Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 4

Problem

Is the set of all irrational real numbers countable?

Answer

No.

Suppose that the set of all irrational real numbers is countable. Then the set of real number is countable since it is a union of two countable sets, which is contradiction.

Comments

Post a Comment

Popular posts from this blog

Elementary Classical Analysis, Marsden, 2nd ed, Chapter 2, Problem 1

Problem Discuss whether the following sets are open or closed: (a) (1,2) in $\mathbb{R}$ (b) [2,3] in $\mathbb{R}$ (c) $\bigcap_{n=1}^\infty [-1,1/n)$ in $\mathbb{R}$ (d) $\mathbb{R}^n$ in $\mathbb{R}^n$ (e) A hyperplane in $\mathbb{R}^n$ (f) $\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ in $\mathbb{R}$ (g) $\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ in $\mathbb{R}^2$ (h) $\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$ in $\mathbb{R}^n$. Answer (a) Open: For $x \in (1,2)$, we can choose $\epsilon = \min(x-1,2-x)$ so that $D(x,\epsilon) \subset (1,2)$. (b) Closed: For $x \in [2,3]^c$, we can choose $\epsilon = \min(2-x,x-3)$ so that $D(x,\epsilon) \subset [2,3]^c$. (c) Closed: We first prove that $[-1,0] = A:=\bigcap_{n=1}^\infty [-1,1/n)$. Since $[-1,0] \subset [-1,1/n)$ for all $n \in \mathbb{N}$, $[-1,0] \subset A$. To prove $A \subset [-1,0]$, we assume that $x \in A$. Then obviously, $x \geq -1$. If $x > 0$, then there exists $n \in \mathbb{N}$ such that $x > 1/n$, which means
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 15

Problem Suppose $a \in \mathbb{R}^1$, $f$ is a twice-differentiable real function on $(a,\infty)$, and $M_0,M_1,M_2$ are the least upper bounds of $|f(x)|,|f'(x)|,|f''(x)|$, respectively, on $(a,\infty)$. Prove that $$ M_1^2 \leq 4M_0M_2.$$ Does $M_1^2 \leq 4M_0M_2$ hold for vector-valued functions too? Answer For $x \in (a,\infty)$ and $h>0$, $f(x+2h)=f(x)+2f'(x)h + \frac{1}{2}f''(\xi)(2h)^2$ for some $\xi$ between $x$ and $x+2h$ by Theorem 5.15. Then $|f'(x)| \leq \frac{|f(x+2h) - f(x)|}{2h} + |f''(\xi)h| \leq \frac{M_0}{h} + M_2h$. This holds for all $x$ and $h$ and $$ \inf_h(\frac{M_0}{h} + M_2h) = 2\sqrt{M_0M_2}$$ Hence, we have $M_1 \leq 2\sqrt{M_0M_2}$  $\Rightarrow$  $M_1^2 \leq 4M_0M_2.$ This inequality also holds for vector-valued functions $f=(f_1,\cdots,f_k)$. We may assume that $M_1 \neq 0$. For a fixed point $y \in (a,\infty)$, define a function $$ g(x):=f_1'(y)f_1(x) + \cdots + f_k'(y)f_k(x). $$ Then we have $$|g'(x)|^2
Read more