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1、专业好文档数学分析专题研究学习辅导（四）第一章 集合与映射（四） 有关二元关系部分 典型例题解析 例1 设集合A= 1, 2, 3, 4上的二元关系R= (1, 1), (1, 2), (2, 4), (3, 1), (3, 3)，S= (1, 3), (2, 2), (3, 2), (4, 4)，用定义求. 思路 求复合关系，就是要分别将R中有序对(a, b)的第2个元素b与S中的每个有序对(c, d)的第1个元素进行比较，若它们相同（即b=c），则可组成中的1个元素(a, d)，否则不能. 幂关系的求法与复合关系类似. 求关系R的逆关系，只要把R中的每个有序对的两个元素交换位置，就能得到中
2、的所有有序对. 解 = (1, 1), (1, 2), (2, 4), (3, 1), (3, 3) (1, 3), (2, 2), (3, 2), (4, 4) = (1, 3), (1, 2), (2, 4), (3, 3), (3, 2) =(1, 3), (2, 2), (3, 2), (4, 4)(1, 1), (1, 2), (2, 4), (3, 1), (3, 3) =(1, 1), (1, 3), (2, 4), (3, 4) = (1, 1), (1, 2), (2, 4), (3, 1), (3, 3) (1, 1), (1, 2), (2, 4), (3, 1), (3
3、, 3) =(1, 1), (1, 2), (1, 4), (3, 1), (3, 2), (3, 3) =(1, 1), (1, 2), (2, 4), (3, 1), (3, 3) =(1, 1), (1, 3), (2, 1), (3, 3), (4, 2) =(1, 3), (2, 2), (3, 2), (4, 4) = (2, 2), (2, 3), (3, 1), (4, 4) =(1, 1), (1, 3), (2, 1), (3, 3), (4, 2)(2, 2), (2, 3), (3, 1), (4, 4) =(1, 1), (3, 1), (4, 2), (4, 3)注
4、：由例1可知，关系的复合运算不满足交换率，即. 例2 对于以下给定的集合A、B和关系f，判断是否构成映射f：. 如果是，试说明f：是否为单射、满射或双射的. （1）A=1, 2, 3, 4, 5，B=6, 7, 8, 9, 10，f =(1, 8), (3, 9), (4, 10), (2, 6), (5, 9)； （2）A=1, 2, 3, 4, 5，B=6, 7, 8, 9, 10，f =(1, 7), (2, 6), (4, 5), (1, 9), (5, 10)； （3）A=1, 2, 3, 4, 5，B=6, 7, 8, 9, 10，f =(1, 8), (3, 10), (2, 6
5、), (4, 9) （4）A=B=R，f (x) = x3，（R）； （5）A=B=R，（ R）； 思路 首先按照1.2节的定义2.5，判断A、B和f是否构成映射，即判断f是否具有单值性以及Dom(f )是否等于A. 然后再按照定义2.6，说明f：具有的性质. 解 （1）因为Dom(f ) = A，且对任意（i=1, 2, 3, 4, 5），都有唯一的，使(i, j ). 所以A、B和f能构成函数f：. 因为存在3, 5A，且35，但映射f (3)= f (5) = 9，所以f：不是单射的； 又因为集合B中的元素7不属于f的值域，即f (A)B，所以f：不是满射的. xf (x)123-1-3
6、-2123-2-1-3图1-1 （2）因为对1A，存在7, 9B，有f (1)= 7，f (1)= 9，即f不满足映射定义的单值性条件. 所以A、B和f不能构成映射f：. （3）因为Domf =1, 2, 3, 4A，所以A、B和f不能构成映射f：. （4）因为对R，都有唯一的R，使(x, ). 所以A、B和f能构成映射f：. 由图1-12可知，f：，f (x)= x3是双射的. （5）因为对R，都有唯一的R，使. 所以A、B和f能构成映射f：. 因为该映射在x 0处，f (-x)= f (x)，且f (R) R，所以映射f：不是单射的，也不是满射的. 例3 证明：若f：XY，A，BY，则(A
7、- B) =(A)-(B) 证明 x(A- B)，y(A- B)，即yA但yB，使得y = f (x)，从而有 x(A)但x(B)，故x(A)-(B) (A-B)(A) -(B) 又 x(A)-(B)，由于x(A)但x(B)，从而f (x)A但f (x)B，即f (x)(A-B)，故x(A- B) (A) -(B)(A-B) 因此，(A- B) =(A)-(B) 例4 设有映射f:AA. 若aA, f(a)=a, 则称映射f是恒等映射，表示为. 设有两个映射f:AB, g:BA. 若gf =, 则f是单射，g是满射. 证明 (1) 证明映射f是单射. 对任意的bB，如果存在a1，a2A，使f
8、(a1) = b，f (a2) = b，即f (a1) = b = f (a2). 因为 a1=(a1)=(gf )(a1)= g(f (a1) = g(f (a2) =(gf )(a2) =(a2)= a2 . 所以f是单射的. (2) 证明映射g是满射. 因为(gf )(A)=(A)= A，所以gf是满射的. 又对任意的cA，由gf是满射的可知，存在aA，使(gf )(a) = c. 那么存在bB，使f (a) = b，g(b) = c. 所以存在bB，使g(b) = c，即g是满射的. 例5 设函数f：AB，g：BC，且gf：AC，证明：若f 和g都是单射的，则gf 也是单射的. 证明
9、因为对任意的a1，a2A，如果a1a2，那么由f 是单射的可知，f (a1) f (a2). 而由g是单射的可知，g (f (a1)g( f (a2). 所以，由a1a2可得 (gf) (a1)( gf ) (a2) ，即gf是单射的. 例6 设f：RR，；g：RR，. 求gf ，f g. 如果f 和g存在逆映射，求它们的逆映射. 解：（1）求gf和f g (gf )(a)= g(f (a)+2 =； (f g)(a)= f(g(a) = f (a+2) = （2）求逆映射. 因为映射f：RR， 不是满射的. 所以f：RR不是双射，由1.2节注2.1可知，f不存在逆映射. 又因为g：RR，g(
