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 Post subject: Function and partitionPosted: Fri Jan 01, 2016 3:28 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{E}$ be a non-empty set and $\displaystyle{A\,,B\in\mathbb{P}(E)-\left\{\varnothing\right\}}$ .

We define $\displaystyle{f:\mathbb{P}(E)\longrightarrow \mathbb{P}(A)\times \mathbb{P}(B)}$ by

$\displaystyle{X\mapsto f(X)=\left(X\cap A,X\cap B\right)}$ .

Prove that $\displaystyle{\left\{A,B\right\}}$ is a partition of the set $\displaystyle{E}$ if, and only if, the function $\displaystyle{f}$

is one to one and onto $\displaystyle{\mathbb{P}(A)\times \mathbb{P}(B)}$ .

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 Post subject: Re: Function and partitionPosted: Fri Jan 01, 2016 3:30 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Good evening y'all.

$\left ( \Rightarrow \right )$We have that $B=A^c$. Suppose $f(X) = f(Y)$, then:
$$X\cap A= Y\cap A \;\;\; {\rm and} \;\;\; X\cap A^c = Y \cap A^c$$

Thus :
$$X=(X \cap A) \cup (X \cap A^c)=(Y \cap A) \cup (Y \cap A^c)=Y$$

so $f$ is $1-1$.

Also
$$X \subset A, Y \subset B \Rightarrow X=A \cap (X \cup Y), Y=B \cap (X \cup Y) \Rightarrow (X,Y)=f(X \cup Y)$$

meaning that is onto also proving the first part.

$\left ( \Leftarrow \right )$ We have that $f(A \cup B) =f(E)$ and since $f$ is $1-1$ we also have $A \cup B=E$. The function $f$ is also onto meaning that there exists $Z \subset E$ such that

$$f(Z)=(A, \varnothing) \Rightarrow A \cap Z=A \wedge B \cap Z=\varnothing \\ \Rightarrow A \subset Z \wedge Z \subset B^c \Rightarrow A \subset B^c \Rightarrow A \cap B=\varnothing$$

_________________
Imagination is much more important than knowledge.

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 Post subject: Re: Function and partitionPosted: Fri Jan 01, 2016 3:34 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426

Here is another solution that $\displaystyle{f}$ is one to one according to the hypothesis

that $\displaystyle{\left\{A,B\right\}}$ is a partition of $\displaystyle{E}$, that is

$\displaystyle{E=A\cup B\,\,,A\cap B=\varnothing}$ . Let $\displaystyle{X\,,Y\in\mathbb{P}(E)}$ such

that $\displaystyle{f(X)=f(Y)}$. Then, $\displaystyle{X\cap A=Y\cap A\,\,\,,X\cap B=Y\cap B}$.

We'll prove that $\displaystyle{X=Y}$ . For this purpose, let $\displaystyle{x\in X}$. Then, $\displaystyle{x\in E}$

and $\displaystyle{x\in A}$ or $\displaystyle{x\in B}$ . If $\displaystyle{x\in A}$, then :

$\displaystyle{x\in X\cap A\implies x\in Y\cap A\implies x\in Y\,\land x\in A}$, so : $\displaystyle{x\in Y}$ . If $\displaystyle{x\in B}$

then $\displaystyle{x\in X\cap B\implies x\in Y\cap B\implies x\in Y}$. In any case, $\displaystyle{X\subseteq Y}$.

Similarly, $\displaystyle{Y\subseteq X}$ and finally, $\displaystyle{X=Y}$ .

Here is another solution that $\displaystyle{A\cup B=E}$according to the hypothesis that $\displaystyle{f}$ is

one to one an onto $\displaystyle{\mathbb{P}(A)\times \mathbb{P}(B)}$ .

Suppose that $\displaystyle{A\cup B\neq E}$. Then, there exists $\displaystyle{x\in E}$ such that

$\displaystyle{x\notin A\,\,,x\notin B}$. Therefore,

$\displaystyle{f(\left\{x\right\})=\left(\left\{x\right\}\cap A,\left\{x\right\}\cap B\right)=\left(\varnothing,\varnothing\right)=f\,(\varnothing)}$

and since the function $\displaystyle{f}$ is one to one, we get $\displaystyle{\left\{x\right\}=\varnothing}$, a contradiction,

so $\displaystyle{A\cup B=E}$ .

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