solutions to exercises - springer978-1-4612-0083... · 2017. 8. 25. · solutions to exercises 157...

Solutions to Exercises

The solutions here are intended to be instructive. They are therefore sprinkled with comments, alternative approaches, and questions. Except for leaving such questions to be pondered by the reader, these answers are complete.

Many of them were worked out by Sera Cremonini, summa cum laude graduate of City College, whose knowledge of mathematics is all the more wonderful because it was not her major.

Chapter 1 Section 1.1

1. (a) The graph is a rectangular hyperbola--one whose asymptotes Y = ±x are at right angles-crossing the x-axis at (±1, 0). See the figure for this solution.

(b) No. (2, J3) and (2, -J3) both satisfy the equation, but J3 and

-J3 cannot both be g(2). (Note that an example is essential, and that many exist.)

2. This is familiar from the definition of function. Keep Exercise 1 in mind.

(a) It must pass the "vertical-line test": No vertical line can have as many as two points from the set. Observe that the graph in (1) fails.

(b) For any value of x, if (x, Yl) and (x, Y2) are both in the set, then

Yl = Y2·

156 Solutions to Exercises

y z

---+-"111--1-+ ...... x

Exercise la Exercises 3a, c

3. (a) Algebraic experience tells us that the graph is a plane. Without worrying too much about proof (one is suggested by Exercise 8 in Section 1.2), we can give a good idea of the graph. First, the intercepts x = 4, y = 2, and z = 1 are evident. Second, the lines joining the intercepts are on the graph. For example, points with x + 2y = 4 in the z = 0 plane necessarily satisfy the equation. Hence the edges of the triangle in the figure are in the solution set. The plane of that triangle is the graph of the equation.

(b) Since (a) is a plane, this is the intersection of two planes. The planes are evidently unequal, and setting Z = 0 gives xy-plane lines of unequal slope. Therefore, the planes have nontrivial intersection; that is, they intersect in a line.

(c) The system's augmented matrix reduces to

1 0 -1 -2 o 2 3

giving z = arbitrary t, Y = 3 - 2t, x = -2 + t. Those equations are the parametric description of a line, shown dashed in the figure.

4. If a, b, and c are all 0, then R2 is the solution set. If a = b = 0 and c i= 0, then the set is empty. Anything else-if either a or b is nonzero-allows us to solve for either x or y and recognize the equation of a line. To put it differently, if a or b is nonzero, then the set is a one-parameter family that we can recognize as a Cartesian line.

5. (a) Same idea as in (4): a = b = c = d = 0 => all solutions; a = b = c = 0 i= d => none; otherwise, solution has two free variables, parametrizes a plane.

(b) If a through h are all zero, then R3 is the set. If either a, b, and care zero with d i= 0, or if e, j, g are zero with h i= 0, then one of the equations has no solutions; the system cannot have any. Otherwise, three possibilities remain. Case 1: One equation is a multiple of the

Solutions to Exercises 157

other; either they both represent the same plane, and that plane is the set, or the "one" says Ox + Oy + Oz = 0, and the plane of the "other" is the solution. Case 2: One of (a, b, c) and (e, j, g) is a multiple of the other, but d and h are not in the same ratio; in this case, the planes are parallel, and the system has no solution. Case 3: Neither of (a, b, c) and (e, j, g) is a multiple of the other, in which case ("one-parameter solution") the planes intersect in a line, irrespective of d and h.

Section 1.2

1. (a) Slope 4 tells us that the line is in the direction of the vector c : = (1, 4), with b := (3,5).

(b) We calculate slope = ~. Then c has to be (2,1) or some multiple. We can take b := (-2, -1),thencheckthat(4,2) = b+3c.Alternatively, set b2 := (4,2); then (-2, -1) = b2 + (-3)c.

(c) The slope points us along c := (1, m), and the intercept locates b := (0, b).

2. (a) Point-slope form, where the slope has to be -1: y - 1 = -2(x + 2). Alternatively, parametrically: x = -2 + 2t, y = 1 - 4t.

(b) y - 1 = (-~) (x + 2), which goes through the origin.

3. (a) Yes and yes. The line joining (1, 2) and (3, 4) consists of vectors of the form (1- a)(l, 2) +a(3, 4). Is (5, 6) one such? Solve (1- a)(1, 2) + a(3,4) = (5,6): a = 2, answering the first question. Then we need not answer the second separately: Since (5, 6) = -1(1,2) + 2(3, 4), we have (1, 2) = 2(3,4)+( -1)(5,6), putting (1,2) on the line joining the other two.

(b) The argument is within (a): c is on the line joining a and b <:} c = (1 - a)a + ab for some a <:} a = -a/(l - a)b + 1/(1 - a)b. (How come a -:p I?) The last says that a is on the line joining band c, because the coefficients add up to 1.

4. For comparison, calculate b - a and c - a for the vectors in Exercises 3a, 5, and 6.

:::} Assume that a, b, and c are on the line d + ((e)); that is, a = d + ae, b = d + {3e, c = d + yeo Then b - a and c - a are mUltiples of e; they are dependent.

