Munkres - Topology - Chapter 1 Solutions
Munkres - Topology - Chapter 1 SolutionsSection 3Problem 3.2. Let C be a relation on a set A. If A0 A, define the restriction of C to A0 to be the relation C (A0 A0 ).Show that the restriction of an equivalence relation is an equivalence relation.Solution: Let C 0 be the restriction of C to A0 . As an initial matter, clearly if (a, b) C 0 , then (ab ) C. Further, if(a, b) C and a, b A0 , then (a, b) C 0 . We will now show that the C 0 is an equivalence relation.Let x0 A0 . Since (x0 , x0 ) C, it follows that (x0 , x0 ) C 0 . Therefore C 0 is reflexive.Let (x1 , y1 ) C 0 . It follows that (y1 , x1 ) C, so (y1 , x1 ) C 0 . Therefore, C 0 is symmetric.Let (x2 , y2 ), (y2 , z2 ) C 0 . It follows that (x2 , z2 ) C, so (x2 , z2 ) C 0 . Therefore, C 0 is transitive. Since all thenecessary properties are met, C 0 is an equivalence relation.Problem 3.4. Let f : A B be a surjective function. Let us define a relation on A by setting a0 a1 if f (a0 ) f (a1 ). (a)Show that this is an equivalence relation. (b) Let A be the set of equivalence classes. Show there is a bijective correspondenceof A with B.Solution: Part (a) Suppose R is the relation of interest. Let a0 A. Clearly f (a0 ) f (a0 ), so a0 a0 . It follows thatR is reflexive. Now let a0 , a1 A where a0 a1 . Since f (a0 ) f (a1 ) implies that f (a1 ) f (a0 ), we infer that a1 a0 .It follows that R is symmetric. Finally, let a0 , a1 , a2 A where a0 a1 and a1 a2 . We see that f (a0 ) f (a0 ) f (a2 ),so a0 a2 . Therefore, R is transitive. Accordingly, A is an equivalence relation.Part (b) Let A be the collection of equivalence classes of A of R, where if x A, there is an Ex A such thatEx {y A : x y}. Let g : A B where g(Ex ) f (x). First we will show that g is injective. Let Ex , Ey A andg(Ex ) g(Ey ). It follows that f (x) f (y), so x Ex and x Ey . Since Ex and Ey are not disjoint, we infer fromLemma 3.1 that Ex Ey . Therefore, g is injective.Now let b B. Since f is surjective, there is an a A such that f (a) b. Therefore, g(Ea ) f (a) b. Accordingly,g is surjective, from which it follows that g is a bijection from A to B.Problem 3.10. (a) Show that the map f : ( 1, 1) R of Example 9 is order preserving. (b) Show that the equationg(y) 2y/[1 (1 4y 2 )1/2 ] defines a function g : R [ 1, 1] that is both a left and a right inverse for f .Solution: Part (a) Let A ( 1, 1). The function f is order preserving if a0 a1 implies that f (a0 ) f (a1 ) for alla0 , a1 A. Now let x, y A. We may assume without loss of generality that x y. We then have:f (x) f (y) x y 0,(1 x2 )(1 y 2 )because 1 x2 , 1 y 2 0 for all x, y A. It follows that f (x) f (y). The function f is therefore order preserving.Part (b) This problem a straightforward (but annoying) substitution of f and g into each other. let x A. We thenhave: 2x12x(g f )(x) x,1 x2(1 x2 ) (1 x2 )1 [1 4x2 /(1 x2 )2 ]1/2showing that g is a left inverse of f .For (f g), we can again do the substitution and show (through a lot of algebra) that (f g)(y) y. For y R, wehave: 2y12y(1 (1 4y 2 )1/2 )(f g)(y) y,21/2221/221 (1 4y )1 4y /[1 (1 4y ) ]2(1 (1 4y 2 )1/2 )showing that g is a right invoice for f .Problem 3.11. Show that an element in an ordered set has at most one immediate successor and at most one immediatepredecessor. Show that a subset of an ordered set has at most one smallest element and at most one largest element.
