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Math Help - Indicator function properites

  1. #1
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    Indicator function properites

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

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

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

    Proof:
    1_{A \cap B} = \left\{ \begin{array}{rcl}<br />
1 & \mbox{for} & x \in A\cap B \\<br />
0 & \mbox{for} & x \notin A\cap B<br />
\end{array}\right. which implies that 1_{A \cap B} = 1_A \cdot \left\{ \begin{array}{rcl}<br />
1 & \mbox{for} & x \in B \\<br />
0 & \mbox{for} & x \notin B<br />
\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}<br />
1 & \mbox{for} & x \in A\cup B \\<br />
0 & \mbox{for} & x \notin A\cup B<br />
\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}<br />
1 & \mbox{for} & x \in A\cup(B-A\cap B) \\<br />
0 & \mbox{for} & x \notin A\cup(B-A\cap B)<br />
\end{array}\right. = 1_A +\left\{ \begin{array}{rcl}<br />
1 & \mbox{for} & x \in B -A\cap B \\<br />
0 & \mbox{for} & x \notin B -A\cap B<br />
\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
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  2. #2
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    \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.
    Last edited by Plato; October 8th 2010 at 02:19 PM.
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