Let's break the answer into **two steps** with explanations and conclude with a statement about the converse. --- ### **Step 1: Prove that every singleton set is closed in a Hausdorff space** **Proof:** Let \( X \) be a Hausdorff space and let \( x \in X \). We want to show that the singleton set \( \{x\} \) is closed, i.e., its complement \( X \setminus \{x\} \) is open. **Explanation:** By the Hausdorff property, for any \( y \in X \) with \( y \neq x \), there exist open sets \( U_y \) containing \( x \) and \( V_y \) containing \( y \) such that \( U_y \cap V_y = \emptyset \). Now, for each \( y \in X \setminus \{x\} \), choose \( V_y \) as above. Consider the union \( V = \bigcup_{y \neq x} V_y \). Since each \( y \) is in some \( V_y \), \( V \) covers all of \( X \setminus \{x\} \), and \( x \notin V_y \) for any \( y \). So, \( V = X \setminus \{x\} \) is open. **Conclusion:** Thus, \( \{x\} \) is closed in \( X \). --- ### **Step 2: Prove that every finite set is closed** **Proof:** Let \( A = \{x_1, x_2, \ldots, x_n\} \) be a finite subset of \( X \). **Explanation:** From Step 1, each singleton \( \{x_i\} \) is closed. The finite union of closed sets is closed in any topological space. Therefore, \( A = \{x_1\} \cup \{x_2\} \cup \cdots \cup \{x_n\} \) is closed. --- ### **Converse:** **Is the converse true?** No, the converse is **not** true. There exist spaces where every finite set is closed, but the space is **not** Hausdorff. For example, in the **finite complement topology** on an infinite set, every finite set is closed, but the space is not Hausdorff. --- ### **Final Answer:** **In any Hausdorff space, every finite point set is closed. However, the converse is not true: the fact that every finite set is closed does not imply the space is Hausdorff.**