0

Suppose $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ are sets with $\mathcal{A} \cap \mathcal{B} \cap \mathcal{C} = \emptyset$. Then $| \mathcal{A} \cup \mathcal{B} \cup \mathcal{C} |$ = $|\mathcal{A}|$ + $|\mathcal{B}|$ + $|\mathcal{C}|$ Prove or disprove. If false, modify statement to make it true.

  • You are right that the given statement is false. "If false, modify statement to make it true" is exceedingly vague; there are an infinite number of ways to do that. – bof Dec 16 '15 at 09:08
  • what would be one possible way? @bof –  Dec 16 '15 at 09:47
  • One way would be to change the hypothesis from $A\cap B\cap C=\emptyset$ to $A\cap B=A\cap C=B\cap C=\emptyset.$ Another would be to change the conclusion to $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|+|A\cap C|+|B\cap C|$. – bof Dec 16 '15 at 09:55

1 Answers1

1

For any three sets $$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|+|A\cap B\cap C|.$$ If $A\cap B\cap C=\phi$ then $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C| -|A\cap C|\neq |A|+|B|+|C|. $

For example $A=\{1,2\}, B=\{2,3\}$ and $C=\{3,1\}$ then $A\cap B\cap C=\phi$ and $ |A\cup B\cup C|=3\neq |A|+|B|+|C|=6$

K_user
  • 2,438