Which function equals $f(x,A \cap B)$?

Question

Let $X$ be a nonempty set and $\mathcal{P}(X)$ be the set of all subsets of $X$. Define a function $f~:~X\times \mathcal{P}(X)\mapsto \{0,1\}$ such that for $x\in X$ and $A\subseteq X$,

$$f(x,A)=\begin{cases} 1& \text{if}~x\in A\\ 0&\text{if}~x\notin A\end{cases}$$

Then, for $x\in X$ and $A,B\subseteq X,~f(x,A\cap B)$ is not equal to which of the following:

• (A) $f(x,A)\cdot f(x,B)$
• (B) $f(x,A\cup B) - |f(x,A)-f(x,B)|$
• (C) $f(x,A^c)+f(x,B^c) - f(x,A^c\cup B^c)$
• (D) $f(x,A)+f(x,B)-f(x,A\cup B)$

Let $X =\{1,2\}$

Then, $\mathcal P(X) = \{ \emptyset, \{1\}, \{2\}, \{1,2\}\}$

Lets suppose we take, $A = \emptyset$ and $B = \{1,2\}$ Then $A' = \{1,2\}$ and $B' = \emptyset$

So, $f(x, A \cap B) = 0$. Since Intersection of $A$ and $B$ is the null set.

$f(x, A) = 0,~ f(x, B)=1,~ f(x, A \cup B)=1$

$f(x, A') =1,~ f(x, B') = 0,~ f(x, A' \cup B')= 1$

Proceeding as per the above values I find that all the options are equal to the function $f(x, A \cap B)$. But the answer says Option C is correct.

Where am I making the mistake?

Answer 1

Don't bother with coming up with a specific example yet.

There are effectively three cases we need to check for each expression: what happens when $x\notin A$ and $x\notin B$, what happens when $x\in A$ and $x\in B$, and what happens when $x\in A$ but $x\notin B$. (The remaining fourth case of $x\in B$ but $x\notin A$ would be effectively the same as the third and so does not need to be considered so long as the expression is symmetric with respect to $A$ and $B$)

If the expressions were equivalent to $f(x,A\cap B)$ then we should have expected to see the results, $0,1,0$ respectively.

For the first expression $f(x,A)\cdot f(x,B)$, in the first case that would result in $0\cdot 0 = 0$ as we had hoped. In the second case that would result in $1\cdot 1 = 1$ as we had hoped. Finally in the third case that would have resulted in $1\cdot 0 = 0$, again as we had hoped, proving that $f(x,A)\cdot f(x,B) = f(x,A\cap B)$.

For the second expression, $f(x,A\cup B) - |f(x,A)-f(x,B)|$, in the first case we have $0-|0-0|=0$ as we had hoped. In the second case that would be $1 - |1-1|=1$. In the third we have $1-|1-0|=0$, showing that this too is equivalent.

You can continue in this fashion for the third and fourth expressions.

Supposing $x\notin A$ and $x\notin B$, we would have expected $f(x,A\cap B) = 0$. However $f(x\in A^c)+f(x\in B^c)-f(x\in A^c\cup B^c) = 1+1-1=1\neq 0$

Now that we see this, we can come up with a specific example that works.

Let $x=1,A=B=\emptyset$ and $X=\{1\}$

Answer 2

I will prove here that $f \left( A \cap B \right)$ can be written as (A) and (D). However, I am not able to come up with how to write it as (B) or (C). Based on similar ideas, you may proceed.

Looking at the definition of $f$,

$$f \left( x, A \cap B \right) = \begin{cases} 1 & x \in A \cap B \\ 0 & x \notin A \cap B \end{cases}$$

From the definition of intersection, we can have

$$f \left( x, A \cap B \right) = 1 \Leftrightarrow x \in A \wedge x \in B \Leftrightarrow f \left( x, A \right) = 1 \wedge f \left( x, B \right) = 1$$

In other words,

$$f \left( x, A \cap B \right) = f \left( x, A \right) \cdot f \left( x, B \right)$$

when $x \in A \cap B$.

The case, where $x \notin A \cap B$ can be written as $x \notin A$ or $x \notin B$. Therefore, we get $f \left( x, A \right) = 0$ or $f \left( x, B \right) = 0$. Therefore,

$$f \left( x, A \cap B \right) = f \left( x, A \right) \cdot f \left( x, B \right)$$

for both the conditions and we have the representation (A).

As for (D), consider the sum $f \left( x, A \right) + f \left( x, B \right)$. Clearly, each term acknowledges the presence of $x$ in the corresponding set. Therefore, the acknowledgement of presence in the union must be this sum. However, if the element is present in both (i.e., intersection), then we have acknowledged the presence twice in the union, which we need to take care of. Therefore,

$$f \left( x, A \cup B \right) = f \left( x, A \right) + f \left( x, B \right) - f \left( x, A \cap B \right)$$

or, what is the same

$$f \left( x, A \cap B \right) = f \left( x, A \right) + f \left( x, B \right) - f \left( x, A \cup B \right)$$

thereby getting the representation (D).