~ Assume b - a = ae and c - a = {3e. Then a = a + Oe, b = a + ae, and c = a + (3e are all on the line a + ((e)).

5. Yes and no. They have to be dependent; they are three vectors in R2. For collinearity, Exercise 4 tells us to look at b - a = (2,2) and c - a = (5,3), which are independent.


6. Yes and yes. Tum the vectors into rows in a matrix and do row reduction: first row times -6 added to second, first row times -11 added to third, resulting second row times - 2 added to third, third row is now all zeros. This means thate-lla-2(b-6a) = la-2b+ Ie = 0, and the vectors are dependent. Rewrite the last as e = 2b - la, and we see that e is on the line joining the other two, because the coefficients sum to 1.

7. Algebraically: If three or more vectors are on the line d + ((e)), then they are linear combinations of d and e; three combinations of two vectors are necessarily dependent. Geometrically: If vectors lie along a line, then they are on the plane (or line) determined by that line and the origin; that plane being a two-dimensional subspace, the span of the vectors has at most dimension 2; they are dependent. For the converse, see Exercise 5.

8. Say a f.= O. The solutions are given by y = arbitrary = t,x = cia - (bla)t, which can be written (x, y) = (cia, 0) + t( -bla, 1). (Check that the right side is what the hint refers to.) The right side describes the line (cia, 0) + (((-bla, 1»)).

9. Assume a f.= b and that they are both on the line e + ((d)). Thus, a = e + ad, b = e + f3d. We solve: e = (f3a - ab)/(f3 - a), d = (b - a)/(f3 - a). (Why is f3 f.= a?)

Now, suppose p := ka + lb is on the line joining a and b. We check that p = (k + l)e + (ka + 1f3)d = e + md, which is on the line e + ((d)). (Why is k + I = I?) That shows that "the line joining a and b" is a subset of e + ((d)). Conversely, suppose q := e + rd is on the line e + ((d)). We check that q = (f3 - r)/(f3 - a)a + (-a + r)/(f3 - a)b, a combination that is on the line joining a and b. (Why?) Hence e + ((d)) is a subset of the line joining a and b. This proves equality of the lines.

Section 1.3

1. To minimize the distance between (6, 8) and (x, xI2), minimize instead its square, d2 := (x - 6)2 + (x 12 - 8)2 = 5x2 14 - 20x + 100. By calculus: Its derivative 5x 12 - 20 is zero when x = 8, where its second derivative is positive. By algebra: d 2 = 5([X_~2+16) is least when x = 8.

2. (a) By definition, the angle is cos-1 (ae b/llall2l1bll2) = cos-1 ([ J3J3 + 1( -1) + O]/[ A.J4]) = rr 13.

(b) The (your) sketch should show that the angle between a and -b is the supplement of the angle between a and b. This is borne out by the definition: because a • (-b) = - (a • b), we have (using the values here) cos-1 (-i) = rr - cos- 1 (i) = 2rr 13.

(c) From a + b = (2J3, 0, 0), the calculation is cos-1 (6/4J3) = cos-1 (J312) = rr 16. That is predictable. Vectors a and b, being


equally long, are sides of a rhombus. In the sketch, a + b is the diagonal of the rhombus. Therefore, it bisects the angle between a and b.

3. This is vector-space material.

(a) x. 0 = x • (0 + 0) (vector identity) = x • 0 + x • 0 (Theorem 1.1(b». Subtract x • 0 from both sides and you get x. 0 = O.

(b) Yes. (1,2) • (-2,1) = 0 is an example in R2.

(c) No. If x is orthogonal to all, then in particular, x • x = 0, forcing x = O. (Reason?)

(d) In R 1, yes vacuously; there are no "vectors that are not .... " In dimension > 1, no: given x f= 0 and another vector y that is not a multiple of x, it is impossible for x to be orthogonal to both y and to x + y, because then x • x = x • (x + y) - x • y = o.

4. Angle is -IT {} x. y/llxl1211yll2 = -1 =} x. Y = -llxll2l1yll2 =} one of x and y is a multiple of the other (Cauchy's inequality). Since they are nonzero, we write x = ay. Then a(y • y) = (ay) • y = -liay 11211y 112 = -lalllyll2l1yll2 implies a = -Ial, and a is negative. Conversely, x = ay with a negative gives x • y IIIxll211yll2 = a/lal = -1, making the angle IT.

5. (a) By definition, IIx ± ylll = (x ± y) • (x ± y), which by numerous uses of Theorem 1.1 match x • x ± 2x • y + y • y (all respectively). If x and yare orthogonal, then we have Ilx + ylll = x. x + y. y = IIx - yll{

(b) In the figure, we see that the hypothesis says that x and yare the sides of a rectangle. The equation says that in a rectangle, the diagonals are equally long.

I

y

Exercise 5b Exercise 6b Exercise 8c

Ix-ay I

\

\ ay - proj

6. (a) By multiple uses of Theorem 1.1, we show that (x + y) • (x - y) = x • x - y • y = Ilxlll- lIyll{ If those last two norms match, then x + y and x - yare orthogonal.

(b) In the figure, IIxll2 = IIyll2 says that x and yare the sides of a rhombus. The equation says that in a rhombus, the diagonals are perpendicular. (Compare Exercise 2c.)

