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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 \(\displaystyle{\mathbb{P}(A)\times \mathbb{P}(B)}\) .

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

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

Thank you for your solution.

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 \(\displaystyle{A\cup B=E}\) .