4. Sketch the sets \( X=\left\{(x, y) \in \mathbb{R}^{2} \mid x^{2}+y^{2} \leq 1\right\} \) and \( Y=\left\{(x, y) \in \mathbb{R}^{2} \left\lvert\, x-\frac{1}{2} \leq y \leq x+\frac{1}{2}\right.\right\} \) on \( \mathbb{R}^{2} \). On separate drawings, shade the sets \( X \cup Y, X \cap Y, X-Y \), and \( Y-X \). 5. Let \( A=\{1,3,5\} \) and \( B=\{0,2,4,6\} \) have univeral set \( U=\{0,1,2,3, \ldots, 6\} \). Find the following: (a) \( \bar{A} \) (b) \( A-\bar{B} \) (c) \( \overline{\bar{A} \cap B} \)

Let's solve the problems step by step. ### Problem 4: Sketching Sets \( X \) and \( Y \) 1. **Set \( X \)**: - The set \( X \) is defined by the inequality \( x^2 + y^2 \leq 1 \). This represents a filled circle (disk) centered at the origin (0,0) with a radius of 1. 2. **Set \( Y \)**: - The set \( Y \) is defined by the inequalities \( x - \frac{1}{2} \leq y \leq x + \frac{1}{2} \). This represents a region between two lines: - The line \( y = x - \frac{1}{2} \) (a line with slope 1, shifted down by 0.5). - The line \( y = x + \frac{1}{2} \) (a line with slope 1, shifted up by 0.5). - The region between these two lines extends infinitely in the x-direction. 3. **Shading the Sets**: - **\( X \cup Y \)**: Shade the area of the disk and the area between the two lines. - **\( X \cap Y \)**: Shade the area where the disk overlaps with the region between the two lines. - **\( X - Y \)**: Shade the area of the disk that does not overlap with the region between the two lines. - **\( Y - X \)**: Shade the area between the two lines that does not overlap with the disk. ### Problem 5: Set Operations Given: - \( A = \{1, 3, 5\} \) - \( B = \{0, 2, 4, 6\} \) - Universal set \( U = \{0, 1, 2, 3, 4, 5, 6\} \) Let's find the required sets: #### (a) \( \bar{A} \) - The complement of \( A \) in the universal set \( U \) is: \[ \bar{A} = U - A = \{0, 2, 4, 6\} \] #### (b) \( A - \bar{B} \) - First, we need to find \( \bar{B} \): \[ \bar{B} = U - B = \{1, 3, 5\} \] - Now, calculate \( A - \bar{B} \): \[ A - \bar{B} = A \cap \bar{B} = \{1, 3, 5\} \cap \{1, 3, 5\} = \{1, 3, 5\} \] #### (c) \( \overline{\bar{A} \cap B} \) - First, find \( \bar{A} \cap B \): \[ \bar{A} \cap B = \{0, 2, 4, 6\} \cap \{0, 2, 4, 6\} = \{0, 2, 4, 6\} \] - Now, find the complement of this intersection: \[ \overline{\bar{A} \cap B} = U - \{0, 2, 4, 6\} = \{1, 3, 5\} \] ### Summary of Results - \( \bar{A} = \{0, 2, 4, 6\} \) - \( A - \bar{B} = \{1, 3, 5\} \) - \( \overline{\bar{A} \cap B} = \{1, 3, 5\} \)