Let's solve the question step by step as requested: --- ### **Step 1: Write the equation in the form \( f(x) = 0 \)** Given equation: \[ x^2 = p \] Here, \( p = 60 \). To write it in the form \( f(x) = 0 \): \[ x^2 - 60 = 0 \] So, \[ f(x) = x^2 - 60 \] **Explanation:** We rearranged the equation so that all terms are on one side, which is the required form for Newton-Raphson: \( f(x) = 0 \). --- ### **Step 2: Newton-Raphson Method to find \( \sqrt{60} \) correct to 7 decimals** The Newton-Raphson iteration formula is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] For \( f(x) = x^2 - 60 \): \[ f'(x) = 2x \] The iteration becomes: \[ x_{n+1} = x_n - \frac{x_n^2 - 60}{2x_n} = \frac{2x_n^2 - (x_n^2 - 60)}{2x_n} = \frac{x_n^2 + 60}{2x_n} = \frac{1}{2}\left(x_n + \frac{60}{x_n}\right) \] Choose a starting value \( x_0 \). Since \( \sqrt{60} \approx 7.75 \), let’s use \( x_0 = 8 \). #### **First Iteration:** \[ x_1 = \frac{1}{2}\left(8 + \frac{60}{8}\right) = \frac{1}{2}(8 + 7.5) = \frac{1}{2}(15.5) = 7.75 \] #### **Second Iteration:** \[ x_2 = \frac{1}{2}\left(7.75 + \frac{60}{7.75}\right) \approx \frac{1}{2}(7.75 + 7.741935484) = \frac{1}{2}(15.49193548) \approx 7.74596774 \] #### **Third Iteration:** \[ x_3 = \frac{1}{2}\left(7.74596774 + \frac{60}{7.74596774}\right) \] \[ \frac{60}{7.74596774} \approx 7.74596671 \] \[ x_3 = \frac{1}{2}(7.74596774 + 7.74596671) = \frac{1}{2}(15.49193445) \approx 7.74596722 \] #### **Fourth Iteration:** \[ x_4 = \frac{1}{2}\left(7.74596722 + \frac{60}{7.74596722}\right) \] \[ \frac{60}{7.74596722} \approx 7.74596722 \] \[ x_4 = \frac{1}{2}(7.74596722 + 7.74596722) = \frac{1}{2}(15.49193444) = 7.74596722 \] The value has stabilized to 7 decimal places: **7.7459672** **Explanation:** By applying the Newton-Raphson formula iteratively, the value converges rapidly to the true square root. After 3-4 iterations, the value is stable to 7 decimal places. --- ## **Final Answer** **The square root of 60, accurate to 7 decimals using the Newton-Raphson method, is:** \[ \boxed{7.7459672} \]

Let's analyze and answer this question step by step. --- ### **Step 1: Prove that every finite point set in a Hausdorff space is closed** **Explanation:** In a Hausdorff space, all singleton sets are closed. This is a fundamental property because, in a Hausdorff space, for any two distinct points, there exist disjoint open neighborhoods separating them. Specifically, for a singleton \( \{x\} \), we can show it is closed by expressing it as the intersection of complements of open neighborhoods around other points. **Detailed reasoning:** Let \( X \) be a Hausdorff space and \( \{x_1, x_2, \dots, x_n\} \) be a finite subset of \( X \). For each \( x_i \), since \( X \) is Hausdorff, the singleton \( \{x_i\} \) is closed, or equivalently, its complement is open. - For each \( x_i \), define: \[ U_i = X \setminus \{x_i\} \] which is open because singleton points are closed in Hausdorff spaces, so their complements are open. - The finite set can be written as: \[ S = \{x_1, x_2, \dots, x_n\} \] - The complement of \( S \) is: \[ X \setminus S = \bigcap_{i=1}^n U_i \] which is an intersection of finitely many open sets, hence open. - Therefore, the finite set \( S \) is the complement of an open set, making it **closed**: \[ S = X \setminus \left( \bigcap_{i=1}^n U_i \right) \] **Conclusion:** Since the complement of a finite set in a Hausdorff space is open, **every finite subset of a Hausdorff space is closed**. --- ### **Step 2: What can be said about the converse?** **Explanation:** The converse would be: *If every finite subset of a space \( X \) is closed, then \( X \) is Hausdorff.* This is **not necessarily true**. The property that all finite sets are closed is quite weak and does not imply the space is Hausdorff. For example, in **finite spaces** (spaces with finitely many points), all subsets are closed, and such spaces are trivially \( T_2 \) (Hausdorff), but in infinite spaces, the property that all finite subsets are closed is weaker than being Hausdorff. **Counterexample:** - Consider the **cofinite topology** on an infinite set: all cofinite sets are open, and finite sets are closed. But the space is **not** Hausdorff because you cannot separate two distinct points with disjoint open neighborhoods in general. - In particular, many infinite topological spaces have all finite sets closed but are not Hausdorff. **Summary:** - **If a space is Hausdorff**, then **every finite set is closed**. - **However, if every finite set is closed**, the space need **not** be Hausdorff. --- ## **Final answer:** **Every finite point set in a Hausdorff space is closed because singleton sets are closed and finite sets are finite unions of singleton closed sets. The converse is not true: having all finite sets closed does not imply the space is Hausdorff, as there exist non-Hausdorff spaces where all finite sets are closed (e.g., cofinite topology spaces).**