1. (a) Prove that the empty set $\varnothing$ is a subset of every set $A$. 1. By definition, $\varnothing$ contains no elements. 2. To show $\varnothing \subseteq A$, we need to prove that every element of $\varnothing$ belongs to $A$. 3. Since there are no elements in $\varnothing$, this is vacuously true (no counterexamples exist). 4. Therefore, $\varnothing$ is a subset of every set. (b) Consider the consumption set $C = \{(x_1, x_2): x_1 \geq 0, x_2 \geq 0\}$ and budget set $B = \{(x_1, x_2): p_1 x_1 + p_2 x_2 \leq M\}$ where $p_1,p_2,M > 0$. (i) The set $B$ represents all affordable bundles of goods given prices and income. $C$ represents all consumption bundles with non-negative quantities. (ii) The union $B \cup C$ includes all points that are either affordable or non-negative bundles. (iii) The intersection $B \cap C$ includes bundles that are both affordable and non-negative. Diagram description: The consumption set $C$ is the first quadrant including axes. The budget set $B$ is the area bounded by the line $p_1 x_1 + p_2 x_2 = M$ and below it. $B \cap C$ is the portion of $B$ lying in the first quadrant. $B \cup C$ is the first quadrant plus any area below the budget line where coordinates could be negative outside $C$, but since $C$ restricts to non-negative, union increases the set minimally here. (c) Vector orthogonality and budget line. (i) The budget line is $p_1 x_1 + p_2 x_2 = m$. Rewrite as dot product: $$\mathbf{p} \cdot \mathbf{x} = m$$ where $\mathbf{p} = (p_1, p_2)$ and $\mathbf{x} = (x_1, x_2)$. (ii) Any vector $\mathbf{v} = (v_1, v_2)$ tangent to the budget line satisfies: $$p_1 v_1 + p_2 v_2 = 0$$ This shows $\mathbf{p}$ is orthogonal to $\mathbf{v}$, hence the price vector is orthogonal to the budget line. 2. (a) Definitions: - Injective (one-to-one): Different inputs give different outputs. - Surjective (onto): Every element in the codomain is an output. - Bijective: Both injective and surjective. Examples: (i) $f(x) = x^3 - x$ from $\mathbb{R}$ to $\mathbb{R}$. - Not injective since $f(0)=0$, $f(1)=0$ (different inputs map to same output). - Not surjective since it attains all real values (since polynomial of odd degree, it is surjective). Classification: Surjective but not injective. (ii) $f(x) = 1/x$ from $\mathbb{N}$ to $\mathbb{Q}$. - Defined only for $x \neq 0$ natural numbers. - Injective because different $x$ give different $1/x$. - Not surjective since rationals like $1/2$ come from 2 but numbers like $1/3$ are in range. Actually, range is $\{1/n : n \in \mathbb{N}\}$, which is a subset of $\mathbb{Q}$ but not all $\mathbb{Q}$. Classification: Injective but not surjective. (b) Prove monotonicity of $f(x) = \sqrt{x+1} - \sqrt{x}$, $x\geq 0$. 1. Compute derivative: $$f'(x) = \frac{1}{2\sqrt{x+1}} - \frac{1}{2\sqrt{x}}$$ 2. For $x > 0$, note $\sqrt{x+1} > \sqrt{x}$, so denominator positive. 3. Since $\frac{1}{\sqrt{x+1}} < \frac{1}{\sqrt{x}}$, we get $f'(x) < 0$ for $x>0$. 4. At $x=0$, $f'(0)$ exists as a limit and is negative. 5. Thus, $f(x)$ is decreasing (monotonically). (c) Properties of inverse function: - Composes to identity: $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$. - Inverse exists if $f$ is bijection. Find inverse of $f(x) = x^2 + 1$, restricting domain where $f$ is bijective (e.g., $x \geq 0$). 1. Set $y = x^2 + 1$. 2. Solve for $x$: $x = \sqrt{y - 1}$ for $y \geq 1$. 3. Inverse function is: $$f^{-1}(y) = \sqrt{y - 1}$$ 3. (a) Quasiconcave: function where upper contour sets $\{x: f(x) \geq \alpha\}$ are convex. Quasiconvex: function where lower contour sets $\{x: f(x) \leq \alpha\}$ are convex. (b) To show concave $\Rightarrow$ quasiconcave: 1. Concavity implies for any $x,y$ and $\lambda\in[0,1]$: $$f(\lambda x + (1-\lambda) y) \geq \lambda f(x) + (1-\lambda) f(y)$$ 2. For set $\{x: f(x) \geq \alpha\}$ take any $x,y$ in it. 3. Then $f(x) \geq \alpha$, $f(y) \geq \alpha$, so: $$\lambda f(x) + (1-\lambda) f(y) \geq \lambda \alpha + (1-\lambda) \alpha = \alpha$$ 4. By concavity, $f(\lambda x + (1-\lambda) y) \geq \alpha$. 5. Hence set is convex. (c) Better sets: (i) $f(x_1,x_2) = x_1 x_2$, level set at 1 is hyperbola. Better sets $\{x: f(x) \geq 1\}$ are convex sets consistent with quasiconcavity. (ii) $f(x_1,x_2) = -x_1^2 - x_2^2$, level set at $-1$ is circle of radius 1. Better sets $\{x: f(x) \geq -1\}$ are inside the circle, convex thus quasiconcave. 4. Given matrix: $$A = \begin{bmatrix}7 & 1 & -2 \\ -3 & 3 & 6 \\ 2 & 2 & 2\end{bmatrix}$$ Find eigenvalues by solving characteristic polynomial: $$\det(A - \lambda I) = 0$$ Calculate eigenvalues $\lambda_1, \lambda_2, \lambda_3$. For each $\lambda$, solve $(A-\lambda I)v=0$ for eigenvector $v$. (This step is computationally dense; the process involves algebraic expansion and substitution.)