Solution: First we will show that an element in an ordered set has at most one immediate successor. Let A be an orderedset and x A. If x is the largest element of A, then it has no immediate successor. Now suppose instead that x hastwo immediate successors y, z A. If we assume y 6 z, we may further assume without loss of generality that y z.Since each is an immediate successor of x, by definition (x, y) (x, z) . However, since x y z, it must be thaty (x, z), which is a contradiction. Therefore, y z, establishing that any element in an ordered set has at most oneimmediate successor. A similar argument shows that any element of A has at most one immediate predecessor.Now we will show that A has at most one smallest element. Clearly A may not have any smallest element (for example,the empty set or {x R : x 0}). Suppose x0 , x1 A are its two smallest elements. By definition, if y A, then x0 yand x1 y. We infer then that x0 x1 and x1 x0 . It follows that x0 x1 . Accordingly, A has at most one smallestelement. A similar argument shows that A has at most one largest element.Problem 3.13. Prove the following: Theorem. If an ordered set A has the least upper bound property, then it has the greatestlower bound property.Solution: Let A0 be a non-empty subset of A that has a lower bound in A, and let B be the set of all elements of A thatare lower bounds of A0 (which is plainly non-empty). It follows that for each b B, if a A0 then b a. The set Bis therefore bounded above by the elements of A0 . Since A has the least upper bound property, it follows that B has asupremum x A. It follows that if x0 B, then x0 x, so x is at least as large as any lower bound of A0 . We now needto show that x is a lower bound of A0 . Let a A0 . Since x sup B, it cannot be that a x; otherwise, x could not bethe supremum of A0 because a b for all b B. Accordingly, x inf A0 .Since every non-empty subset of A with a lower bound in A has an infimum, it follows that A has the greatest lowerbound property.Following an analogous argument, we can show that if A has the greatest upper bound property, then it has the leastupper bound property.Problem 3.14. If C is a relation on a set A, define a new relation D on A by letting (b, a) D if (a, b) C. (a) Show thatC is symmetric if and only if C D. (b) Show that if C is an order relation, then D is also an order relation. (c) Prove theconverse of the theorem in Exercise 13.Solution: In all my solutions, I assume that D {(b, a) A A : (a, b) C}. The prompt isn’t entirely clear on this(there is no ”if and only if”).Part (a) Suppose C is symmetric. If (a, b) C, then (b, a) C. Therefore, (a, b) D, so C D. If (b, a) D, then(a, b) C, from which it follows that (b, a) C. Therefore, D C, so C D.Now suppose C D. If (a, b) C, then (b, a) D, from which it follows that (b, a) C. Therefore, C is symmetric.Part (b) Assume C is an order relation. Since (a, a) / C, it follows that (a, a) / D. Now let a, b A. Either (a, b) Cor (b, a) C, but not both. Therefore, either (a, b) D or (b, a) D, but not both. Now suppose (a, b), (b, c) C. Itfollows (a, c) C. Therefore, (c, b), (b, a), (c, a) D. With all the properties met, D is an order relation.Part (c) Let C be an order relation on A. Let D be an order relation on A where D {(b, a) A A : (a, b) C}.Let A0 be a subset of A. Observe that any upper bound of A0 by C is a lower bound of A0 by D. Suppose x A is anupper bound of A0 by C, in which case if a A0 , then a C x. It follows that x D a, so x is a lower bound of A0 .There is an analogous result for the suprema and infima under C and D. Suppose x0 sup A0 by C. As just shown, x0is a lower bound of A0 by D. If x00 is any upper bound of A0 by C, then x0 C x00 , from which it follows that x00 D x0 .Therefore, x0 inf A0 by D.Now suppose A have the greatest lower bound property by D. Therefore, A has the least upper bound property by C.Applying the result in Exercise 3.13, we infer that A has the greatest lower bounded property by C. But this means thatA has the least upper bound property by D. Accordingly, if A has the greatest lower bound property, it also has the leastupper bound property.Section 4Problem 4.1. [See question.]Page 2