7. (a) Symmetry: x. y := X1YI + ... + XnYn = Y1Xl + ... + YnXn = Y. x by commutativity of real multiplication.


(b) Distributivity: (x + y) e Z := (Xl + Yl)ZI + ... + (Xn + Yn)Zn = (XI ZI + X/lZn) + ... + (YIZI + YnZ/l) by several uses of the real-field axioms; the last is x e Z + y e z. Similarly for the other half.

(c) Homogeneity: (ax)ey:= (axI)YI + .. ·+(axn)Yn = a(xIYI + ... + xnYn) = a(x e y). The next-to-Iast quantity also matches XI (aYI) + ... + Xn (aYn) = x e (ay).

(d) Positive definiteness: x ex := XIXI + ... + X/lXn = x? + ... + xn2. Squares being nonnegative, the last is a sum of nonnegative terms, so x e x 2: O. Moreover, if any of the squares is positive, then the sum is at least that positive. Hence if x e X = 0, then all the Xi have to be zero, and x = o.

8. (a) [x-proj(x, y)]ey = xey-proj(x, y)ey := xey-[xey jyey]yey = 0 by parts (b) and (c) of Theorem 1.1.

(b) (See the answer to (5).) IIx - yll} := (x - y) e (x - y) = x e x -2x e y + Y e Y (symmetry, homogeneity). If x is orthogonal to y, then IIx - ylll = x e X + Y e Y = II x lll + lIyllt

(c) See the figure for this solution. Let ay be a multiple of y. Because x - proj(x, y) is perpendicular to y (part (a)), it is perpendicular to all mUltiples ofy, and therefore to ay -proj(x, y). (Why is x -proj(x, y) perpendicular to multiples of y? Why is ay - proj(x, y) a multiple of y?) By the Pythagorean theorem, IIx - aylll = IIx - proj(x, Y)lIl + Ilay - proj(x, Y)lIt Clearly, the shortest length Ilx - aYl12 occurs precisely for ay = proj(x, y).

9. We have seen that IIx - ylll = x e x - 2x e y + Y e y. On the right, x e x and y e yare Ilxlll and lIylll by definition of norm, x ey is Ilxll2 lIy 112 cos() by definition of angle.

Section 1.4

1. (a) lIaIl2:= )(12+22+42) =,J2f, IIbIl2:= )((-3)2+42 +52) =

5.J2, lIa + bll2 := ) (( _2)2 + 62 + 92) = 11.

(b) It suffices to check that the smaller two add up to more than the third: ,J2f + 5.J2 > 4.5 + 7> 11.

(c) By properties of norm, 113al12 = 311al12 = 3,J2f and 113a + 3bll2 = 311a + bl12 = 33.

2. (a) The calculations are direct: lIell2 = J84 = 2.J21, IIdll2 = 3,J2f, lie + dll2 = ,J2f.

(b) Clearly, two of them sum to the third, so the vectors form a degenerate triangle. More specifically: lIell2+ Ile+dll2 = IIdll2 = lI-dll2 = lIe(e + d) 112 forces (Theorem 1.6(a)) e + d = -,Be, and d = (-1 - ,B)e. Indeed, we see d = -1.5e.


(c) In view of (b), we have II 30c+40dll2 = 1130c-60c1l2 = 11-30c1l2 = 30llcll2 = 30.J84.

3. (a) (x, y) has unit maxnorm iff Ixi ~ 1, lyl ~ 1, and one of them equals 1. Hence the "unit circle" is {(x, y): x = ±1 and -1 ~ y ~ 1, or y = ± 1 and -1 ~ x ~ I}. In our usual language, that set is the square with vertices at the four points (±1, ±1).

(b) If (x, y) is on the "circle," then 11(2,0) - (x, y)lIo := max{12 - xl, I y I} = 2 - x. This length is 1 iff x = 1; all the points on the right side of the square fit the description.

4. Clearly, 11(1, 0) - (-1, 0) II 0 = 2, so this is the length we are seeking. We noted that for (x, y) on the "circle," Ix I ~ 1 and Iy I ~ 1. Hence 11(1, 0) -(x, y)lIo := max{11 - xl, Iyll can be 2 iff x = -1. Thus the chord from (1, 0) to any point on the left side of the square is of length 2; no uniqueness.

5. (a) By Theorem l.l(c), x ex 2: 0, with equality iff x = O. The same holds for (x. x)I/2.

(b) lIaxll2 := (ax. ax)I/2 = (aa)1/2(x. x)I/2 (Theorem l.l(c)) =

lalllxll2 (definitions).

6. Thelawofcosinessaysthatllx-ylll = Ilxlll+llylll-2I1xll2l1yll2cose. Since cose 2: -1, we have Ilx - Yill ~ IIxlll + IIYlIl + 211xll211yl12 = (lIxll2 + lIyIl2)2. The inequality follows.

7. Set x := a, y := -b.

(a) => (b): From lIa - bll2 ~ lIall2 + IIbll2, we get IIx + yll2 ~ IIxll2 + 11- yll2 = IlxIl2+llyI12, with equality iffy := -b = -(aa) = (-a)x for an a ~ O.

