几何学 - 作业 2

T1

Consider the function on $[0,1]$ defined by the formula:

$$f(x)=\{\begin{array}{ll}x\sin \frac{1}{x},& \text{if }0<x⩽1\\ 0,& \text{if }x=0\end{array}$$

(a) Show that $f$ is right continuous at $0$ . (b) Show that the path

$$\gamma :[0,1]\to {\mathbb{R}}^2,t\mapsto (t,f(t))$$

is not rectifiable.

Proof

(a) Set $t=\frac{1}{x}$ :

$$\underset{x\to 0^{+}}{\lim }x\sin \frac{1}{x}=\underset{t\to +\infty }{\lim }\frac{\sin t}{t}=0=f(0)$$

Then $f$ is right continuous at $0$ .

(b) Consider

$$A_n=\left\{x_k|x_k=\frac{1}{k\pi +\frac{\pi }{2}},k=1,2,\cdots ,n;x_0=0\right\}$$

And $A_n$ is a kind of partition of $\gamma$ . Then

$$\begin{array}{rl}l(\gamma )& ⩾\sum _{i=0}^{n}d(\gamma (x_i),\gamma (x_{i+1}))\\ & ⩾\sum _{i=0}^n\frac{1}{k\pi +\frac{\pi }{2}}\to \infty (n\to \infty )\end{array}$$

because harmonic series is divergent, which means $l(\gamma )$ is not rectifiable. $\square$

T2

Prove that the Koch’s snowflake is not rectifiable.

Proof

Assume the original length is $a_0$ and the number of edges is $b_0=1$ . The total length will be $L_0=a_0{b}_0$ .

According to the process of the generation of Koch’s snowflake:

• $a_n=\frac{1}{3^n}{a}_0$ .
• $b_n=4^n{b}_0=4^n$ .
• Then the total length will be $L_n={\left(\frac{4}{3}\right)}^n{L}_0\to \infty (n\to \infty )$ .

In conclusion, Koch’s snowflake will not be rectifiable. $\square$

T3

Consider any $f\in \mathrm{Isom}{\mathbb{R}}^2$ . (a) Show that $f$ sends circles to circles with same radius.

(b) Show that $f$ sends straight lines to straight lines.

Proof

(a) For a circle whose radius is $r$ and the center is $a\in {\mathbb{R}}^2$ :

$$C=\left\{x\in {\mathbb{R}}^2\mid d(x,a)=r\right\}$$

For any $x\in C$ , consider

$$d(f(x),f(a))=r$$

Then

$$f(C)=\left\{y\in {\mathbb{R}}^2\mid d(y,f(a))=r\right\}$$

is a circle with radius $r$ .

(b) For a straight line that crosses $x$ and $y$ :

$${\mathbb{L}}_{x,y}=\left\{tx+(1-t)y\mid t\in \mathbb{R}\right\}$$

for $z\in {\mathbb{L}}_{x,y}$ , $f(z)\in f({\mathbb{L}}_{x,y})$ and

$$d(f(z),f(x))=d(z,x),d(f(z),f(y))=d(z,y),d(f(y),f(x))=d(x,y)$$

Then as $x,y,z\in {\mathbb{L}}_{x,y}$ ,

$$d(f(z),f(x))+d(f(z),f(y))=d(f(x),f(y))$$

The triangle equality holds iff $f(x),f(y),f(z)$ are on the same line. $\square$

T4

Prove that for any $\underline{x},\underline{y}\in {\mathbb{R}}^2$ , there is a unique $\overrightarrow{v}\in \overrightarrow{{\mathbb{R}}^2}$ , such that $\underline{y}=\underline{x}+\overrightarrow{v}$ .

Proof

Consider the mapping:

$$T:{\mathbb{R}}^2\to \overrightarrow{{\mathbb{R}}^2},\underline{x}\mapsto \overrightarrow{x}$$

Define

$$\overrightarrow{v}=T\underline{y}-T\underline{x}$$

It is unique since $\overrightarrow{{\mathbb{R}}^2}$ is a vector space and by defining translation:

$$T_{\overrightarrow{v}}:{\mathbb{R}}^2\to {\mathbb{R}}^2,x\mapsto x+\overrightarrow{v}$$

Then we get $\underline{y}=\underline{x}+\overrightarrow{v}$ . $\square$