Solution: I only show solutions to some of the more interesting problems here. To be true to the problem, I work out thesteps (particularly the distributive and commutative properties of R) in special detail.Part (c) By (4) there is a 0 such that 0 0 0. It follows from (1), (2), and (4) that: 0 0 0 0 0 0.Part (d) By (4), for x R, there is a x such that x x 0. Applying (4) again, there is a ( x) such that x ( x) 0. Putting these together with (1) and (3):(x x) ( x) x ( x ( x)) x 0 ( x) ( x) 0 ( x).Part (e) For x, y R, for xy there is a xy by (4). Since x · 0 x(y y) 0, we have from (1), (2), and (5):x(y y) xy (xy x( y)) xy x( y) (xy xy) x( y) 0 xy xy 0 xy.Following the same argument, since y · 0 0 · y (x x) · y 0, we have:(x x)y xy (xy ( x)y) xy ( x)y (xy xy) ( x)y 0 xy xy.It follows that x( y) ( x)y xy.Part (f) Applying part (e) and (2), we have x( y) ( y)x ( x)y. Plugging in y 1 and using (4), we have( 1)x x(1) x.Part (i) There is a 1/x such that 1/x · x x/x 1.Part (k) From exercise (j), 1/1 1. Therefore:x1 .11Part (l) Because x 6 0 and y 6 0, there are multiplicative inverses 1/x and 1/y by (4). As a result, by (1) and (2):x x·1 x·(xy)(1/x · 1/y) x · (y · (1/x · 1/y)) x · (1/x · (y · 1/y)) x · (1/x · 1) x · 1/x 1.It cannot be that xy 0 because, if it were, the foregoing expression would equal zero by exercise (b). Therefore,xy 6 0.Part (m) By exercise (l), yz 6 0, so it has multiplicative inverse 1/(yz) by (4). By (1) and (2), we have: 1 1111·1 y·z· (yz)· 1.yzy zIt follows that (1/y)(1/z) is a multiplicative inverse of yz. But 1/(yz) is also a multiplicative inverse of yz, and by (4)the inverse is unique. We infer then that (1/y)(1/z) 1/(yz).Part (n) From exericse (m) and properties (1) and (2), we have: xw11111xw x· .w· (xw) (xw) ·yzyzyzyzyqPart (o) From exercise (j), y/y z/z 1. From (1), (2), and (5) and exercise (m), we have: x wxwx zw y1 11 1 · 1 · 1 · · (xz)· (wy)·yzyzy zz yy zz y (xz)111xz wy (wy) (xz wy) .yzyzyzyzPart (p) There is a 1/x by (4). Since 1/x · x 1, it must be that 1/x 6 0; otherwise, by exercise (b) 1/x· 0, acontradiction.Part (q) Since z 6 0, it follows from (4) and exercise (p) that 1/z exists and is non-zero. Since w 6 0, we inferthat w · 1/z w/z 6 0 from exercise (l). Applying (4), we see that w/z has multiplicative inverse (1/(w/z). Sincew/z · 1/(w/z) 1, we have from (1) and (2) and exercises (j) and (n): zw1z w 111·· ·· (wz)wz w/zw zw/zwz w/z wz11zz· ·1 .wz w/zw/zwwPage 3
Solution: Part (r) From exercises (n) and (q):x/yx zxz · .w/zy wywPart (s) From (1):ax (ax)y 11x a x· a.yyyPart (t) From exercises (f) and (s), ( x)/y [( 1)(x)]/y ( 1)(x/y) (x/y).The multiplicative inverse of y is 1/( y). Starting with ( y) · 1/( y) 1, we then have from (1) and (2) andexercise (d) and (g): 1 x11111 x( y) · ( 1)(y) · ( 1)( x) ·y· x·y·y yy yy yy y 1111x x x ·xy· ·yx· ·1 .y yy y yyyIt follows that ( x)/y x/( y) (x/y).Problem 4.2. [See question.]Solution: Part (a) From (6), x z y z. Relying on (4) and (6), we see that w x x w z x x z. Bythe transitivity property of order relations, x w y z.Part (b) From exercise 2(a), x y 0 0 0.Part (c) Suppose x 0. Applying (6), we have x ( x) 0 0 ( x) x. Conversely, suppose x 0. Wethen have x x 0 0 x x.Part (d) Suppose x y. From (6), x y y y 0. From exercises 1(h) and 2(c), (x y) x y 0, so x y y x 0 y y. Conversely, suppose x y. From (2) and (6), x x 0 y x x y. Fromexercises 1(b) and 2(c), (x y) x y 0, so x x y y x 0 x.Part (e) If z 0, then z 0 by exercise 2(d). Applying (6) and exercise 1(e), we have x( z) (xz) y( z) (yz). From exercise 1(d), xz yz.Part (f) If x 0, then x · x x2 0 · x 0 by (6). If x 0, then x 0 by exercises 2(c) and (d). Applying (6)and exercises 1(b), (e), and (f), we have:( x)( x) [x · ( x)] [ (x · x)] x2 0 · ( x) 0.Part (g) Let x R where x 0. It follows that x has multiplicative inverse 1/x. From (1) and exercise 1(f), sincex2 0, we have:x2 · (1/x · 1/x) 1 0 · (1/x · 1/x) 0.From exercise 2(c), 1 0. By the transitivity of the order relation, 1 0 1.Part (h) Suppose xy 0. We can prove the implication by cases. Clearly from exercise 2(b), neither x nor y may beequal to zero. If x, y 0, then by (6) and exercise 1(b) x · y 0 · y 0. If x, y 0, then x, y 0 by exercise 2(c).Applying (6) and exercises 1(b) and (e):( x)( y) [x( y)] [ (xy)] xy 0 · ( y) 0.Now assume that x 0 and y 0. From exercise 2(e), we infer from x 0 that xy 0 · y 0, which is acontradiction. Therefore, it cannot be that x 0 and y 0. Having exhausted all possibilities for x and y, we infer thatx and y are either both positive or both negative.Conversely, suppose x and y are either both positive or both negative. As shown above, in either of those cases xy 0.Part (i) Since x 0, it has multiplicative inverse 1/x by (4). Because x · 1/x 1 0 by exercise 2(g), it follows fromexercise 2(h) that 1/x 0.Part (j) Since x, y 0, there are multiplicative inverses 1/x and 1/y, each of which is greater than zero by exercise2(i). By exercise 2(h), 1/x · 1/y 0. Since x y by hypothesis, by (1) and (2) and exercise (h): 1 111 11x·· y .x yyx yxPage 4
