# Math Help - Indicator function properites

1. ## Indicator function properites

I just wanted to make sure my proofs are correct.
Given the indicator function.

$1_A = \left\{ \begin{array}{rcl}
1 & \mbox{for} & x \in A \\
0 & \mbox{for} & x \notin A
\end{array}\right.$

show that:
i) $1_{A \cap B} = 1_A \cdot 1_B$:

Proof:
$1_{A \cap B} = \left\{ \begin{array}{rcl}
1 & \mbox{for} & x \in A\cap B \\
0 & \mbox{for} & x \notin A\cap B
\end{array}\right.$
which implies that $1_{A \cap B} = 1_A \cdot \left\{ \begin{array}{rcl}
1 & \mbox{for} & x \in B \\
0 & \mbox{for} & x \notin B
\end{array}\right. = 1_A \cdot 1_B$

ii) $1_{A \cup B} = 1_A + 1_B - 1_A \cdot 1_B$

Proof:
$1_{A \cup B} = \left\{ \begin{array}{rcl}
1 & \mbox{for} & x \in A\cup B \\
0 & \mbox{for} & x \notin A\cup B
\end{array}\right.$
now $A\cup B$ can be written as two disjoint sets so we have

$A\cup B \Rightarrow A\cup(B-A\cap B)$ so we have :

$1_{A \cup B} = \left\{ \begin{array}{rcl}
1 & \mbox{for} & x \in A\cup(B-A\cap B) \\
0 & \mbox{for} & x \notin A\cup(B-A\cap B)
\end{array}\right. = 1_A +\left\{ \begin{array}{rcl}
1 & \mbox{for} & x \in B -A\cap B \\
0 & \mbox{for} & x \notin B -A\cap B
\end{array}\right. = 1_A+1_B-1_A\cdot 1_B$
where the terms are expanded out.

iii) $1_{A \cap B} =\min(1_A,1_B)$

I have no idea

2. $\min\{1_A,1_B\}=0$ if and only if $1_A=0\text{ or }1_B=0$.

Does that work?

P.S. Here is another ‘fun’ one for symmetric difference.

$1_{A\Delta B}=1_A+1_B-2\cdot 1_A\cdot 1_B$.