###### tags: `108上助教` `數學`
# 數學導論HW8
## ch5-6
#### solution

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**a)**
Suppose $x \in f(A)-f(B)$
$i.e.\ x\in f(A)$ and $x \notin f(B)$
$\therefore \exists\ y \in A$ such that $f(y) = x$
and $\because\ x\notin f(B)\ ,\ \therefore\ y \notin B$
$\Rightarrow y\in A-B$
$\Rightarrow x\in f(A-B)$
$\Rightarrow f(A) - f(B) \subseteq f(A - B)\ \Box$

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**b)**
Suppose $x\in f^{-1}(A) - f^{-1}(B)$
$i.e.\ x\in f^{-1}(A)$ and $x \notin f^{-1}(B)$
$\Rightarrow f(x)\in A$ and $f(x)\notin B$
$\Rightarrow f(x) \in A - B$
$x \in f^{-1}(A - B)$
$\Rightarrow f^{-1}(A) - f^{-1}(B) \subseteq f^{-1}(A - B)\ \Box$

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**c)**
For (a), we give a counter example.
Consider $f:\mathbb{R} \rightarrow \mathbb{R}\ ,\ f(x) = x^2$
Let $A = [-1,1]\ ,\ B = [-1,0]$
then $A - B = (0,1]$
$\Rightarrow f(A-B) = (0,1]$
Let $y = \dfrac{1}{4} \in f(A - B)$
We can find $x_{1} \in A - B$
such that $f(x_{1}) = \dfrac{1}{4}$
$\Rightarrow x_{1} = \dfrac{1}{2}$
But let $x_{2} = \dfrac{-1}{2} \in B$
$f(x_{2}) = \dfrac{1}{4}$
which belongs to $f(A - B)$
$\Rightarrow y \in f(B)$
$\Rightarrow y \notin f(A) - f(B)$
$\Rightarrow f(A - B) \nsubseteq f(A) - f(B)$
Hence $f(A - B) \neq f(A)- f(B)\ \Box$

For (b), we give a proof.
We need to show that
$f^{-1}(A - B) \subseteq f^{-1}(A) - f^{-1}(B)$
(以下是證明)
Let $x \in f^{-1}(A - B)$
$\Rightarrow f(x) \in A - B\ ,\ i.e.\ f(x) \in A$ and $f(x) \notin B$
$\Rightarrow x \in f^{-1}(A) - f^{-1}(B)$
$\Rightarrow f^{-1}(A - B) \subseteq f^{-1}(A) - f^{-1}(B)\ \Box$

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**d)**
We have showed (b) above.
Now we want to show the equality holds in (a) if $f$ is one-to-one.
(以下是證明)
Suppose $f: X \rightarrow Y$ is one-to-one
We want to show that $f(A - B) \subseteq f(A) - f(B)$
Let $x \in f(A - B)$
$\Rightarrow \exists y \in A - B$ such that $f(y) = x$
$\because\ y \in A - B\ ,\ \therefore\ y \in A$
$\Rightarrow x \in f(A)$
Since $f$ is one-to-one, there exists **NO** $y_{2} \in B$ such that $f(y_{2}) = x$
$\Rightarrow x \notin f(B)$
$\Rightarrow x \in f(A) - f(B)$
$\Rightarrow f(A - B) \subseteq f(A) - f(B)$

Hence we prove that $f(A - B) = f(A) - f(B)\ \Box$