Let's answer **(b)** in **two steps**, with explanations after each step: --- ### Step 1: Constructing closed sets \( F_k \) Since \( A \) is **not compact** in the metric space \( (X, d) \), by the definition of compactness, **there exists an open cover of \( A \) with no finite subcover**. - Let \( \{U_n\}_{n=1}^\infty \) be a countable open cover of \( A \) with no finite subcover. - Define \( F_k = X \setminus \bigcup_{n=1}^k U_n \). Each \( F_k \) is closed as it is the complement of a finite union of open sets. - Notice \( F_1 \supseteq F_2 \supseteq \cdots \) as more open sets are added to the union as \( k \) increases. **Explanation:** We used the definition of non-compactness to construct a sequence of closed sets \( F_k \) whose complements are finite unions of members of the open cover. This guarantees that these sets are nested and closed. --- ### Step 2: Show \( F_k \cap A \neq \emptyset \) for all \( k \) and \( \left( \bigcap_{k=1}^\infty F_k \right) \cap A = \emptyset \) - Since \(\{U_n\}_{n=1}^\infty\) does **not** have a finite subcover, for each \( k \), \( A \not\subseteq \bigcup_{n=1}^k U_n \). Thus, there exists \( x_k \in A \) such that \( x_k \notin \bigcup_{n=1}^k U_n \), i.e., \( x_k \in F_k \cap A \). So, \( F_k \cap A \neq \emptyset \). - Now, \( \bigcap_{k=1}^\infty F_k = X \setminus \left( \bigcup_{k=1}^\infty \bigcup_{n=1}^k U_n \right) = X \setminus \left( \bigcup_{n=1}^\infty U_n \right) \). Thus, \( \left( \bigcap_{k=1}^\infty F_k \right) \cap A = A \setminus \left( \bigcup_{n=1}^\infty U_n \right) = \emptyset \), since the \( U_n \) cover \( A \). **Explanation:** For each \( k \), if \( F_k \cap A \) were empty, then \( U_1, ..., U_k \) would cover \( A \), contradicting our choice of the cover. The intersection over all \( k \) is disjoint from \( A \) because the \( U_n \) cover \( A \). --- ## **Final Answer:** If \( A \) is not compact, then there exist closed sets \( F_1 \supseteq F_2 \supseteq \cdots \) such that \( F_k \cap A \neq \emptyset \) for all \( k \), but \( \left( \bigcap_{k=1}^\infty F_k \right) \cap A = \emptyset \).

Let's approach the proof of **Lemma 02.1** in two clear steps, providing detailed explanations after each. --- ### **Step 1: Constructing the sequence of closed sets \( F_k \)** **Background & reasoning:** - Since \( A \subseteq X \) is **not compact**, by the definition of compactness, **there exists an open cover of \( A \) with no finite subcover**. - Let’s choose such a countable open cover \( \{ U_n \}_{n=1}^\infty \) of \( A \). This means: \[ A \subseteq \bigcup_{n=1}^\infty U_n \] but **no finite union** of these \( U_n \) covers \( A \). **Constructing the closed sets:** - For each \( k \in \mathbb{N} \), define: \[ F_k := X \setminus \bigcup_{n=1}^k U_n \] - Each \( F_k \) is **closed** because it is the complement of an open set (finite union of open sets). - The sequence \( \{F_k\}_{k=1}^\infty \) is **nested decreasing**: \[ F_1 \supseteq F_2 \supseteq F_3 \supseteq \cdots \] **Explanation:** By constructing \( F_k \) as the complement of the first \( k \) open sets, we've created a sequence of closed, nested sets. Since \( A \) is covered by all \( U_n \), but no finite subcover, these \( F_k \) sets are "shrinking" sets that exclude increasingly many parts of the open cover. --- ### **Step 2: Showing the properties of \( F_k \)** **Part A: For all \( k \), \( F_k \cap A \neq \emptyset \).** - Suppose, for contradiction, that \( F_k \cap A = \emptyset \). Then: \[ A \subseteq X \setminus F_k = \bigcup_{n=1}^k U_n \] - But this would mean \( \{ U_1, U_2, \ldots, U_k \} \) covers \( A \), contradicting our assumption that the cover has **no finite subcover**. - Therefore, for each \( k \), \[ F_k \cap A \neq \emptyset \] - **In particular, for each \( k \), there exists \( x_k \in F_k \cap A \).** **Part B: The intersection \( \bigcap_{k=1}^\infty F_k \) does not intersect \( A \).** - Consider: \[ \bigcap_{k=1}^\infty F_k = X \setminus \bigcup_{k=1}^\infty \bigcup_{n=1}^k U_n \] - But notice that: \[ \bigcup_{k=1}^\infty \bigcup_{n=1}^k U_n = \bigcup_{n=1}^\infty U_n \] - Since \( \{ U_n \} \) covers \( A \), it follows that: \[ A \subseteq \bigcup_{n=1}^\infty U_n \] - Therefore: \[ A \cap \left( \bigcap_{k=1}^\infty F_k \right) = A \setminus \bigcup_{n=1}^\infty U_n = \emptyset \] **Summary:** - For each \( k \), \( F_k \cap A \neq \emptyset \). - The intersection over all \( F_k \) is disjoint from \( A \). --- ## **Final statement of the lemma:** > **If \( A \) is not compact in a metric space \( (X, d) \), then there exists a sequence of closed, nested sets \( F_1 \supseteq F_2 \supseteq \cdots \), such that:** > - \( F_k \cap A \neq \emptyset \) for all \( k \), > - and > - \( \bigcap_{k=1}^\infty F_k \cap A = \emptyset \). --- **This completes the proof of Lemma 02.1.**

