Function and partition
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Function and partition
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 \(\displaystyle{\mathbb{P}(A)\times \mathbb{P}(B)}\) .
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 \(\displaystyle{\mathbb{P}(A)\times \mathbb{P}(B)}\) .
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: Function and partition
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
$$f(Z)=(A, \varnothing) \Rightarrow A \cap Z=A \wedge B \cap Z=\varnothing \\
\Rightarrow A \subset Z \wedge Z \subset B^c \Rightarrow A \subset B^c \Rightarrow A \cap B=\varnothing$$
\( \left ( \Rightarrow \right ) \)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.
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Function and partition
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 \(\displaystyle{A\cup B=E}\) .
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 \(\displaystyle{A\cup B=E}\) .
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 6 guests