Let's solve **1-2** step by step as requested: --- ### 1-2. Let \( E \subset \mathbb{R} \) a non-empty set. Show that \( \mathcal{F} = \{ \emptyset, E, E^c, \mathbb{R} \} \) is the σ-algebra of subsets of \( \mathbb{R} \) generated by \( \{E\} \). --- #### **Step 1: Check if \( \mathcal{F} \) is a σ-algebra** - A σ-algebra must satisfy: 1. \( \emptyset \in \mathcal{F} \) and \( \mathbb{R} \in \mathcal{F} \). 2. If \( A \in \mathcal{F} \), then \( A^c \in \mathcal{F} \). 3. If \( \{A_n\} \subset \mathcal{F} \), then \( \bigcup_{n=1}^{\infty} A_n \in \mathcal{F} \). **Explanation:** - \( \emptyset \) and \( \mathbb{R} \) are in \( \mathcal{F} \) by definition. - \( E \in \mathcal{F} \implies E^c \in \mathcal{F} \) (closure under complements). - Any union of sets from \( \mathcal{F} \) is either \( \emptyset, E, E^c, \mathbb{R} \), so \( \mathcal{F} \) is closed under countable unions. --- #### **Step 2: Show that \( \mathcal{F} \) is the smallest σ-algebra containing \( E \)** - The σ-algebra generated by \( \{E\} \) is the intersection of all σ-algebras containing \( E \). - Any σ-algebra containing \( E \) must also contain \( E^c \), \( \emptyset \), and \( \mathbb{R} \) (since σ-algebras are closed under complements and must include \( \emptyset \) and \( X \)). - Thus, \( \mathcal{F} \) is the smallest σ-algebra containing \( E \). **Explanation:** Any σ-algebra containing \( E \) must contain all four sets in \( \mathcal{F} \), and no other subsets are forced by closure properties. --- **Final Answer:** The σ-algebra generated by \( \{E\} \) is \( \mathcal{F} = \{\emptyset, E, E^c, \mathbb{R}\} \). --- Would you like the step-by-step for 1-3 as well?

Let's proceed with **1-3** step by step: --- ### 1-3. Let \( X = \{1, 2, 3, 4\} \) and consider \( C = \{\{1\}, \{2, 3\}\} \subseteq \mathcal{P}(X) \). Determine \( \sigma(C) = o(C) \), the σ-algebra generated by \( C \). --- #### **Step 1: Find the minimal σ-algebra containing \( C \)** - The σ-algebra generated by \( C \) must include: - The sets in \( C \): \( \{1\} \) and \( \{2, 3\} \). - The complements of these sets. - All countable unions, intersections, and complements generated from these. **Explanation:** Since \( C \) contains specific subsets, the generated σ-algebra \( o(C) \) includes all sets that can be formed by countable operations starting from \( \{1\} \) and \( \{2, 3\} \). --- #### **Step 2: Compute the basic sets and their complements** - \( \{1\} \) and \( \{2, 3\} \): - Complement of \( \{1\} \): \( \{2, 3, 4\} \). - Complement of \( \{2, 3\} \): \( \{1, 4\} \). - Generate sets through unions and intersections: - \( \{1\} \cup \{2, 3\} = \{1, 2, 3\} \) - \( \{1\} \cap \{2, 3\} = \emptyset \) - \( \{2, 3\} \cup \{1\} = \{1, 2, 3\} \) - \( \{2, 3\} \cap \{1\} = \emptyset \) - Also, possible combinations: - \( \{1\} \cup \{4\} = \{1, 4\} \) - \( \{2, 3\} \cup \{4\} = \{2, 3, 4\} \) **Explanation:** Using the initial sets and their complements, we can generate all the relevant subsets. --- #### **Step 3: List all elements of \( o(C) \)** From the above, the σ-algebra \( o(C) \) includes: - \( \emptyset \) (empty union) - \( \{1\} \) - \( \{2, 3\} \) - \( \{4\} \) (since \( \{1, 4\} \) is complement of \( \{2, 3\} \)) - \( \{1, 2, 3\} \) - \( \{2, 3, 4\} \) - \( \{1, 4\} \) - \( X = \{1, 2, 3, 4\} \) **Explanation:** All sets generated via countable unions, intersections, and complements from the initial sets are listed. Since \( X \) is finite, the σ-algebra is finite and consists of these particular subsets. --- ### **Final answer:** \[ o(C) = \left\{\emptyset, \{1\}, \{2, 3\}, \{4\}, \{1, 2, 3\}, \{2, 3, 4\}, \{1, 4\}, \{1, 2, 3, 4\}\right\} \] --- Would you like further clarification or assistance with anything else?