page 10 1. Suppose that S 0, 1 1,. a. What is the set of iterior poits of S? The set of iterior poits of S is 0, 1 1,. b. Give that U is the set of iterior poits of S, evaluate U. 0, 1 1, 0, 1 1, S. The purpose of parts a ad b is to exhibit a set S such that, if U is the set of iterior poits of S the U S. c. Give a example of a set S of real umbers such that if U is the set of iterior poits of S the U S. We could take S to be a sigleto like 3 or it could be the set of all itegers. It could also be the set of all ratioal umbers betwee 0 ad 1. d. Give a example of a subset S of the iterval 0, 1 such that S 0, 1 but if U is the set of iterior poits of S the U 0, 1. Oce agai, take the set of all ratioal umbers betwee 0 ad 1.. Give that evaluate S. Hit: Show that S 1 Z, S 0 1 Z. First show that 0 S. The observe that every egative umber belogs to the set,0 ad that if x is ay positive umber the x belogs to the iterval 1 1, 1 for some positive iteger. 3. Give that S is a set of real umbers, that H is a closed set ad that S H, prove that S H. We could argue that S H ad that, because H is closed, H H. 4. Give two sets A ad B of real umbers, prove that A B A B. Solutio: Sice A A ad B B we have A B A B. ad therefore, sice the uio of the two closed sets A ad B is closed we have A B A B. O the other had, sice A is icluded i the closed set A B we have ad, similarly we ca see that ad so Therefore 5. Give two sets A ad B of real umbers, prove that A A B B A B A B A B. A B A B. 1
A B A B. Do the two sides of this iclusio have to be eual? What if A ad B are ope? What if they are closed? Sice A B A we have A B A ad similarly that A B B. Thus A B A B. Now observe that if A 0, 1 ad B 1, the A B 0, 1 1, 1 ad A B. Of course, we could give more spectacular examples like A Q ad B R Q. 6. Prove that if S is ay set of real umbers the the set R S is the set of iterior poits of the set R S. Most studets should be ecouraged to write two separate argumets here. The first task is to show that every member of the set R S must be a iterior poit of R S. The oe should show that every iterior poit of R S must belog to R S. O the other had, a strog studet could be permitted to observe that if x is ay give umber the the statemet that x does ot belog to S is the statemet that there exists a umber 0 such that x, x S, ad that the latter euatio is just the coditio that x, x R S. 7. Give that is a upper boud of a give set S of real umbers, prove that the followig two coditios are euivalet: a. We have sups. b. We have S. To prove that coditio a implies coditio b we assume that sups. We eed to show that S. Suppose that 0. Usig the fact that is the least upper boud of S ad that we choose a member x of S such that x. Sicex, S we have, S. To prove that coditio b implies coditio a we assume that S. We eed to show that is the least upper boud of S. Suppose that p. Sice the set p, is a eighborhood of we have p, S. Thus, sice is a upper boud of S ad sice o umber p ca be a upper boud of S we coclude that is the least upper boud of S. 8. Is it true that if A ad B are sets of real umbers ad A B R the A B R? The aswer is o. Look at A Q ad B R Q. 9. Prove that if A ad B are ope sets ad A B R the A B R. What if oly oe of the sets A ad B is ope? Solutio: All we eed to kow is that at least oe of the sets A ad B is ope. Suppose that A ad B are sets of real umbers, that A B R ad that the set A is ope. To prove that A B R, suppose that x is ay real umber ad that 0. Sice x A we kow that the set x, x A is oempty ad we also kow that this set is ope. Therefore, sice B R we kow that x, x A B. We have therefore show that every real umber must belog to A B.
