###### tags: `108上助教` `數學`
# Ch7習題
>作者：啊綸

{%hackmd PO2taSqBSV2DzTe-rjWuJQ %}

[TOC]

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## 7-1
### 3

**proof**
Let $A$ be the set of all perfect squares and $B$ be the set of all perfect cubes.

We know that $A$ is countably infinite.
$\Rightarrow A \sim \mathbb{N}$   ($A$ and $\mathbb{N}$ have the same cardinality)

==Goal:== We want to show that $B \sim \mathbb{N}$
(proof of Goal)
Consider $f: B \rightarrow \mathbb{N}$ with $f(n) = \sqrt[3]{n}$
then $f$ is 1-1 and onto
so $B \sim \mathbb{N}$
We prove our goal

$\because\ B \sim \mathbb{N}\ ,\ \therefore\ \mathbb{N} \sim B$
and $\because\ A \sim \mathbb{N}\ ,\ \mathbb{N} \sim B$
$\Rightarrow A \sim B\ \Box$

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### 4

**proof**
Consider $f:\mathbb{N} \rightarrow \mathbb{Z}$ by
\begin{equation} f(n) =
\begin{cases}
\dfrac{n}{2} & \mbox{ if $n$ is even}\\
-\dfrac{n-1}{2} & \mbox{ if $n$ is odd}
\end{cases}
\end{equation}
We want to show that $f$ is 1-1 and onto.

**(proof of 1-1)**
Let $n, m \in \mathbb{N}$ such that $f(n)=f(m)$
We have two cases:

==case 1:== $f(n),f(m)>0$
that is, $\dfrac{n}{2} = f(n) = f(m) = \dfrac{m}{2}$
$\Rightarrow n = m$

==case 2:== $f(n),f(m)\leq 0$
that is, $-\dfrac{n-1}{2} = f(n) = f(m) = -\dfrac{m-1}{2}$
$\Rightarrow n = m$

Hence we proved that $f$ is 1-1.

**(proof of onto)**
We want to show that $\forall a \in \mathbb{Z}\ , \exists\ n \in \mathbb{N}$ such that $f(n) = a$
Suppose $a\in \mathbb{Z}$

==case 1:== $a = 0$
It is easy to find that $f(1) = 0$

==case 2:== $a > 0$
Note that $a = \dfrac{2a}{2}$
Let $n = 2a$ , then $n$ is even
that is, we find $n\in \mathbb{N}$ such that $f(n) = \dfrac{n}{2} = a$

==case 3:== $a < 0$
Note that $a = -\dfrac{(-2a+1)-1}{2}$
Let $n = -2a + 1$
$\because a < 0$ , we have $n > 0$
and $n$ is odd
that is, we find $n\in\mathbb{N}$ such that $f(n) = -\dfrac{n-1}{2} = a$
By the case above, we proved that $f$ is onto

Since $f$ is 1-1 and onto, we proved that $\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality.$\Box$

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### 8

**proof**
Let $f: [a,b] \rightarrow [c,d]$
be a function with $f(x) = \dfrac{d-c}{b-a}\ x + \dfrac{bc-ad}{b-a}\ ,\ \forall x \in [a,b]$

then $f$ is 1-1 and onto (You can check by yourself)
$\Rightarrow [a,b] \sim [c,d]$


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### 10

**proof**
Consider $f: (-\dfrac{\pi}{2}\ ,\ \dfrac{\pi}{2})\rightarrow \mathbb{R}$
with $f(x) = \tan{x}\ ,\ \forall x \in (-\dfrac{\pi}{2}\ ,\ \dfrac{\pi}{2})$

then since $\tan{x}$ is 1-1 and onto,
$\Rightarrow (-\dfrac{\pi}{2}\ ,\ \dfrac{\pi}{2}) \sim \mathbb{R}\ \Box$

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### 11

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## 7-4
### 25

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### 29

**proof**
Let $A$ be a countable set , and let $B$ be a subset of $A$.

Since $A$ is countable
$\exists\ f:A \rightarrow \mathbb{N}$ , $f$ is 1-1 and onto

Define $i:B \rightarrow A$
as $i(x) = x$
(this is called inclusion map)

since $i$ is 1-1
so the function $h:B \rightarrow \mathbb{N}$ , $h = f \circ i$
is 1-1

Hence we prove that $B$ is countable.$\Box$

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### 30

**proof**
==法一==
By (29)
$\because\ A-\{x\} \subset A$ , and $A$ is countable
$\therefore\ A-\{x\}$ is countable

==法二==

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### 34