Part (k) Since x y by hypothesis, from (6) and exercise 2(a) we have x x 2x y x x y. Similarly,y y 2y x y. From exercises 2(a) and (g), we infer that 1 1 2 0 0 0. From exercise 2(i), and 2(j), wehave 1/2 0. From exercise 2(b) and the transitivity of order relations, we have:2x ·11x y1 x (x y) · 2y · y.2222Problem 4.3. (a) Show that if A is a collection of inductive sets, then the intersection of the elements of A is an inductive set.(b) Prove the basic properties (1) and (2) of Z 0 .TSolution: Part (a) Let A be the collection of all inductive sets and A A0 A A0 . Since each A0 A is an inductive set,1 A0 , so 1 A. If x A, then x A0 for all A0 A. Again, since A0 is inductive, x 1 A0 . Therefore, x 1 A.Consequently, A is an inductive set.TPart (b) Property (1): Because Z 0 A0 A A0 where A0 is the collection of all inductive sets of R, from part (a)Z 0 is an inductive set.Property (2): Suppose A is an inductive set of positive integers. If x A, then x Z 0 by the definition of A.Therefore A Z 0 . Conversely, we know 1 A and if x A, then x 1 A. As a result, 1 1 2 A, 2 1 3 A,and so on such that every positive integer is a member of A. It follows that Z 0 A and A Z 0 .Problem 4.4. (a) Prove by induction that given n Z 0 , every nonempty subset of {1, . . . , n} has a largest element.(b) Explain why you cannot conclude from (a) that every nonempty subset of Z 0 has a largest element.Solution: Note that N Z 0 (by my definition, where zero is not a natural number), so everything true for one is true ofthe other. I will generally use N instead of Z 0 .Part (a) We will show by induction that for n N, any nonempty subset Sn 1 {1, . . . , n} of N has a largest element.For n 1, the set S2 {1} has only one nonempty subset ({1}), which has the largest element 1. Now assume theinductive hypothesis is true for some n N. Let S 0 Sn 2 {1, . . . , n 1}. If n 1 S 0 , then n 1 is the largestelement of S 0 . If n 1 / S 0 , then S 0 is some non-empty subset of Sn 1 . By the inductive hypothesis, S 0 has a largestelement. Since this is true for all n N, the proposition is true for any non-empty subset Sn 1 of N.Part (b) The result in (a) is limited to bounded subsets of N; it doesn’t hold true for unbounded subsets. LetTn {x N : x n} for n N. Obviously Tn N and is non-empty. Suppose Tn has a largest element M . It followsthat M N. Since N is an inductive set, M 1 N. But M 1 M n, so M 1 Tn , which contradicts that M isthe largest element of Tn . Therefore Tn does not have a largest element.Problem 4.9. (a) Show that every nonempty subset of Z that is bounded above has a largest element. (b) If x / Z, showthere is exactly one n Z such that n x n 1. (c) If x y 1, show there is at least one n Z such that y n x.(d) If y x, show there is a rational number z such that y z x.Solution: Part (a) Let A Z that is nonempty and bounded above by some M Z. Let A0 {M 1 a : a A}.Note that we can map between corresponding a A and a0 A0 as follows: a0 M 1 a and a M 1 a0 .Since M a 0 for all a A, it follows that M 1 a 1. As a result, a0 1 for all a A, so A0 N. From theWell-Ordering Principle, A0 has a least element m0 . The corresponding element of A is m M 1 m0 .Simple arithmetic shows since m0 a0 , we have M 1 m M 1 a, so m a. Because this implies thatm a for all a A, we conclude that A has a largest element. Since our choice of A was arbitrary, we conclude that anynon-empty subset of Z that is bounded above has a largest element.Part (b) Let x R\Z and Sx {n Z : n x}. Since Sx is a non-empty subset of Z (because Z is not boundedbelow), it has a greatest element n Z that is less than x. Because n 1 / Sx , we infer that n 1 x. It cannot bethat n 1 x because x is not an integer, so n 1 x.Now suppose there is an m Z where m x m 1. It follows that m Sx , so m n. If m n, thenm 1 n x, which is not possible. Therefore m n, from which it follows that n is unique.Part (c) Let x, y R where x y 1. From part (b), there is an n Z such that n x n 1. It follows thatx 1 n. Because y x 1, we conclude that y n x for some n Z.Part (d) Let x, y R where y x. If x y 1, by part (c) there is an n Z such that y n x. Since n is arational number as well, we’re done.Page 5
