Let's walk through the solution **step by step** and explain each part clearly, as if teaching it aloud. --- ### **Step 1: Lemma 04.1** #### **Statement** Let \(A\) be a set and \(S\) be a countable set. If there is a one-to-one function \(g: A \to S\) mapping \(A\) into \(S\) (not necessarily onto), then \(A\) is countable. #### **Proof** - \(f[A]\) is the image of \(A\) under \(f\), and is thus a subset of \(S\). - Since \(S\) is countable, every subset of \(S\) is at most countable. - The function \(f: A \to f[A]\) is a one-to-one onto map (bijection). - So, \(A\) is countable because it is in one-to-one correspondence with a countable set. ✅ **Conclusion:** If you can injectively map \(A\) to a countable set, \(A\) must itself be countable. --- ### **Step 2: Theorem 04.10** #### **Statement** The product set \(\mathbb{N} \times \mathbb{N}\) is countable. #### **Proof** - Define a function \(f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) by: \[ f(m, n) = 2^m \cdot 3^n \] - This function is **one-to-one** because every natural number has a unique factorization into primes (Fundamental Theorem of Arithmetic). - By **Lemma 04.1**, since we can injectively map \(\mathbb{N} \times \mathbb{N}\) into \(\mathbb{N}\), \(\mathbb{N} \times \mathbb{N}\) is countable. ✅ **Conclusion:** The cartesian product of countable sets is countable. --- ### **Step 3: Disjointification of Sets** Let \(X_1, X_2, \ldots\) be nonempty sets. - For each \(n \in \mathbb{N}\), define: \[ \widetilde{X}_n = \{(x, n) \mid x \in X_n\} \] - So, for \(k \neq m\): \[ \widetilde{X}_k \cap \widetilde{X}_m = \emptyset \] because the second component (the index) is different. - The sets \(\widetilde{X}_n\) are **mutually disjoint**. #### **Claim** \[ \text{card}\left(\bigcup_{n=1}^{\infty} X_n\right) \leq \text{card}\left(\bigcup_{n=1}^{\infty} \widetilde{X}_n\right) \] - For every \(z \in \bigcup_{n=1}^{\infty} X_n\), let \(i(z)\) be the smallest positive integer such that \(z \in X_{i(z)}\). - Define \(\beta: \bigcup_{n=1}^{\infty} X_n \to \bigcup_{n=1}^{\infty} \widetilde{X}_n\) by: \[ \beta(z) = (z, i(z)) \] - \(\beta\) is **one-to-one** (injective). - Therefore, the union of the \(X_n\) is at most as large as the union of the disjointified sets \(\widetilde{X}_n\). --- ### **Step 4: Mapping Unions of Disjoint Sets** Let \(S_1, S_2, \ldots\) be disjoint sets, and for each \(n\), let \(g_n: S_n \to T_n\) be a function. Define \(\overline{g}: \bigcup_{n=1}^{\infty} S_n \to \bigcup_{n=1}^{\infty} T_n\) as: - For \(x \in \bigcup_{n=1}^{\infty} S_n\), there is **exactly one** \(k\) such that \(x \in S_k\). - Set \(\overline{g}(x) = g_k(x)\). - \(\overline{g}\) is **well-defined** and, if each \(g_n\) is injective, so is \(\overline{g}\). --- ## **Summary (Human Voice Style)** 1. **If you can inject a set into a countable set, then your set is countable.** 2. **The set of pairs of natural numbers is countable.** 3. **You can "disjointify" a sequence of sets by tagging each element with its set index to avoid overlap.** 4. **If you have disjoint sets and functions on each, you can piece them together into a function on their union.** Would you like a spoken explanation (audio file) or more examples?

The key conclusions are: - If a set \(A\) can be injectively mapped into a countable set \(S\), then \(A\) is countable. - The Cartesian product \(\mathbb{N} \times \mathbb{N}\) is countable. - Any countable union of disjoint sets can be embedded into the union of their disjointified versions. - Functions defined on each disjoint set can be combined into a single function on the union, preserving injectivity if each original function is injective.