(b) => (a): From Ilx + yll2 ~ IIxll2 + lIyll2, we get lIa - bib ~ lIall2 + II - bll2 = lIall2 + Ilbll2, with equality iff a := x = f3y = (-f3)b with f3 2: O.

8. (a) Positive definiteness: Ifx = (0, ... , 0), then IIxlio := max{O, ... , O} = o. If instead x =1= 0, then some Xi =1= 0, so IXi I > 0; in this case, IIxlio := max{lxII,···, IXnll2: Ixil > O.

(b) Radial homogeneity: lIaxlio := max{laxII, ... , laxnll = max{lallxll, ... , la Ilxn Il. Because the common factor la I is nonnegative, the biggest IXi I gives the biggest la Ilxi I. Hence

Ilaxlio = lal max{lxII, ... , IXnll = lalllxllo.

(c) Triangle inequality: IIx + Yllo := max{lxI + YII,.··, IXn + Ynll. For each i, we have IXi + Yi I ~ IXi 1+ IYi I ~ IIxlio + lIyllo. (Why?) Hence max{lxi + Yi Il ~ IIxllo + lIyllo.


Section 1.5

1. Clearly, y - x = 52(1, 1, 1) and z - x = 16(1, 1, 1) are dependent, so x, y, and z are on a line (Exercise 4 in Section 1.2). In fact, the sizes make clear that z is between x and y. Therefore, it should work out that d(x, z) + d(z, y) = d(x, y). Verify!

2. (a) It is the perpendicular bisector of their segment: the y-axis. The distances from (x, y) to (±1, 0) are lI(x, y) - (±1, 0)112 = (x ± 1)2 +

y2)1/2. Setting their squares equal, we obtain 2x = -2x, equivalent tox = O.

(b) Not just a line: the y-axis is part of it, but it also contains the vee-shaped region above y = 1 between the lines y = x + 1 and y = -x + 1, plus its mirror image below the x-axis.

To see that, view only quadrant I and its edges. There, the two distances are do «x , y), (±1, 0)) := max{lx =f lI, Iyj} = max{lx =f lI, y}. Clearly, x = 0 makes the two distances match, max{l, y}. For x > 0, Ix + 11 = x + 1 exceeds Ix - 11 = ±(x - 1); therefore, if y :=: x + 1, then the two maxes are both y, and if y < x + 1, then one max is x + 1 and the other is less. We conclude that in this quarter-plane, the required points satisfy y :=: x + 1. By symmetry, the entire set satisfies Iyl :=: Ixl + 1.

3. The three parts of Theorem 1.9 are named after three properties of norms, and those norm properties are the explanations:

(a) d(x, y) := IIx - yll = d(y, x) because norms are symmetric;

(b) d(x, y) := IIx - yll > 0, except when x = y, because norms are positive definite;

(c) d(x, z) := IIx - zll :s d(x, y) + d(y, z) because norms satisfy the triangle inequality.

4. (a) d(ax, ay) := lIax-ayll = lalllx-yil (radial homogeneity of norms) = lald(x, y).

(b) d(x, y) := IIx - yll = II (x + z) - (y + z) II (vector algebra) = d(x + z,y+z).

Section 1.6

1. As x goes from 0 to 1, f decreases from 1 to -1, g increases from 1 to 2, and f - g decreases from 0 to -3. Hence max If I = 1, max Igl = 2, and max If - gl = 3; we have IIf - gllo = IIfllo + IIgllo· But IIf - gll2 = IIfll2 + IIgll2 is impossible. It would violate Theorem 1.6, because f and g are not multiples: f(,J'i/2) = 0 =F g(,J'i/2).


2. The cosines C2 and CI are 2 apart: At x = ~,cos 2rr x = -1 and cos 4rr x = 1; therefore, 2 :s sup IC2 - cII :s sup IC21 + sup ICJ I = 2, and IIc2 - cIllo = 2. Cosine CI and sine SI are ,J2 distant: cos 2rr x - sin 2rr x = ,J2 sin(rr /4 -2rrx), so that llel - sllio := sup ICJ - sll = ,J2.

3. (a) From trigonometry, H = F - 2G.

(b) We cannot have f = ago Algebraic evidence: f has three zeros, but g, being quadratic, cannot have that many; in fact, g ~ i~. Calculus evidence: If f were ag, we would have f' = ag', f" = ag", ... ; but gill is already 0, fill not.

4. Positive definite: IIfllo := sup If I ~ 0, and 111110 = 0 * If(x)1 = 0 for every x * f = O.

Radially homogeneous: one way to proceed is to recall that sup If(x)1 is some value If(b)l. Then laf(x)1 = lallf(x)1 :s lallf(b)1 for all x, with equalityatx = b. Hence sup laf(x)1 = lallf(b)l, which says that II alii 0 = laillfilo.

5. Both parts follow easily from properties of inner products. We have seen that IIx + Yill ± IIx - Yill := (x + y) • (x + y) ± (x - y) • (x - y) = Ilxlll + 2x. Y + lIylll ± Ilxlll =f 2x. y ± lIylll- The upper signs give (a), the lower (b).

