【数学A】例題1.1.4：集合の要素の個数の最大・最小（One More）★★★

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% 例題A1.1.4：集合の要素の個数の最大・最小（One More）★★★
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集合$U$とその部分集合$A,B$に対して，$n(U)=120,n(A)=70,n(B)=55$とする．このとき，次の値を求めよ．

(1) $n(A\cap B)$の最大値

(2) $n(A\cap B)$の最小値

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% 例題A1.1.4の解答（One More）★★★
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\[
n(A\cap B)=n(A)+n(B)-n(A\cup B)=70+55-n(A\cup B)=125-n(A\cup B)
\]

(1) $n(A)>n(B)$であるから，$n(A\cup B)$が最小となるのは$A\supset B$のときである．

したがって，右の図より，
\[
n(A\cup B)=n(A)=70
\]

よって，$n(A\cap B)$の最大値は，$125-70=55$

(2) $n(A)+n(B)>n(U)$であるから，$n(A\cup B)$が最大となるのは$A\cup B=U$のときである．

したがって，右の図より，
\[
n(A\cup B)=n(U)=120
\]

よって，$n(A\cap B)$の最小値は，$125-120=5$

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% 問題A1.1.4：集合の要素の個数の最大・最小（One More）★★★
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集合$U$とその部分集合$A,B$に対して，$n(U)=200,n(A)=90,n(B)=120$とする．このとき，次の値を求めよ．

(1) $n(A\cap B)$の最大値

(2) $n(A\cap B)$の最小値

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% 問題A1.1.4の解答（One More）★★★
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\[
n(A\cap B)=n(A)+n(B)-n(A\cup B)=90+120-n(A\cup B)=210-n(A\cup B)
\]

(1) $n(B)>n(A)$であるから，$n(A\cup B)$が最小となるのは$B\supset A$のときである．

したがって，右の図より，$n(A\cup B)=n(B)=120$

よって，$n(A\cap B)$の最大値は，$210-120=90$

(2) $n(A)+n(B)>n(U)$であるから，$n(A\cup B)$が最大となるのは$A\cup B=U$のときである．

したがって，右の図より，$n(A\cup B)=n(U)=200$

よって，$n(A\cap B)$の最小値は，$210-200=10$

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% 例題A1.1.4の側注：集合の要素の個数の最大・最小（One More）★★★
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% 側注1
このとき，$\overline{A\cup B}=\overline{A}\cap\overline{B}$の要素の個数が最大である．

% 側注2
$n(A\cap B)=125-n(A\cup B)$

% 側注3
$A\cup B=U$，すなわち，$\overline{A}\cap\overline{B}=\varnothing$

% 側注4
$n(A\cap B)=125-n(A\cup B)$

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% 問題A1.1.4の側注：集合の要素の個数の最大・最小（One More）★★★
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% 側注1
このとき，$\overline{A\cup B}=\overline{A}\cap\overline{B}$の要素の個数が最大である．

% 側注2
$n(A\cap B)=210-n(A\cup B)$

% 側注3
$A\cup B=U$，すなわち，$\overline{A}\cap\overline{B}=\varnothing$

% 側注4
$n(A\cap B)=210-n(A\cup B)$

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% 例題A1.1.4の考え方・補足など：集合の要素の個数の最大・最小（One More）★★★
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% 考え方
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$n(A)+n(B)=70+55=125$であり，$n(A\cap B)=n(A)+n(B)-n(A\cup B)$であるから，$n(A\cup B)$が最大となるとき，$n(A\cap B)$は最小となり，$n(A\cup B)$が最小となるとき，$n(A\cap B)$は最大となることを利用する．