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11.If(A,R) is a poset but not a total order, and B(不等於空集合) 包含於 A,
does it follow that (B x B)交集 R makes B into a poset but not a total order?
解答舉了一個反例
u = {1,2} A=P(u) let R be the inclusion relation.
Then (A,R) is poset but not a total order.
Let B={空集合 , {1}}. then (B x B)交集 R is a total order
我想知道(B x B)交集 R會是什麼樣子
我無法想像
請大家幫忙
謝謝
1 則留言:
把 R 想成是最早將relation定義為cartesian product形式, i.e., R = {({},{1}),({},{2}),({},{1,2}),({1},{1,2}),({2},{1,2}), 再加上4個自己包含自己的反身性}; BxB = {({},{}),({},{1}),({1},{}),({1},{1})}, 所以 R ∩ BxB = {({},{}),({},{1}),({1},{1})}
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