Is Φ a Bijective Map from Power Set to Characteristic Functions?

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In summary: You probably need to say a bit more. Start by considering an element of $\{0,1\}^X$, namely a function $f:X\to\{0,1\}$. Can you find an element of $\mathcal P(X)$, namely a subset $A$ of $X$, such that $\Phi(A)=f$?
  • #1
mathmari
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Hey! :eek:

Let $X$ be a set. We consider the map \begin{equation*}\Phi : \ \mathcal{P}(X)\rightarrow \{0,1\}^X, \ \ A\mapsto 1_A\end{equation*} that maps a subset $A\subset X$to its characteristc function $1_A$.

I want to show in the following two ways that $\Phi$ is bijective:
  1. verify directly that $\Phi$ is injective and surjective
  2. give explicitly an inverse map

Could you give me a hint how we can show that? I don't really have an idea. (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

Let $X$ be a set. We consider the map \begin{equation*}\Phi : \ \mathcal{P}(X)\rightarrow \{0,1\}^X, \ \ A\mapsto 1_A\end{equation*} that maps a subset $A\subset X$to its characteristc function $1_A$.

I want to show in the following two ways that $\Phi$ is bijective:
  1. verify directly that $\Phi$ is injective and surjective
  2. give explicitly an inverse map

Could you give me a hint how we can show that? I don't really have an idea. (Wondering)

Hey mathmari!

Let's look at an example.
Suppose $X=\{1,2,3\}$, can we find the answers then? (Wondering)
 
  • #3
I like Serena said:
Let's look at an example.
Suppose $X=\{1,2,3\}$, can we find the answers then? (Wondering)

Using the first way:

$\Phi$ is surjective because for every element of in the range, i.e. $0$ and $1$ there is a preimage in $\mathcal{P}(X)$ because either one element is contained in the set $A$ or not.

$\Phi$ is injective because every element of $\Phi (X)$ has an image in $\{0,1\}$.

So, $\Phi$ is bijective.

Is everything correct so far?
Using the second way:

Can we determine the inverse? (Wondering)
 
  • #4
mathmari said:
$\Phi$ is surjective because for every element of in the range, i.e. $0$ and $1$ there is a preimage in $\mathcal{P}(X)$ because either one element is contained in the set $A$ or not.
You probably need to say a bit more. Start by considering an element of $\{0,1\}^X$, namely a function $f:X\to\{0,1\}$. Can you find an element of $\mathcal P(X)$, namely a subset $A$ of $X$, such that $\Phi(A)=f$?

mathmari said:
$\Phi$ is injective because every element of $\Phi (X)$ has an image in $\{0,1\}$.
No, not that way. Consider two elements of $\mathcal P(X)$, namely subsets $A$ and $B$ of $X$. Suppose $\Phi(A)=\Phi(B)$, i.e. their characteristic functions are equal, $1_A=1_B$. Can you show from this that the subsets must be equal, $A=B$? To do so, let $x\in A$ and show that $x\in B$, which will prove that $A\subseteq B$. The reverse inclusion will follow by symmetry.

mathmari said:
Can we determine the inverse?
Here’s a hint: the method used in proving surjectivity can come in handy here. It will help you find a function $\Psi:\{0,1\}^X\to\mathcal P(X)$ such that $\Psi\circ\Phi=\Phi\circ\Psi=I_{\mathcal P(X)}$ (the identity map on the power set of $X$). (Happy)
 
  • #5
Olinguito said:
You probably need to say a bit more. Start by considering an element of $\{0,1\}^X$, namely a function $f:X\to\{0,1\}$. Can you find an element of $\mathcal P(X)$, namely a subset $A$ of $X$, such that $\Phi(A)=f$?

We set $A$ the subset of $X$ such that $f(x)=1$ for $x\in X$, i.e. $A=\{ x\in X\mid f(x)=1$, or not? (Wondering)
Olinguito said:
No, not that way. Consider two elements of $\mathcal P(X)$, namely subsets $A$ and $B$ of $X$. Suppose $\Phi(A)=\Phi(B)$, i.e. their characteristic functions are equal, $1_A=1_B$. Can you show from this that the subsets must be equal, $A=B$? To do so, let $x\in A$ and show that $x\in B$, which will prove that $A\subseteq B$. The reverse inclusion will follow by symmetry.