10. Two sets A ad B are said to be separated from each other if A B A B. Which of the followig pairs of sets are separated from each other? a. 0, 1 ad, 3. Yes. b. 0, 1 ad 1,. Yes. c. 0, 1 ad 1,. No because 0, 1 1, 1. d. Q ad R Q. No. 11. Prove that if A ad B are closed ad disjoit from oe aother the A ad B are separated from each other. Suppose that A ad B are closed ad disjoit from oe aother. Sice A A ad B B, the fact that A B A B follows at oce from the fact that A B. 1. Prove that if A ad B are ope ad disjoit from oe aother the A ad B are separated from each other. Suppose that A ad B are ope ad disjoit from oe aother. Give ay umber x A, we deduce from the fact that A is a eighborhood of x ad A B that x is ot close to B. Therefore A B ad we see similarly that A B. 13. Suppose that S is a set of real umbers. Prove that the two sets S ad R S will be separated from each other if ad oly if the set S is both ope ad closed. What the do we kow about the sets S for which S ad R S are separated from each other? Suppose that S ad R S are separated from each other. To show that S is ope, suppose that x S. Sice S R S we kow that x is ot close to R S. Choose 0 such that x, x R S ad observe that x, x S. Thus S is ope ad a similar argumet shows that R S is also ope. We therefore kow that if the sets S ad R S are separated from oe aother the S is both ope ad closed. Now suppose that S is both ope ad closed. Sice the two set S ad R S are closed ad disjoit from oe other they are separated from oe aother. 14. This exercise refers to the otio of a subgroup of R that was itroduced i a earlier exercise. That exercise should be completed before you start this oe. a. Give that H ad K are subgroups of R, prove that the set H K defied i the sese of a earlier exercise is also a subgroup of R. To prove that H K is a subgroup of R we eed to show that H K is oempty ad that the sum ad differece of ay members of H K must always belog to H K. To show that H K is oempty we use the fact that H ad K are oempty to choose x H ad y K. Sice x y H K we have H K. Now suppose that w 1 ad w are ay members of the set H K. Choose members x 1 ad x of H ad members y 1 ad y of K such that w 1 x 1 y 1 ad w x y. Sice the umbers x 1 x 1 ad x 1 x belog to H ad the umbers y 1 y ad y 1 y belog to K, ad sice w 1 w x 1 x y 1 y ad w 1 w x 1 x y 1 y we see at oce that w 1 w ad w 1 w belog to H K. b. Prove that if a, b ad c are itegers ad if the a b c 0. Solutio: From the euatio we see that 3
a 3b bc 3 c. Therefore, uless bc 0wehave 3 a 3b c bc which cotradicts the fact that the umber 3 is irratioal. Therefore at least oe of the umber b ad c must be zero. I the evet that c 0, the euatio becomes a b 3 ad, uless a 0, the latter euatio gives us 3 b a which cotradicts the fact that is irratioal. So i the case c 0wealsohavea 0 ad we see at 3 oce that b 0 as well. I the evet that b 0, the euatio becomes a c ad, uless a 0, the latter euatio gives us a c which cotradicts the irratioality of. So, oce agai, a 0 ad we see at oce that c 0 as well. c. Prove that if m,, p ad are itegers the it is impossible to have ad deduce that if is ay real umber ad if H Z the the subgroup H Z caot cotai both of the umbers ad 3. Solutio: The euatio implies that 3 p m which, by part b, tells us that 0 m p which is clearly impossible sice ad appear deomiators of the fractios i the euatio. Now, to obtai a cotradictio, suppose that the subgroup H Z cotais both of the umbers ad 3. Choose itegers m ad such that m ad choose itegers p ad such that 3 p. Sice is irratioal, we kow that m ad so 0; ad we kow similarly that 0. Thus 4
which we kow to be impossible. d. Suppose that G is a subgroup of R other tha 0, that p ifx G x 0 ad that the umber p is positive. Prove that the set G is closed. Solutio: We kow from a earlier exercise that G p Z. e. Prove that if G is a subgroup of R other tha 0 ad that G has o least positive member the G R. Solutio: This fact was established i a earlier exercise. f. Suppose that is a irratioal umber, that H Z ad that G H Z (i the sese of this exercise). Prove that although the sets H ad Z are closed subgroups of R ad although the set G is also a subgroup of R, the set G is ot closed. Solutio: Sice G caot cotai both of the umbers ad 3 we kow that G R. To show that G is ot closed we shall make the observatio that G R ad, for this purpose, all we have to show is that if p ifx G x 0 the p 0. Suppose that p is defied i this way ad, to obtai a cotradictio, suppose that p 0. We kow that G p Z ad, usig the fact that both of the umbers 1 ad belog to G, we choose itegers m ad such that 1 mp ad p. From the fact that p 1/m we see that p is ratioal but from the fact that p / we see that p must be irratioal. Thus we have arrived at the promised cotradictio. 5