题目

设R1和R2是集合A上的等价关系,证明R1交R2是A上的等价关系

提问时间：2021-01-09

答案

证明 由交集的定义r1∩r2={(a,b)|(a,b)Îr1且(a,b)Îr2}.
对任意一个aÎA,因为r1和r2都是自反的,所以有(a,a)Îr1且(a,a)Îr2,因而有(a,a)Îr1∩r2,故r1∩r2是自反的.
对任意a,bÎA,若(a,b)Îr1∩r2,则有(a,b)Îr1且(a,b)Îr2,由r1和r2的对称性有(b,a)Îr1且(b,a)Îr2,因而有(b,a)Îr1∩r2,故r1∩r2是对称的.
对任意a,b,cÎA,若(a,b)Îr1∩r2,(b,c)Îr1∩r2,则有(a,b)Îr1,(b,c)Îr1；(a,b)Îr2,(b,c)Îr2.由r1和r2的传递性有(a,c)Îr1,(a,c)Îr2,因而有(a,c)Îr1∩r2,故r1∩r2是传递的.
由以上三方面知r1∩r2是A上的等价关系.证毕