I got it! (Smile)
Olinguito said:
Here’s a hint: the method used in proving surjectivity can come in handy here. It will help you find a function $\Psi:\{0,1\}^X\to\mathcal P(X)$ such that $\Psi\circ\Phi=\Phi\circ\Psi=I_{\mathcal P(X)}$ (the identity map on the power set of $X$). (Happy)

We have that $\Phi (A)=f$ then the inverse is of the form $\Psi (f)=A$, right? But how is this defined? (Wondering)
 
  • #6
mathmari said:
We set $A$ the subset of $X$ such that $f(x)=1$ for $x\in X$, i.e. $A=\{ x\in X\mid f(x)=1$, or not? (Wondering)
$A$ should be the subset of $X$ such that for all $x\in X$, $x\in A$ if and only if $f(x)=1$. Can you see what happens next?

mathmari said:
I got it! (Smile)
Great! (Happy)
mathmari said:
We have that $\Phi (A)=f$ then the inverse is of the form $\Psi (f)=A$, right? But how is this defined? (Wondering)
Given a function $f:X\to\{0,1\}$, let $A\subseteq X$ be as defined above in the surjectivity proof, and set $\Psi(f)=A$. Now work out $\Psi\circ\Phi$ and $\Phi\circ\Psi$.
 
  • #7
Olinguito said:
$A$ should be the subset of $X$ such that for all $x\in X$, $x\in A$ if and only if $f(x)=1$. Can you see what happens next?

With this $A$ we get that $1_A=f$, right? (Wondering)


Olinguito said:
Given a function $f:X\to\{0,1\}$, let $A\subseteq X$ be as defined above in the surjectivity proof, and set $\Psi(f)=A$. Now work out $\Psi\circ\Phi$ and $\Phi\circ\Psi$.

We have that $\Phi (A)=1_A$. Using the definition of $A$ as aboce we have that $\Phi (A)=f$ and then $\Psi (f)=A$ is the inverse which maps elements from $\{0,1\}^X$ to $\mathcal{P}(X)$.

Is everything correct? (Wondering)
 
  • #8
That’s basically it. (Nod)

This exercise proves the following. While every characteristic function is a function from $X$ to $\{0,1\}$ (by definition), the converse is also true: every function from $X$ to $\{0,1\}$ is a characteristic fuction.
 

FAQ: Is Φ a Bijective Map from Power Set to Characteristic Functions?

What does it mean for a function to be bijective?

A function is bijective if it is both injective (one-to-one) and surjective (onto). This means that every element in the range of the function is mapped to by exactly one element in the domain, and every element in the range has at least one element in the domain that maps to it.

How can you show that Φ is bijective?

To show that Φ is bijective, you must prove that it is both injective and surjective. This can be done by showing that for every element in the range, there is a unique element in the domain that maps to it, and that every element in the range is mapped to by at least one element in the domain.

What is the process for proving that Φ is bijective?

To prove that Φ is bijective, you must first show that it is injective. This can be done by assuming that there are two elements in the domain that map to the same element in the range, and then showing that this leads to a contradiction. Next, you must show that Φ is surjective by demonstrating that for every element in the range, there is at least one element in the domain that maps to it.

Why is it important to show that Φ is bijective?

Showing that Φ is bijective is important because it guarantees that the function has a well-defined inverse. This means that the inverse function will also be injective and surjective, and thus bijective. Bijective functions are also useful in many mathematical and scientific applications, such as in cryptography and data compression.

Can a function be bijective if it is not both injective and surjective?

No, a function cannot be bijective if it is not both injective and surjective. If a function is not injective, there are elements in the range that are mapped to by more than one element in the domain, which violates the definition of a bijective function. Similarly, if a function is not surjective, there are elements in the range that do not have any elements in the domain that map to them, also violating the definition of bijectivity.

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