Partition and function

Let $E$ be a non empty set and $A, B$ be two non empty subsets of $E$. Consider the function $f:P(E) \rightarrow P(A) \times P(B)$ that is given by $f(X)=\left(X\cap A,X\cap B\right)$. Show that $\{A, B\}$ is a partition of $E$ if-f then function $f$ is $1-1$ and onto $P(A) \times P(B)$.

Solution



$ \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

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