6. (a) Yes. The product of integrable functions is integrable.

(b) Yes, yes, no. It is symmetric, because fg = gf. Linear, by properties of integrals. Not positive definite: F(x) := 0 for 0 :s x < 1, := 1 for x = 1, defines an integrable function that is not the zero of the space 1[0, 1], but has F • • F = o.

7. (a) Induction: f(1) = If(I); and f(k) = kf(l) forces f(k + 1) = f(k) + f(l) (additivity) = kf(l) + f(1) = (k + l)f(I).

(b) f(O) = f(O + 0) = f(O) + f(O) forces f(O) = Of (1). Likewise, f(k) + f( -k) = f(k + -k) = 0 (just proved) forces f( -k) = - f(k) = -(kf(I)) = -kf(I).

(c) We may assume m > O. For any s, mf(s) = f(s) + ... + f(s) = f(s + ... + s) = f(ms). Hence mf(k/m) = f(k) = kf(I). Divide by m to get f(k/m) = k/mf(l).

(d) (s-8 argument) Assume that f is continuous at x = b. Choose any s > O. There is a corresponding 8 such that Ix - bl < 8 implies If(x) - f(b)1 < s.

Now suppose Iy - 01 < 8. The point y + b is within 8 of b, so If(y + b) - f(b)1 < s. On the other hand, fey + b) - feb) = fey). We conclude that If (y) - f (0) I = If (y) I < s. We have proved that f is continuous at y = o.


(e) (sequence argument) Assume that f is continuous at x = b. For any sequence (Xi) converging to b, we have f(xi) -+ feb).

Now let Yi -+ c. We see that f(Yi) = f(Yi + [b - cD + fCc -b) -+ feb) + fCc - b), because Yi + [b - c] -+ b. Since the sum feb) + fCc - b) matches fCc), we have shown that f(Yi) -+ fCc), proving that f is continuous at Y = c.

(t) Let z be irrational. By density of the rationals, there is a sequence (qi) of rationals converging to z. By part (e), f is continuous at x = z. Therefore, fez) = lim f(qi) = lim(qi f(l)) (by part (c)) = f(1) lim(qi) = f(1)z.

Chapter 2

Section 2.1

1. Write z := f3x + ay. We discussed in Section 1.2 that z is on the segment from x toy. Thelengthsare IIx-zll = 11(1-f3)x-aYIl = allx-yll (because a ~ 0) and liz - yll = lIf3x + (a - l)YII = lIf3x - (1- a)yll = f3l1x - YII· The ratio is clear.

2. Write d := IIx - yll. The hypothesis YEN says that d < r.

(a) Pick a fixed s ~ r - d. Then r ~ d + s, and we are in the situation of Theorem 2.2(c). Hence N(y, s) £; N.

(b) It suffices to show that N £; N(y, 2r). That zEN implies liz - yll ~ liz - xII + IIx - yll < r +d < 2r, which in turn implies Z E N(y,2r). The inclusion follows.

3. (a) If r + t exceeded d, then the neighborhoods would intersect (Theorem 2.2(b)), and so would the balls. If r + t equaled d, then the point t f(r + t)x + r /(r + t)y would be on both spheres (Verify!), and would therefore be common to the balls. Hence r + t < d.

(b) and (c) Using the triangle inequality, we getd := IIx - yll ~ IIx - zll + liz - wil + IIw - yll ~ r + liz - wll + t. Hence liz - wll ~ d - r - t, and the inequality is strict if either IIx - zll < r or IIw - yll < t.

These parts prove that the least liz - wll might be is d - r - t, and that the only way this minimum can be achieved is if z and ware on the surfaces of the respective balls.

(d) For any point (1-a)x+ay on the line joining x andy, the distance tox IIx - «l-a)x+ay) II = lalllx-Yil is r iff a = ±r /d. That is, theline meets the sphere in just two points, of which u := (1 - r / d)x + r / d y is the one on the segment (a > 0).


(e) Either reasoning as in (d), or observing that the point we want breaks the segment from x to Y into the ratio (d - t) : t (compare Exercise 1), we get the same answer: v := t /d x + (1 - t /d)y.

(f) lJu-vlJ = 11(I-r/d)x+r/dy-t/dx-(I-t/d)ylJ = IJ(I-r/dt/d)(x - y)1J = (1 - rid - t/d)d.

Parts (b) and (c) established the smallest conceivable distance d - r - t. Parts (d)-(f) show that this minimum is actually achieved along the segment, and maybe other places.

(g) Assume that Z and ware on the two spheres. In (b), we saw that d := Ilx - yll :::: IJx - zlJ + IJz - wlJ + IJw - ylJ = r + IJz - wlJ + t. In an inner product space, the inequality is strict, unless z is on the segment from x to wand w is on the segment from z to y (Theorem 1.10). In that case, z = (1 - a)x + aw and w = (1 - (3)z + (3y. Solving those equations for z and w in terms of x and y, we deduce that z and w are on the segment from x to y.

(h) By (g), this situation guarantees that z and ware on the segment from x to y. By (d) and (e), the only such points on the spheres are z = u and w = v.

