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MATH 4181 001 Fall 1999 Problem Set 2 Solutions 1. (Exercise 1, page 34) Let(X;d)be a metric space and let k > 0be any positive real number. Let dk(x;y) =k
d(x;y)for any two points x;y2X. Show that (X;dk)is a metric space. We have to check the three conditions for a metric space. (1)dk(x;y) =k
d(x;y)k
0 = 0 since k>0, and ifdk(x;y) = 0, then k
d(x;y) = 0 which implies that d(x:y) = 0. This means that x=ysincedis a metric. (2)dk(x;y) =k
d(x:y) =k
d(y;x) =dk(y:x). (3) Letx;y;z2X, dk(x;z) =k
d(x;z)k
(d(x;y) +d(y;z)) =kd(x;y) +kd(y;z) =dk(x;y) +dk(y;z) Thus,dkis a metric. 2. (Exercise 8, page 35) LetZbe the set of integers. Let pbe a positive prime integer. Given distinct integers m;n2Zthere is a unique integer t=t(m;n)such thatmn= pt
k, wherekis an integer not divisible by p. Dene a function d:ZZ!Rby d(m;n) =( 0 ifm=n 1 ptifm6=n Prove that (Z;d)is a metric space. Let p= 3. What is the set of elements n2Zsuch thatd(0;n)<1? What is the set of elements m2Zsuch thatd(0;m)<1 3? First, we need to prove that it is a metric space. (1) From its denition d(m;n) = 0 ord(m;n) = 1=pt>0. Thus,d(m;n)0. Again, according to the denition, the only time that d(m;n) = 0 is ifm=n. Thus,dsatises the rst condition on being a metric. (2) Ifd(m;n) = 0 then m=nandd(n;m). Now, ifd(m;n)6= 0, thenmn=ptk, k6= 0. Thus, nm=pt
(k) and hence d(m;n) = 1=pt=d(n;m). (3) Ifm;n;r2Z, then letmn=ptk,nr=ps`. Hence, mr= (mn) + (nr) =ptk+ps` Without loss of generality we may assume that s<t =ps(ptsk+`) =ps MATH 4181 Problem Set 2 Solutions 2 andpdoes not divide . Therefore, d(m;r) =1 ps 1 pt+1 ps =d(m;n) +d(n;r) The argument is similar if t<s . Thus,dis a metric on Z. Now, letp= 3. Now, if m;n2Z, then write mn= 3tkand ifm6=nthen d(m;n) = 1=3t. LetA=fn2Zjd(n;0)<1g. Now, this means that we write n= 3tk and we want 1 =3t<1. Thus, we want t>0. Thus,A=fn2Zj3jngwhich are the multiples of 3. LetB=fm2Zjd(m;0)<1=3g. With our analysis above that means that we need to nd all numbers m= 3tkso thatt >1. Thus, we must have that t2 andB consists of all of the multiples of 9. 3. (Exercise 1, page 39) LetXbe the set of continuous functions f: [a;b]!R. Letdbe the metric on Xgiven by d(f;g) =Zb ajf(t)g(t)jdt; forf;g2X. For each element f2Xset I(f) =Zb af(t)dt: Prove that this function I: (X;d)!(R;d)is continuous. Let>0 be given. We need to nd >0 so that if d(f;g)<thend(I(f);I(g))<. From a theorem in calculus, we know thatR f(x)dxR jf(x)jdx. Thus, we have d(I(f);I(g)) =Zb af(x)g(x)dx Z jf(x)g(x)jdx =d(f;g) Thus, if we take =, thend(I(f);I(g))d(f;g)<=. Thus,Iis continuous. c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 2 Solutions 3 4.ForP= (2;1)andQ= (3;4)inR2, compute the distance from PtoQin each of the following metrics: (a)usual; d(P;Q) =p (3(2))2+ (41)2=p 34: (b)taxicab; d0(P;Q) =j3(2)j+j41j= 8. (c)maximum; d00(P;Q) = maxfj3(2)j;j41jg= 5. (d) discrete. SinceP6=Q,d(P;Q) = 1. 5.LetB=fP= (x1;x2)2R2jx2 1+x2 21g. Compute the diameter of Bin each of the following metrics: (a)usual; The diameter is the least upper bound of the distances apart of two points in this set. In the usual metric, we know that the greatest distance will be diametrically opposed points. This gives a diameter of 2. (b)taxicab; LetP= (x;y);Q= (a;b)2B. Thenjxj+jyj1 andjaj+jbj1. Now, d0(P;Q) =jxaj+jybj jxj+jaj+jyj+jbj (jxj+jyj) + (jaj+jbj) = 2 Thus, the least upper bound of these numbers is 2. The diameter of the set is 2. (c)maximum; LetP= (x;y);Q= (a;b)2B. Then maxfjxj;jyjg 1 and maxfjaj;jbjg 1. Now, d00(P;Q) = maxfjxaj;jybjg maxfjxj+jaj;jyj+jbjg 2 Thus, the least upper bound of these numbers is 2. The diameter of the set is 2. (d)discrete. In the discrete metric the greatest distance apart two points can be is 1. Thus the diameter must be 1. c David C. Royster Introduction to Topology For Classroom Use Only