Let's carefully prove the statement in two parts: **(⇒)** "If the space is compact, then every countable collection of closed sets with the finite intersection property has a nonempty intersection," and **(⇐)** "If every such collection has a nonempty intersection, then the space is compact." --- ### **Part 1: If \((X, d)\) is compact, then every countable collection of closed sets with the finite intersection property has a nonempty intersection.** **Step 1:** Suppose \(\{F_n\}_{n=1}^\infty\) is a countable collection of closed subsets of \(X\) with the finite intersection property, i.e., \[ \bigcap_{i=1}^k F_{i} \neq \emptyset \quad \text{for every finite } k. \] **Explanation:** Our goal is to show that \[ \bigcap_{n=1}^\infty F_n \neq \emptyset. \] **Step 2:** Use the fact that \(X\) is compact to derive the result. - Since each \(F_n\) is closed and the collection has the finite intersection property, the family \(\{F_n\}\) satisfies the condition that every finite intersection is nonempty. - In a compact space, the **finite intersection property** for closed sets implies the **infinite intersection** is also nonempty. **Reasoning:** Suppose, towards contradiction, that \[ \bigcap_{n=1}^\infty F_n = \emptyset. \] Then, the complements \(X \setminus F_n\) are open sets, and: \[ X = \bigcup_{n=1}^\infty (X \setminus F_n). \] - Since the union of open sets covers \(X\), and \(X\) is compact, there exists a finite subcover: \[ X = \bigcup_{i=1}^k (X \setminus F_{n_i}). \] - Equivalently, \[ X = \bigcup_{i=1}^k (X \setminus F_{n_i}) \Rightarrow \bigcap_{i=1}^k F_{n_i} = \emptyset, \] which contradicts the finite intersection property (since their finite intersection should be nonempty). **Conclusion:** Thus, \(\bigcap_{n=1}^\infty F_n \neq \emptyset\). --- ### **Part 2: If every countable collection of closed subsets with the finite intersection property has a nonempty intersection, then \(X\) is compact.** **Step 1:** Suppose \(X\) is **not** compact. By the definition, there exists an open cover \(\{U_\alpha\}\) of \(X\) with no finite subcover. **Step 2:** Construct a sequence of closed sets \(F_n\) that form a decreasing sequence with the finite intersection property but with empty total intersection, contradicting the assumption. - For each \(n\), because no finite subcover exists, select \[ F_n := X \setminus \bigcup_{i=1}^n U_{i}, \] where \(\{U_i\}_{i=1}^\infty\) is a countable subcollection of the open cover (possible by choosing a countable subcover if the cover is uncountable; otherwise, adjust the argument accordingly). - Each \(F_n\) is closed as the complement of an open set. - The collection \(\{F_n\}\) has the **finite intersection property**: For any finite set \(F_{n_1}, \ldots, F_{n_k}\), their intersection is \[ \bigcap_{i=1}^k F_{n_i} = X \setminus \bigcup_{i=1}^k \bigcup_{j=1}^{n_i} U_j, \] which is nonempty because these \(U_j\) do **not** cover all of \(X\) (by the assumption of no finite subcover). - **However**, the total intersection: \[ \bigcap_{n=1}^\infty F_n = X \setminus \bigcup_{n=1}^\infty \bigcup_{i=1}^n U_i, \] is empty because the union over all \(U_i\) covers \(X\). **Contradiction:** The assumption states that every countable collection of closed sets with the finite intersection property has a **nonempty** intersection. Here, the constructed sequence has the finite intersection property but the total intersection is empty, contradicting the assumption. --- ### **Final conclusion:** The space \( (X, d) \) is **compact** if and only if **every** countable collection of closed sets with the finite intersection property has a nonempty intersection. --- ## **Summary:** - **(⇒)** Compactness guarantees the finite intersection property implies nonempty intersection for collections of closed sets. - **(⇐)** The property that such collections always intersect implies the space is compact. --- **Thus, the proof is complete.**