Let's solve the problem step by step: Given: \( A = \left\{ (x, y) : \frac{1}{4} < x^2 + y^2 \leq 1 \text{ and } y > 0 \right\} \) We are to find the **interior** (\(\operatorname{Int}(A)\)), **exterior** (\(\operatorname{Ext}(A)\)), and **boundary** (\(\partial A\)) of \(A\). --- ### **Step 1: Find the Interior of \(A\), \(\operatorname{Int}(A)\)** - **Explanation:** The interior consists of all points where you can draw a small open ball around the point that is completely inside \(A\). - The set \(A\) is the region in the upper half-plane (\(y>0\)), between the circles \(x^2 + y^2 = \frac{1}{4}\) (not included) and \(x^2 + y^2 = 1\) (included). - **Interior:** For the interior, we must **exclude** the boundaries: - \(x^2 + y^2 = \frac{1}{4}\) is already excluded (strict inequality). - \(x^2 + y^2 = 1\) is included (but points exactly on this circle may not have a neighborhood fully inside \(A\), because any neighborhood will include points with \(x^2 + y^2 > 1\), which are not in \(A\)). - \(y=0\) is not included (since \(y>0\)). So, the interior is: \[ \operatorname{Int}(A) = \left\{ (x, y) : \frac{1}{4} < x^2 + y^2 < 1, \ y > 0 \right\} \] **End of Step 1 Explanation:** The interior excludes the "outer" circle (\(x^2 + y^2 = 1\)), the "inner" circle (\(x^2 + y^2 = \frac{1}{4}\)), and \(y=0\). --- ### **Step 2: Find the Exterior (\(\operatorname{Ext}(A)\)) and Boundary (\(\partial A\))** #### **Exterior (\(\operatorname{Ext}(A)\)):** - **Explanation:** The exterior consists of all points where you can draw a small open ball around the point that does **not** intersect \(A\) at all. - So, \(\operatorname{Ext}(A)\) is: - Points with \(x^2 + y^2 \leq \frac{1}{4}\) and \(y \geq 0\) - Points with \(x^2 + y^2 > 1\) and \(y \geq 0\) - All points with \(y < 0\) (since all of \(A\) is in \(y > 0\)) So, \[ \operatorname{Ext}(A) = \left\{ (x, y) : x^2 + y^2 \leq \frac{1}{4}, \ y \geq 0 \right\} \cup \left\{ (x, y) : x^2 + y^2 > 1, \ y \geq 0 \right\} \cup \left\{ (x, y) : y < 0 \right\} \] --- #### **Boundary (\(\partial A\)):** - **Explanation:** The boundary consists of points where every neighborhood contains points in \(A\) and points not in \(A\). - The boundary will be: - The arc of the circle \(x^2 + y^2 = \frac{1}{4}\) for \(y > 0\) (not in \(A\), but arbitrarily close points are in \(A\)) - The arc of the circle \(x^2 + y^2 = 1\) for \(y > 0\) (included in \(A\), but arbitrarily close points outside the circle are not in \(A\)) - The segment along \(y = 0\), for \(\frac{1}{4} < x^2 < 1\) (since points just above \(y=0\) are in \(A\), but \(y=0\) itself is not) So, \[ \partial A = \left\{ (x, y) : x^2 + y^2 = \frac{1}{4}, \ y > 0 \right\} \cup \left\{ (x, y) : x^2 + y^2 = 1, \ y > 0 \right\} \cup \left\{ (x, 0) : \frac{1}{4} < x^2 < 1 \right\} \] --- ## **Final Answers:** ### **(a) Interior** \[ \boxed{ \operatorname{Int}(A) = \left\{ (x, y) : \frac{1}{4} < x^2 + y^2 < 1, \ y > 0 \right\} } \] ### **(b) Exterior** \[ \boxed{ \operatorname{Ext}(A) = \left\{ (x, y) : x^2 + y^2 \leq \frac{1}{4}, \ y \geq 0 \right\} \cup \left\{ (x, y) : x^2 + y^2 > 1, \ y \geq 0 \right\} \cup \left\{ (x, y) : y < 0 \right\} } \] ### **(c) Boundary** \[ \boxed{ \partial A = \left\{ (x, y) : x^2 + y^2 = \frac{1}{4}, \ y > 0 \right\} \cup \left\{ (x, y) : x^2 + y^2 = 1, \ y > 0 \right\} \cup \left\{ (x, 0) : \frac{1}{4} < x^2 < 1 \right\} } \] Let me know if you'd like a diagram or further explanation!