neter of a subset A mks) ii) A bounded subset A.of a metric space (X,d) (3 mks) b) Prove that the function, f(x)=x^2-3x is continuous at x x=Iunder the usual metric din't f'(x)-f'(y) 1-3 (+mks Q2 i) Show that the distance function d(x,y)=vert x_(1)-y_(1)vert +vert x_(2)-y_(2)vert for all x.y in R^2(x^2-3x) defines a metric in 9?^2 (4 mks) on X. ii) Let X=[0,1.2] with the standard metric don y_(i) Prove that f(x)=(1)/(4)x^2+(3)/(4) a contraction (4 mks) iii) Verify Banach's fixed theorem using the map in ii) (2 mks) (2) i) Explain what is meant by a compact subset A of metric space, (X, d) (4 mks) ii) Show that the set A=(0,2) is not compact in H_(1) (4 mks) iii) Determine whether the set of integers 2 is of first category or not (2 mks) f'(x)-f(y)

It seems like there are multiple parts to this question. Let's break it down and address each part one by one.<br /><br />### Part 1: Bounded Subset<br /><br />**Statement:**<br />A subset \( A \) of a metric space \( (X, d) \) is bounded if there exists a number \( M > 0 \) such that for every point \( x \in A \), the distance between \( x \) and some point \( y \in A \) is less than \( M \).<br /><br />**Proof:**<br />To prove that \( A \) is bounded, we need to show that there exists an \( M > 0 \) such that for every pair of points \( x, y \in A \), the distance \( d(x, y) < M \).<br /><br />### Part 2: Continuity of the Function<br /><br />**Statement:**<br />Prove that the function \( f(x) = x^2 - 3x \) is continuous at \( x = 1 \) under the usual metric.<br /><br />**Proof:**<br />To prove continuity at \( x = 1 \), we need to show that:<br />1. \( f(1) \) is defined.<br />2. The limit \( \lim_{x \to 1} f(x) \) exists.<br />3. \( \lim_{x \to 1} f(x) = f(1) \).<br /><br />First, calculate \( f(1) \):<br />\[ f(1) = 1^2 - 3 \cdot 1 = 1 - 3 = -2 \]<br /><br />Next, compute the limit as \( x \) approaches 1:<br />\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} (x^2 - 3x) = 1^2 - 3 \cdot 1 = -2 \]<br /><br />Since \( f(1) = -2 \) and \( \lim_{x \to 1} f(x) = -2 \), we have:<br />\[ \lim_{x \to 1} f(x) = f(1) \]<br /><br />Therefore, \( f(x) \) is continuous at \( x = 1 \).<br /><br />### Part 3: Distance Function Defining a Metric<br /><br />**Statement:**<br />Show that the distance function \( d(x, y) = |x_1 - y_1| + |x_2 - y_2| \) for all \( x, y \in \mathbb{R}^2 \) defines a metric in \( \mathbb{R}^2 \).<br /><br />**Proof:**<br />To show that \( d \) is a metric, we need to verify the following properties for all \( x, y, z \in \mathbb{R}^2 \):<br /><br />1. **Non-negativity:** \( d(x, y) \geq 0 \) and \( d(x, y) = 0 \) if and only if \( x = y \).<br />2. **Symmetry:** \( d(x, y) = d(y, x) \).<br />3. **Triangle inequality:** \( d(x, z) \leq d(x, y) + d(y, z) \).<br /><br />1. **Non-negativity:**<br /> - \( |x_1 - y_1| \geq 0 \) and \( |x_2 - y_2| \geq 0 \), so \( d(x, y) = |x_1 - y_1| + |x_2 - y_2| \geq 0 \).<br /> - If \( x = y \), then \( |x_1 - y_1| = 0 \) and \( |x_2 - y_2| = 0 \), so \( d(x, y) = 0 \).<br /><br />2. **Symmetry:**<br /> - \( d(x, y) = |x_1 - y_1| + |x_2 - y_2| = |y_1 - x_1| + |y_2 - x_2| = d(y, x) \).<br /><br />3. **Triangle inequality:**<br /> - For any \( x, y, z \in \mathbb{R}^2 \),<br /> \[<br /> d(x, z) = |x_1 - z_1| + |x_2 - z_2| \leq |x_1 - y_1| + |y_1 - z_1| + |x_2 - y_2| + |y_2 - z_2| = d(x, y) + d(y, z)<br /> \]<br /> - Hence, \( d(x, z) \leq d(x, y) + d(y, z) \).<br /><br />Since \( d \) satisfies non-negativity, symmetry,