AuBuC

ashleyum · #1 $Old$ September 7th, 2008, 10:26 AM

Prove that m(AuBuC) + m(AnBnC) >= m(AnB)+m(AnC)+m(BnC)
Please!

Soroban · #2 $Old$ September 7th, 2008, 02:16 PM

Hello, ashleyum!

Is there a typo? . . . The statement is not true!

Quote:

Prove that: . $m(A \cup B\cup C) + m(A\cap B\cap C)\;\geq \;m(A\cap B)+m(A\cap C)+m(B\cap C)$

Consider the sets $A,B,C$ in this Venn diagrm.

Code:

          *-----------*
        A |  1        |
          |   *-------+---*
          |   |     2 |   |
      *---+---+---*   |   |
      |   | 3 | 4 |   |   |
      |   *---+---+---*   |
      |       | 5 |    1  | C
      |       *---+-------*
    B |  1        |
      *-----------*

$\begin{array}{cccc}\text{We have:} & \underbrace{m(A \cup B\cup C)}_{17} + \underbrace{m(A \cap B\cap C)}_{4} & = & 21 \\ \\[-3mm] \text{and:} & \underbrace{m(A \cap B)}_{7} + \underbrace{m(A \cap C)}_{6} + \underbrace{m(B \cap C)}_{9} &=& 22 \end{array}\quad\cdots\quad See?$