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I have three sets A, B and C satisfying the following conditions:

  • $ \#(A\cap B) = 11$
  • $ \#(A\cap C) = 12$
  • $ \#(A\cap B\cap C) = 5$ What is the minimun cardinality of A?

What I did was this:

(11 - 5) + (12 - 5) = 13.

But I'm not sure if I have to substract the 5 twice or only once. And the fact that I don't know if B and C are disjoint gets me confused =/

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Why did you calculate $(11-5)+(12-5)$? –  Chris Eagle Dec 30 '12 at 20:17
1  
Also, you do know if $B$ and $C$ are disjoint. –  Chris Eagle Dec 30 '12 at 20:18
    
You should really draw this... –  Karolis Juodelė Dec 30 '12 at 20:21
    
I did this: $ ( \#(A\cap B)-\#(A\cap B\cap C)) - ( \#(A\cap C)-\#(A\cap B\cap C))$ –  Victor Jose Arana Rodriguez Dec 30 '12 at 20:23
    
You must subtract the $5$ only once. –  P.. Dec 30 '12 at 20:27

2 Answers 2

up vote 5 down vote accepted

\begin{align} \#(A\cap (B\cup C)) &= \#((A\cap B)\cup (A\cap C))\\ &=\#(A\cap B)+\#(A\cap C)-\#(A\cap B\cap C)\\ &=11+12-5=18. \end{align} Note that $A$ cannot be fewer than $18$ since $A\cap(B\cup C)\subseteq A$, thus $\#(A)\geq18$.

Now, if we can find an example, where $\#(A)=18$, we'll have minimized the cardinality of $A$. For such an example, let $A=\{1,\ldots,18\}$, $B=\{1,\ldots,11\}$, and $C=\{7,\ldots,18\}$.

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Note that $$A = (A\cap B^c \cap C^c) \bigcup (A \cap B \cap C^c) \bigcup (A\cap B^c \cap C) \bigcup (A \cap B \cap C)$$ In the above, we have written $A$ as the union of four disjoint sets. Hence, we get that $$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B \cap C^c) + \#(A\cap B^c \cap C) + \#(A \cap B \cap C) \,\,\,\,\,\,\,\, (\star)$$ Now note that we can write $A \cap B$ and $A \cap C$ as a disjoint union as follows. $$(A \cap B) = (A \cap B \cap C^c) \bigcup (A \cap B \cap C)$$ and $$(A \cap C) = (A \cap B \cap C) \bigcup (A \cap B^c \cap C)$$ Hence, we get that $$\#(A \cap B) = \#(A \cap B \cap C^c) + \#(A \cap B \cap C)$$ and $$\#(A \cap C) = \#(A \cap B \cap C) + \#(A \cap B^c \cap C)$$ Rearranging, we get that $$\#(A \cap B \cap C^c) = \#(A \cap B) - \#(A \cap B \cap C)$$ $$\#(A \cap B^c \cap C) = \#(A \cap C) - \#(A \cap B \cap C)$$ Plugging the above two in $(\star)$, we get that $$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B) + \#(A \cap C) - \#(A \cap B \cap C) \,\,\,\,\,\,\, (\dagger)$$ Plugging in the given values in $(\dagger)$, we get $$\#A = \#(A\cap B^c \cap C^c) + 11 + 12 - 5 = \#(A\cap B^c \cap C^c) + 18$$ Note that $\#X \geq 0$ for any set $X$ and hence, the minimum value of $\#A$ is when $\#(A\cap B^c \cap C^c) = 0$, which gives us $$\#A = 18$$ Note that the minimum is attained when $A \subseteq B \cup C$.

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@Clayton $A = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$, $B = Z^{-} \cup \{0,1,2,3,4,5,6,7,8,9,10,11\}$ and $C = \{7,8,9,10,11,12,13,14,15,16,17,18,19,20,\ldots\}$ –  user17762 Dec 30 '12 at 21:00
    
@Clayton There is no $6$ in $C$. It is enough to start from $7$. Number of numbers between $a$ and $b$ including both is $b-a+1$. –  user17762 Dec 30 '12 at 21:12
    
@Clayton $A \cap B \cap C = \{7,8,9,10,11\}$. –  user17762 Dec 30 '12 at 21:14

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