1. **Problem 12:** Identify \( \operatorname{cl}_{A \cup B}(B) \) where \( A = \{ z \in \mathbb{C} : |z + 1|^2 \leq 1 \} \) and \( B = \{ z \in \mathbb{C} : |z - 1|^2 < 1 \} \). 2. The closure of \( B \) in \( A \cup B \) is given by $$ \operatorname{cl}_{A \cup B}(B) = (A \cup B) \cap \operatorname{cl}_{\mathbb{C}}(B). $$ 3. The closure of \( B \) in \( \mathbb{C} \) is $$ \operatorname{cl}_{\mathbb{C}}(B) = \{ z \in \mathbb{C} : |z - 1|^2 \leq 1 \}. $$ 4. Therefore, $$ \operatorname{cl}_{A \cup B}(B) = (A \cup B) \cap \{ z : |z - 1|^2 \leq 1 \} = B \cup \{0\}. $$ 5. **Problem 13:** Show (i) \( X \setminus \bar{A} = (X \setminus A)^\circ \) and (ii) \( X \setminus A^\circ = \overline{X \setminus A} \). 6. For (i), \( x \in (X \setminus A)^\circ \) means there exists an open ball \( S(x,\varepsilon) \subseteq X \setminus A \), so \( S(x,\varepsilon) \cap A = \emptyset \). 7. This implies \( x \notin \bar{A} \) because \( \bar{A} \) contains all points where every ball intersects \( A \). 8. Hence, \( X \setminus \bar{A} = (X \setminus A)^\circ \). 9. For (ii), replace \( A \) by \( X \setminus A \) in (i) and take complements: $$ X \setminus (X \setminus A)^\circ = \overline{A} \implies X \setminus A^\circ = \overline{X \setminus A}. $$ 10. **Problem 14:** Example where \( (\bar{Y})^\circ \neq \overline{Y^\circ} \). 11. Let \( Y = \mathbb{Q} \subset \mathbb{R} \) with usual metric. 12. The closure \( \bar{Y} = \mathbb{R} \) since rationals are dense. 13. The interior \( (\bar{Y})^\circ = \mathbb{R}^\circ = \mathbb{R} \). 14. The interior of \( Y \) is empty, \( Y^\circ = \emptyset \), so \( \overline{Y^\circ} = \emptyset \). 15. Thus, \( (\bar{Y})^\circ = \mathbb{R} \neq \emptyset = \overline{Y^\circ} \). 16. **Problem 15:** Show properties of boundary \( \partial(Y) = \bar{Y} \cap \overline{X \setminus Y} \). 17. (i) \( \partial(Y) = \partial(X \setminus Y) \) because $$ \partial(Y) = \bar{Y} \cap \overline{X \setminus Y} = \overline{X \setminus Y} \cap \bar{Y} = \partial(X \setminus Y). $$ 18. (ii) \( \bar{Y} = Y^\circ \cup \partial(Y) \) since closure is interior plus boundary. 19. (iii) \( Y^\circ \cap \partial(Y) = \emptyset \) because interior points are not boundary points. 20. (iv) \( (X \setminus Y)^\circ \cap \partial(Y) = \emptyset \) by similar reasoning. 21. (v) \( X = Y^\circ \cup \partial(Y) \cup (X \setminus Y)^\circ \) partitions \( X \) into interior, boundary, and exterior. 22. (vi) \( Y \setminus \partial(Y) = Y^\circ \) since removing boundary leaves interior. **Final answers:** - Q12: \( \operatorname{cl}_{A \cup B}(B) = B \cup \{0\} \). - Q13: (i) \( X \setminus \bar{A} = (X \setminus A)^\circ \), (ii) \( X \setminus A^\circ = \overline{X \setminus A} \). - Q14: Example with rationals \( Y = \mathbb{Q} \subset \mathbb{R} \) shows \( (\bar{Y})^\circ \neq \overline{Y^\circ} \). - Q15: Boundary and interior properties (i) to (vi) as above.