4. By Exercise 2, some neighborhood N(u, p) is contained in N(x, r) and some N(u, q) S; N(y, t). Then N(u, min{p, q}) fits within both.

Section 2.2

1. (a) Yes. Because i sin(I/ i) = sin(1/ i)/(I/ i) ~ 1 and I cos i / i I :::: I/i ~ 0 as i ~ 00, we should expect (Xi) to converge to (1,0). We verify: IJxi - (1, O)lJl = (i sin(1/i) - 1)2 + cos2 i/i2 ~ O.

(b) No. The terms are (0,1), (-1,0), (0, -1), (1,0), (0, 1), .... Clearly, YI, Y5, Y9,··· converges to (0,1), Y2, Y6, YIO,··· to (-1,0). With unlike subsequences, (Yi) diverges (Theorem 2.4(a)).

(c) No. The terms begin (0, 1), (-2,0), (0, -3), .... Verify that IIzi 112 = i. The sequence is unbounded, so it cannot have a finite limit (Theorem 2.7(c)).

(d) No limit. Terms are (0,0), (8, 8), (0, 0), (32, 32), ... ; the subsequence of odd-numbered terms tends to 0, the evens to infinity.

2. Only (c). In the (c) answer, we said that IJzi II = i, which does approach 00.

In the others: (a) is bounded (it converges), so cannot approach 00; (b) is bounded because each norm is 1; (d) has a subsequence not going to 00.

3. (a) Assume (Xi) ~ x and (Yi) ~ y. Then II(axi + (3Yi) - (ax + (3y)11 = lIa(xi - x) + {3(Yi - y)11 :::: allxi - xii + {3I1Yi - yll ~ 0 + 0, which says that aXi + {3Yi ~ ax + {3y.


(b) No. Take (Xi) to be (1, 1), (2,2), ... and Yi := (-l)ixi . They both converge to 00, but (Xi + Yi) behaves like l(d).

4. (a) Since the difference of norms is at most the norm of the difference (Theorem 1.8), we have I II Xi II - IIx III ::: IIxi - X II. If Xi -+ X, then the right side tends to 0; so must the left.

(b) Xi := (_l)i (1,0) has II xiii -+ 11(1,0) II, but not Xi -+ (1,0). (Why?)

(c) Xi -+ 0 {} Ilxi - 011 -+ 0 {} Ilxi II -+ O.

5. By definition, Xi -+ X means that for any 10 > 0, there exists 1(10) such that i ::: 1(10) implies II Xi - xII < e. Just substitute r for 10, then note that IIxi - xII < r {} Xi E N(x, r).

6. (a) (Xi) is bounded {}thereis M with II xiii < Mforalli {} IIxi-OIi < M for all i {} Xi E N(O, M) for all i {} {Xi} ~ N(O, M).

(b) Xi -+ 00 means IIxi II -+ 00. The latter is defined as follows: For any M > 0, there exists I such that i ::: I implies IIxi II ::: M. Thus, the numbers II Xi II have no upper bound. Hence (Xi) is unbounded.

7. (a) Exercise lb.

(b) Exercise 1d. See also the remarks in answer to Exercise 2.

Section 2.3

1. (a) Converges to (~, 1, ~); verify that the coordinate sequences converge accordingly.

(b) Diverges, because the third coordinates diverge: tan 1i /4 = 1, tan 31i/4 = -1, ....

(c) Converges to infinity, because the norm exceeds the second coordinate ei .

2. (a) (1,0), (2,0), (3,0), .... The distance between any two terms is ::: 1. Therefore, no subsequence (Xj(i» can have IIxj(i) - Xj(k) 112 -+ 0; no subsequence can be Cauchy.

(b) A sequence has the property iff it converges to 00. If (Xi) -+ 00, then IIxi - X j 112 -+ 00 as j -+ 00, no matter how you choose i. Conversely, if (Xi) does not tend to 00, then for some M, there is an infinity of terms with norms below M (by definition). Those constitute a bounded subsequence. By BWT, the last must have in tum a subsequence converging to a finite limit. The latter subsequence must be Cauchy.

3. (a) No. The x-coordinates are all 1,0, or -1. A convergent subsequence cannot have an infinity of two of those, because then it would in tum have unlike subsequences. Hence, eventually the x-coordinates become constantly 1, forcing sublimit (1,0); or they become -1, forcing


sublimit = (-1, 0); or they become 0, and we apply similar reasoning to deduce that y stabilizes at either 1 or -1.

(b) Similar to (a): a convergent subsequence cannot have an infinity of odd-numbered terms and an infinity of evens, so eventually it takes the road to 0 or to 00.

4. (a, c) They converge, so all subsequences converge to the parent limit (Theorem 2.4(a)*, on p. 41). That is the only sublimit. (Compare Exercise 8a.)

(b) The coordinates look like x = sins/s -+ 1 (as s -+ 0), y = (1 -cos s) / s2 -+ 1/2, and z = ± 1. Hence (reasoning as in Exercise 3a) a convergent subsequence must approach (1,1/2,1) or (1,1/2, -1).