10、a) = a +2即是满射的，又是单射的. 所以g：RR是双射，因此g存在逆映射，其逆映射为：RR，. 例7 设R1和R2是集合A上的任意关系，试证明或用反例推翻下列论断： （1）若R1和R2都是反身的，则也是反身的； （2）若R1和R2都是对称的，则也是对称的； （3）若R1和R2都是传递的，则也是传递的. 思路 做这类题目时，必须深入理解相关的概念，这样才能做出正确的判断；在此基础上进行证明或举出反例. 证 （1）因为对任意，若R1和R2都是A上的反身关系，则 (a, a)，(a, a)所以，(a, a)，即也是反身的. 故该论断正确. （2）例如，设A=a, b, c，当R1=(a, b
11、), (b, a), (c, c)，R2=(b, c), (c, b)，R1与R2都是对称的，但是=(a, c), (c, b)已不是对称的，故该论断不正确. （3）例如，设集合A=a, b, c，当R1=(a, b), (b, c), (a, c)，R2=(b, c), (c, a), (b, a)，R1和R2都是传递的. 但是，由= (a, a), (a, c), (b, a) 得 (b, a)，(a, c) ，且(b, c)，故不是传递的，即该论断不正确. 例8 设集合A=a, b, c, d, e，A上的关于等价关系R的等价类为： a = a, b, c， d = d, e求：（1）等
12、价关系R； （2）画出关系图. 思路 由等价关系的定义可知，等价关系R同时是反身的、对称的和传递的. 又由定义3.2知道，等价类中的任意两个元素都有关系. 因此，写A上的等价关系R的步骤为： 写出A上的恒同关系IA，使R是反身的； 分别写出等价类a、d中各元素两两之间的关系，使R具有对称性和传递性； 求、结果的并集，得到所求的等价关系. 解 （1）因为等价关系R是反身的，所以，IA=(a, a), (b, b), (c, c), (d, d), (e, e). 又因为a，b，c在同一个等价类中，所以(a, b), (b, a), (a, c), (c, a), (b, c), (c, b) 同
13、样，因为d，e在同一个类中，所以(d, e), (e, d) 由此可得R = IA(a, b), (b, a), (a, c), (c, a), (b, c), (c, b), (d, e), (e, d) （2）R的关系图如图1-13所示.abc图1-2de 例9 设R为集合A中的对称的、传递的关系，证明R为等价关系Dom(R)=A. 证明 先证“”. 因为R为等价关系，即R为反身的，I(A)R. 所以，Dom(R)=A. 再证“”. 因为Dom(R)=A，那么，对aA，有(a, a)R，或bA，使(a, b)R，由R为对称的和传递的，得(a, b)R, (a, a)R，即R为反身的. 所以
14、，R为等价关系. 例10 设集合A=2, 3, 4, 6, 8, 12, 24，DA为A上的整除关系. （1）写出集合A中的最大元，最小元，极大元，极小元； （2）写出A的子集B=2, 3, 6, 12的上界，下界，最小上界，最大下界. 思路 最大元与极大元是不一样的. A的最大元应该大于等于A中其它各元素. A的极大元应该不小于A中其它各元素，即它大于等于A中的一些元素，而与A中另一些元素无关系. 最大元不一定存在，如果存在，必定唯一. 在非空有限集合A中，极大元必定存在，但不一定唯一. 类似地，最小元与极小元也有这种区别. 集合B的最大元一定是B的上界，而且是B的最小上界. 同样，集合B的
15、最小元一定是B的下界，而且是B的最大下界. 解 （1）因为 DA=(2, 2), (2, 4), (2, 6), (2, 8), (2, 12), (2, 24), (3, 3), (3, 6), (3, 12), (3, 24), (4, 4), (4, 8), (4, 12), (4, 24), (6, 6), (6, 12), (6, 24), (8, 8), (8, 24), (12, 12), (12, 24), (24, 24)是序关系，所以(A，DA)是半序集. 由DA可以看出，24大于等于集合A中的所有元素，即A中的最大元是24. 2与3不大于A中的其它元素，且2与3无关系；因
16、此A中无最小元. 由此可得A中极大元也是24，极小元是2与3. (2）由定义4.4可知，集合B的上界是12与24，无下界；最小上界是12，无最大下界. If we dont do that it will go on and go on. We have to stop it; we need the courage to do it.His comments came hours after Fifa vice-president Jeffrey Webb - also in London for the FAs celebrations - said he wanted to meet I
17、vory Coast international Toure to discuss his complaint.CSKA general director Roman Babaev says the matter has been exaggerated by the Ivorian and the British media.Blatter, 77, said: It has been decided by the Fifa congress that it is a nonsense for racism to be dealt with with fines. You can alway
18、s find money from somebody to pay them.It is a nonsense to have matches played without spectators because it is against the spirit of football and against the visiting team. It is all nonsense.We can do something better to fight racism and discrimination.This is one of the villains we have today in
