1. Characteristic Function

Given $\displaystyle A \in \mathcal{P}(X)$ define the characteristic function $\displaystyle \chi_{A}: X \to \{0,1\}$ by $\displaystyle \chi_{A}(x) = \begin{cases} 0&\text{if}\ x \not \in A\\ 1&\text{if}\ x \in A \end{cases}$.

(i) Prove that the function $\displaystyle x \mapsto \chi_{A}(x) \chi_{B}(x)$ is the characteristic function of the intersection $\displaystyle A \cap B$.

So $\displaystyle x \in A \cap B \Rightarrow x \in A \ \text{and} \ x \in B$. By definition, the characteristic function always has values of $\displaystyle 0$ or $\displaystyle 1$. Therefore the characteristic function is $\displaystyle \chi_{A} \chi_{B}$.

(ii) Find the subset $\displaystyle C$ whose characteristic function is given by $\displaystyle \chi_{C}(x) = \chi_{A}(x) + \chi_{B}(x) - \chi_{A}(x) \chi_{B}(x)$. Isn't this just $\displaystyle A \Delta B$?

Thanks

2. (i) Prove that the function $\displaystyle x \mapsto \chi_{A}(x) \chi_{B}(x)$ is the characteristic function of the intersection $\displaystyle A \cap B$.

So suppose that $\displaystyle x \in A \cap B$ then $\displaystyle x \in A$ and $\displaystyle x \in B$; thus by definition,
$\displaystyle \chi_{A}(x)=1$, $\displaystyle \chi_{B}(x)=1$, & $\displaystyle \chi _{A \cap B} = 1$ therefore $\displaystyle \left( {\forall x \in A \cap B} \right)\left[ {\chi _{A \cap B} (x) = \chi _A (x)\chi _B (x)} \right]$

Now if $\displaystyle x \notin A \cap B\quad \Rightarrow \quad x \notin A \vee x \notin B$ which means $\displaystyle \chi_{A}(x)=0$ or $\displaystyle \chi_{B}(x)=0$, & $\displaystyle \chi _{A \cap B} = 0$ therefore $\displaystyle \left( {\forall x \notin A \cap B} \right)\left[ {\chi _{A \cap B} (x) = \chi _A (x)\chi _B (x)} \right]$.

You have this one correct!:
(ii) Find the subset $\displaystyle C$ whose characteristic function is given by $\displaystyle \chi_{C}(x) = \chi_{A}(x) + \chi_{B}(x) - {\color{red}2}\chi_{A}(x) \chi_{B}(x)$. Isn't this just $\displaystyle A \Delta B$?

3. why did you add the 2?

4. Originally Posted by shilz222
why did you add the 2?
Because that is the characteristic function for the symmetric difference operator.
I thought that is what you wanted.
But upon reading the question again, I realize that the answer is $\displaystyle C = A \cup B$ that is $\displaystyle {\chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _A (x)\chi _B (x)}$

5. $\displaystyle \chi_{A}(x)$ corresponds to set $\displaystyle A$

$\displaystyle \chi_{B}(x)$ corresponds to set $\displaystyle B$

$\displaystyle \chi_{A}(x) \chi_{B}(x)$ corresponds to set $\displaystyle A \cap B$

So $\displaystyle (A + B) - (A \cap B) = (A \cup B) - (A \cap B) = A \Delta B$?

How does this correspond to $\displaystyle A \cup B$?

Thanks

6. Suppose that $\displaystyle x \in A \cup B$ then exactly one of these is true:
$\displaystyle \left( {x \in A - B} \right) \vee \left( {x \in B - A} \right) \vee \left( {x \in A \cap B} \right)$.
Consider them one at a time.
1) $\displaystyle \left( {x \in A - B} \right)\quad \Rightarrow \quad \chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _{A \cap B} (x) = 1 + 0 - 0 = 1$.

2) $\displaystyle \left( {x \in B - A} \right)\quad \Rightarrow \quad \chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _{A \cap B} (x) = 0 + 1 - 0 = 1$.

3) $\displaystyle \left( {x \in A \cap B} \right)\quad \Rightarrow \quad \chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _{A \cap B} (x) = 1 + 1 - 1 = 1$.

Moreover, this is clear $\displaystyle \left( {x \notin A \cup B} \right)\quad \Rightarrow \quad \chi _{A \cup B} (x) = 0$.

This is a very well known problem. What you posted above is incorrect.
This is correct: $\displaystyle \chi _{A\Delta B} (x) = \chi _A (x) + \chi _B (x) - 2\chi _{A \cap B} (x).$
Again this is well known.

7. Thanks.

I learned something new today. I see the characteristic function has many applications in other mathematical fields (i.e. probability theory).

8. Originally Posted by shilz222
I learned something new today. I see the characteristic function has many applications in other mathematical fields (i.e. probability theory).
The characteristic function in probability theory is completely unrelated: it is $\displaystyle \phi_X(t) = E(e^{itX})$ and is just another name for the Fourier transform.

Probability theorists often rename standard mathematical objects this way. E.g., convergence with probability 1 is another name for convergence almost everywhere.

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