5. Yes, under our convention that R 1 is not the same as R. In R 1, Xi := [_I]i i has norm (= absolute value) i. This norm tends to infinity, so (Xi) -+ 00,

the only sublimit.

6. (a) Suppose IJr}(xi)1 -+ 00. Then IIxi III = Jrl (Xi)2 + ... + Jrn (Xi)2 ~ Jr j (Xi)2 -+ 00, and (Xi) -+ 00.

(b) (1,0), (0,2), (3,0), (0,4), ... converges to 00, because IIxi 112 = i; but the two coordinate sequences oscillate.

7. (a) Assume (Xi) -+ x. Then for any £ > 0, there exists I such that i ~ I implies II Xi - xii < £/2. By the triangle inequality, II Xi - X} II = IIXi - X + X - x} II S II Xi - xII + IIx - x} II < £/2 + £/2 for every i, j ~ I. Hence (Xi) is Cauchy.

(b) Use £ : = 1 in the definition of Cauchy. There exists a corresponding I suchthati,j ~ I ~ IIXi-X}1I < 1.SetM:= IlxllI+·"+lIxIII+1. Then clearly, II x} II < M for j S I. For j > I, IIx}1I S IIx} - XIII + IIXIIl < 1 + IIxl II SM. This says that (Xi) is bounded.

8. (a) A sublimit is the limit of some subsequence. If the parent sequence is convergent, then (Theorem 2.4(a)* on p. 41) the limit of any subsequence is the parent limit. Hence the parent limit is the only sublimit.

(b) Assume that B is the unique sublimit of (Xi).

The argument is the same whether B is a vector or 00. Let N be any neighborhood of B. There cannot be infinitely many Xi outside N: Such a multitude would constitute a subsequence (x}(i); by the last paragraph of this section, (X) (i) would in turn have a subsequence (Xk(j(i))) converging to some (possibly infinite) C; C could not be B, because then the sequence (Xk(j(i))) would have to enter N (Theorem 2.5*); C would be a second sublimit of (Xi). Hence only finitely many terms are outside N. Thus, starting with some index I, we have Xl,

Xl+l, ... E N. This being true for arbitrary N, Theorem 2.5* tells us that B is the limit of (Xi). The sequence converges. (Why does this argument require Rn?)


9. => Assume that x is a sublimit of (Xi) and N a neighborhood of x. By definition of sub limit, there exists a subsequence (x j (i)) converging to x. By Theorem 2.5, there is I such that i 2: I => Xj(i) E N. Thus, N contains

Xj(l), Xj(I+I),····

{= Assume that every neighborhood of x has infinitely many terms. Then N(x, 1) has some term Xj(l). Also, N (x,~) has terms Xi with i > j(1), because only finitely many terms are numbered j (1) or less; let x j (2) be

such a term. Also N (x, 1) must have Xj(3) with j (3) > j (2), etc. Thus, we recursively define an increasing sequence j (i) with Ilxj(i) - xii < 1/ i. That is, (Xj(i)) -+ x, and x is a sublimit.

10. => Assume that x (or (0) is a sublimit of (Xi), so that some subsequence (Xj(i)) goes to that value. Let 8 and I be chosen. By Theorem 2.5, there is a starting pointl for which Xj(J), x j(1+I), ... , are all in N (x, 8) (respectively, outside N(O, 2/8)). Then i := J + I satisfies i > I and IIxj(i) - xII < 8 (respectively, IIxj(i) II > 1/8).

{= Mimic the argument in Exercise 9. For finite x, set 8 := 1, I := 1, and pick the corresponding index j (1); set 8 := ~, I := 2 + j (1), and pick

the corresponding index j (2); set 8 := 1, I := 3 + j (2), and pick the corresponding index j(3); etc. Then (Xj(i)) is a subsequence with Ilxj(i) -

x II < 1/ i , making x a sublimit.

Section 2.4

y

1. (a) For each graph, the upper horizontal segment is at y = 1, the lower at y =0.

y

, ,

, ,

y

&------+x (9-----40 x

(b) Let 2k + 1 ~ i ~ 2k+1. We know that Hi 2: 0, and Hi is constantly 1

on an interval of length 1 /2k. Hence fOI Hi 2 2: 1 /2k. On the intervals

to either side, the rise/fall of Hi guarantees that f Hi 2 < f Hi = .5 (1 /2k), and there are no more than two such intervals. Since Hi = 0

the rest of the time, we conclude that (1 + .5 + .5)(1/2k ) > fd H?

(c) Part (b) says that II Hi 112 -+ 0, so Hi -+ O. (Recall Exercise 4c in Section 2.2.)


2. (a) Yes. 0 ::::: Hi ::::: 1, and 1 is a value of Hi, so II Hi 110 := sup I Hi! = 1.

(b) No. It does not have Cauchy subsequences, because any two terms are 1 apart: If i < j, then -1 ::::: Hi - Hj ::::: 1 always; also, in part of the subinterval in which Hi = 1, Hj is either constantly 0 or rising from or falling to 0, so that somewhere Hi = 1 and Hj = 0; therefore, IIHi - Hjllo = 1.