If 0 x y 1, we will show that there exists an m Z such that m(x y) 1. Let a x y, so 0 a 1. Itfollows that 2/a 2 1. From part (b), there is an m Z such that m 1 2/a m. Therefore:2 22· a (x y) m(x y) mx my.aaTherefore, mx my 1. From part (c), there is an n Z such that my n mx. Since m 2/a 1, we knowm 6 0. Therefore, y n/m x, where n/m is a quotient of two integers and is therefore a rational number.Problem 4.11. Given m Z, we say that m is even if m/2 Z and m is odd otherwise. (a) Show that if m is odd,m 2n 1 for some n Z. (b) Show that if p and q are odd, so are p · q and pn , for any n Z 0 . (c) Show that if a 0 isrational, then a m/n for some m, n Z where not both m and n are even. (d) Theorem: 2 is irrational.Solution: Part (a) Suppose m is odd so m/2 / Z. Consequently, m R\Z, so by exercise 9(b) there is an n Z suchthat n m/2 n 1. We then have 2n m 2n 2. Since 2n 1 is the only integer between 2n and 2n 2, itfollows that m 2n 1.Part (b) If p, q Z are odd, then from part (a) there are m, n Z such that p 2m 1 and q 2n 1. Therefore:p · q (2m 1)(2n 1) 4mn 2m 2n 1 2(2mn m n) 1 2r 1,where r Z and r 2mn m n. We conclude that p · q is odd.We will prove that pn is odd for n N by induction. Clearly p1 p, which is odd. If the inductive hypothesis is truefor some n N, then pn 1 pn · p, which is the product of two odd numbers by the inductive hypothesis. By the previousresult, pn 1 is odd. We conclude that pn is odd for all n N.Part (c) Let a Q be given where a 0. Clearly a may be expressed as the quotient of two natural numbers. LetBa {x N : x · a N}. By the Well-Ordering Principle, Ba has a smallest element n, and n · a m for some m N.It follows that a m/n, the quotient of two natural numbers. Now suppose both m and n are even, so there are r, s Nsuch that m 2r and n 2s. Th
Munkres - Topology - Chapter 1 Solutions Section 3 Problem 3.2. Let Cbe a relation on a set A. If A 0 A, de ne the restriction of Cto A 0 to be the relation C\(A 0 A 0). Show that the restriction of an equivalence relation is an equivalence relation.
Munkres - Topology - Chapter 2 Solutions Section 13 Problem 13.1. Let Xbe a topological space; let Abe a subset of X. Suppose that for each x2Athere is an open set U containing xsuch that UˆA. Show that Ais open in X. Solution: Let C A the collection of open sets Uwhere x2U Afor some x2A. Suppose U 0 S U2C A U. Since Xis a topological space, U 0 is open in X. Clearly if x2A, then x2U 0, so .
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
most basic notions of combinatorial topology, going back roughly to the 1900-1930 period and it is covered in nearly every algebraic topology book (certainly the “classics”). A classic text (slightly old fashion especially for the notation and terminology) is Alexandrov [1], Volume 1 and another more “modern” source is Munkres [30].
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
set topology, which is concerned with the more analytical and aspects of the theory. Part II is an introduction to algebraic topology, which associates algebraic structures such as groups to topological spaces. We will follow Munkres for the whole course, with some occassional added topics or di erent perspectives.
This course will focus on topology and complex analysis, with a short unit on real analysis. We will be using the following four textbooks as reference. 1. Topology by Munkres. 2. Complex Analysis by Ahlfors. 3. Math 131 Notes by McMullen (Topology). 4. Principles of Mathematical Analysis by Rudin (\Baby Rudin"). 1.2 Metric Spaces
Grouted pile connections shall be designed to satisfactorily transfer the design loads from the pile sleeve to the pile as shown in . Figure K.5-1. The grout packer may be placed above or below the lower yoke plate as indicated in Figure K.5-2. The connection may be analysed by using a load model as shown in Figure K.5-3. The following failure modes of grouted pile to sleeve connections need .