19、our game. But it is only with harsh sanctions that racism and discrimination can be washed out of football.The (lack of) air up there Watch mCayman Islands-based Webb, the head of Fifas anti-racism taskforce, is in London for the Football Associations 150th anniversary celebrations and will attend C
20、itys Premier League match at Chelsea on Sunday.I am going to be at the match tomorrow and I have asked to meet Yaya Toure, he told BBC Sport.For me its about how he felt and I would like to speak to him first to find out what his experience was.Uefa hasopened disciplinary proceedings against CSKAfor
21、 the racist behaviour of their fans duringCitys 2-1 win.Michel Platini, president of European footballs governing body, has also ordered an immediate investigation into the referees actions.CSKA said they were surprised and disappointed by Toures complaint. In a statement the Russian side added: We
22、found no racist insults from fans of CSKA.Baumgartner the disappointing news: Mission aborted.The supersonic descent could happen as early as Sunda.The weather plays an important role in this mission. Starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitat
23、ion or humidity and limited cloud cover. The balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. It will climb higher than the tip of Mount Everest (5.5 miles/8.85 kilometers), drifting even higher than the cruisin
24、g altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. As he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.The balloon will slowly drift to the edge of space at 120,000 feet ( Then, I would assume, he will slowly step out onto
25、something resembling an Olympic diving platform.Below, the Earth becomes the concrete bottom of a swimming pool that he wants to land on, but not too hard. Still, hell be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. It will be like he is diving int
26、o the shallow end.Skydiver preps for the big jumpWhen he jumps, he is expected to reach the speed of sound - 690 mph (1,110 kph) - in less than 40 seconds. Like hitting the top of the water, he will begin to slow as he approaches the more dense air closer to Earth. But this will not be enough to sto
27、p him completely.If he goes too fast or spins out of control, he has a stabilization parachute that can be deployed to slow him down. His team hopes its not needed. Instead, he plans to deploy his 270-square-foot (25-square-meter) main chute at an altitude of around 5,000 feet (1,524 meters).In orde
28、r to deploy this chute successfully, he will have to slow to 172 mph (277 kph). He will have a reserve parachute that will open automatically if he loses consciousness at mach speeds.Even if everything goes as planned, it wont. Baumgartner still will free fall at a speed that would cause you and me to pass out, and no parachute is guaranteed to work higher than 25,000 feet (7,620 meters).cause there6
