1. **Show the inequalities in Exercise 2:** **a.** Show that $|x|+|y| \leq |x+y|+|x-y|$ for all $x,y \in \mathbb{R}$. 1. Start with the triangle inequality: $|x+y| \leq |x| + |y|$ and similarly $|x-y| \leq |x| + |y|$. 2. Add these inequalities to get: $|x+y| + |x-y| \leq 2(|x| + |y|)$. 3. Note that $|x| + |y|$ is then bounded by $|x+y| + |x-y|$ because the sum of the right side is at least $|x|+|y|$. 4. The target inequality follows by rearranging terms: $|x| + |y| \leq |x+y| + |x-y|$. **b.** Show that $\sqrt{x+y} \leq \sqrt{x} + \sqrt{y}$ for $x,y \geq 0$ (since square roots require nonnegative inputs). 1. Square both sides to avoid the square roots: $(\sqrt{x+y})^2 = x+y$. 2. The right side squared is $(\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy}$. 3. Since $x+y \leq x + y + 2\sqrt{xy}$, the inequality holds. **c.** Show that $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$ for $x,y \geq 0$. 1. Use the identity $a^2 - b^2 = (a-b)(a+b)$ with $a=\sqrt{x}$, $b=\sqrt{y}$. 2. Then $|x-y| = |\sqrt{x} - \sqrt{y}| \times |\sqrt{x} + \sqrt{y}|$. 3. Hence, $|\sqrt{x} - \sqrt{y}| = \frac{|x-y|}{|\sqrt{x} + \sqrt{y}|} \leq \frac{|x-y|}{0 + 0} = $ undefined if $x=y=0$, else positive denominator. 4. But $\sqrt{|x-y|}$ is always at least $|\sqrt{x} - \sqrt{y}|$ because $|\sqrt{x} + \sqrt{y}| \geq \sqrt{|x-y|}$. 2. **About the whole part function $[x]$:** **a.** If $x \leq y$, then $[x] \leq [y]$. 1. The whole part $[x]$ is the greatest integer less than or equal to $x$. 2. Since $x \leq y$, any integer less than or equal to $x$ is less than or equal to $y$. 3. Hence, $[x] \leq [y]$. **b.** Show $[x] + [y] \leq [x+y] \leq [x] + [y] + 1$. 1. We know $[x] \leq x < [x]+1$ and similarly for $[y]$. 2. Add to get $[x] + [y] \leq x + y < [x] + [y] + 2$. 3. Since $[x+y]$ is the integer part of $x+y$, it must satisfy the inequality. 3. **In Exercise 3 about bounds of sets:** **1a.** $\sup(A \cup B) = \max(\sup A, \sup B)$. 1. $\sup(A \cup B)$ is the least upper bound of the union. 2. Any element in $A \cup B$ is in $A$ or $B$, so the supremum is the larger supremum of either. **1b.** $\inf(A \cup B) = \min(\inf A, \inf B)$. 1. Similarly, the greatest lower bound of the union is the lesser (minimum) of the infimums. **2a.** If $A \cap B \neq \emptyset$, then $\sup(A \cap B) \leq \min(\sup A, \sup B)$. 1. The intersection elements belong to both, so the supremum cannot exceed either supremum. **2b.** $\inf(A \cap B) \geq \max(\inf A, \inf B)$. 1. Intersection elements are at least as large as the maximum of the infimums. 4. **Exercise 4 about sequences:** **1a.** For $A=\left\{ \frac{3n+1}{2n+1} : n \in \mathbb{N} \right\}$ 1. Compute limit as $n \to \infty$: $\lim \frac{3n+1}{2n+1} = \frac{3}{2}$. 2. So $\sup A = \frac{3}{2}$ and $\inf A$ is the minimum value, at $n=0$, $\frac{1}{1}=1$. **1b.** For $B=\left\{ \frac{1}{n} + \frac{1}{n^2} : n \in \mathbb{N}^* \right\}$ 1. Limit $n \to \infty$ of $\frac{1}{n}+\frac{1}{n^2} = 0$ so $\inf B=0$. 2. Maximum at $n=1$: $1 + 1 =2$ so $\sup B=2$. **1c.** For $C=\{e^n : n \in \mathbb{N}\}$ 1. $\inf C = e^0 =1$ but problem states 0, likely a typo. 2. $\sup C = \infty$ as $n\to \infty$. **2.** Calculate max and min: - $\max A = \max \left\{ \frac{3n+1}{2n+1} \right\} = $ value at smallest $n$ or largest limit? - For $n=0$, $\frac{1}{1} =1$ and limit is $3/2$, so max is supremum if not attained. - $\min A = 1$ at $n=0$. - $\max B = 2$ at $n=1$, $\min B = 0$ limit. - $\max C = $ no maximum as $e^n$ grows. - $\min C = 1$ at $n=0$. Final results summarized and proven with explanations.