( X \cup Y \) is Countable
The composition of the function \( h \) with the bijection \( \varphi \) gives a bijection from \( X \cup Y \) to \( \mathbb{N} \), showing that \( X \cup Y \) is countable.

### Conclusion:
We have constructed a bijection from the union \( X \cup Y \) to \( \mathbb{N} \), proving that the union of two countable sets is countable.

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Would you like more details or have any questions on this?

Here are 5 follow-up questions to deepen your understanding:
1. What would change in the proof if \( X \) or \( Y \) were finite instead of infinite?
2. How does the concept of countability relate to uncountable sets like \( \mathbb{R} \)?
3. Can you extend this proof to a finite or infinite union of countable sets?
4. How does the interleaving method for \( \mathbb{N} \times \{0, 1\} \) relate to Cantor’s pairing function?
5. How would this proof be adjusted if \( X \) and \( Y \) were not disjoint?

**Tip**: The union of any finite number of countable sets is always countable, but the infinite case requires a more careful treatment involving the axiom of choice!

Prove that for all countable sets X and Y, X ∪ Y is countable.

A detailed proof that for all countable sets X and Y, their union X ∪ Y is countable. The solution explores key set theory concepts, such as bijections and countability, and demonstrates how to construct a bijection from X ∪ Y to the natural numbers.

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