3. Nowhere. The upper horizontals for each group (graphs number 2k + 1 to 2k+1) cover the whole segment from (0, 1) to (1, 1), and the lower horizontals cover (0,0) to (1,0) more than once. Therefore, at each x = a, there exist an infinity of i for which Hi(a) = 1, along with even more j with Hj(a) = O. The sequence (Hi (a» diverges.

4. (a) gl (x) == 0 and g2(X) == 1. If g(a) = 0 and g(b) = 1, then (intermediate value theorem) g = ! someplace. Hence if g is always 0 or 1, then it cannot take on both values.

(b) If h is continuous at x = ~, then limi-HlO h(! - t) = h(!) = limi~oo h(! + t). The values of h cannot then be 0 to the left of

x = ~ and 1 to the right.

5. (a) See the solution to Exercise 7b in Section 2.3.

(b) Assume (Xj(i) --* x, and let s > O. By convergence, there is I such that i ~ I implies Ilxj(i) - xii < s/2; by Cauchyness, there is J such that i, j ~ J implies IIxi - Xj II < s/2. Then i ~ 1+ J, which makes j (i) also ~ 1+ J, forces IIxi - xII ::::: Ilxi - Xj(i) II + IIxj(i) - xII < s. We conclude that (Xi) --* X.

(c) Assume that (Xi) is Cauchy. By (a), it is bounded. By BWT, it has a subsequence converging to a vector. By (b), (Xi) itself must converge to that vector.

Chapter 3

Section 3.1

1. => By definition, f := (fl, ... , fm) is of first degree if fi(X) = ailXI + ... + ainXn + bi for each i. Then

[ fl(X)-fl(Y)] [all

f(x) - fey) = . . . =

fm(x) - fm(Y) ami

~ Iff(x) - fey) = A[x - y] for all vectors, then f(x) - Ax = fey) - Ay = constant b, and we have f(x) = Ax + b. (Compare Example 2.) This says that each row fi (x) is of the form ai I XI + ... + ainXn + bi.


2. (a) g\(x, y) := (1/[x2 + y2], 1/[x2 + y2], 1/[(x - 1)2 + (y - 1)2]) is undefined at precisely (0,0) and (1,1).

(b) g2(X, y) := (x, y, 1/[y - ax - b]) is undefined along y = ax + b.

3. No. You would need f(x) = (P\ (x)/q\ (x), ... , Pm(x)/qm(x)) such that at each place in the square, one of the denominators is zero. Then q(x) := q\ (x) ... qm (x) would be zero throughout the square. Organize q as the hint says: q(x, y) = rk(x)l + ... + r\(x)y + ro(x). For a fixed t E [-1,1], q(t, y) would be a polynomial in y with infinitely many zeros. Hence each coefficient r j (t) would be zero. This being true for an infinity of t, each polynomial rj(x) would have to be identically zero. Hence q(x, y) would be identically zero, and f would have empty domain.

4. ae - bd -:j:. O. The hypothesis says that

h(X'Y)=(~ ~)(~)+(;). Clearly, c and! are irrelevant; h is one-to-one iff the matrix is one-to-one, which occurs iff it has nonzero determinant.

5. Yes. The question becomes whether the system represented by F(x, y, z) = (13, 14, 15) has solutions. Reduce the augmented matrix

[1 2 5 6 9 10

The system is consistent.

6. Leti=(j,k, ... ,p),j=(l,m, ... ,q).

(a) (axi)(,Bxj) := (ax/xl ... x/)(,Bx/ x2m ... xnq ) = (a,B)x/+1 x2k+m p+q '+' ••• Xn = (a,B)xl J.

(b) A polynomial is a sum of terms, and the term ax/ x2k ... x/ is ax(j,k, ... ,p).

(c) g(x, y, z) = x 2y3Z4 + 5x6y 7 Z8 has degree 21. In general, ax i + ... + ,Bxj has degree max {j + k + ... + P, ... , I + m + ... + q }.

Section 3.2

1. (a) Yes. !(Xi) = IIXi III converges to 00 as Xi -+ 00.

(b) Yes, similarly. !(Xi) = IIXi III converges to 0 as Xi -+ O.

2. (a) Yes. If (Xi, Yi) -+ 00, then II (Xi , Yi)1I2 > 2 for i ~ some I. Thereafter, max {x/, Yi 2} must exceed 1, so that g(Xi, Yi) := x i4 + Yi 4 ~ max {xi4, Yi4} > max {x/, Y/} ~ (x i2 + y/)/2 -+ 00.


(b) Yes. If (Xi, Yi) ~ 0, then eventually lI(xi, Yi)1I2 < 1. Thereafter, xi2

and Yi 2 are both ~ 1, so that g(Xi, Yi) := xi4 + Yi 4 ~ x/ + Y/ ~ O.

(c) Yes, because g(Xi, Yi) ~ 0+ (part (b)) forces 1/ g(Xi, Yi) ~ 00.

3. (a) No, because G(i, i) = i2 ~ 00, while G(i, 0) = 0 ~ O.

(b) Yes; 0 ~ IXiyi! ~ (X/+y/)/2 (property of rea Is) = II (Xi, Yi)lIl/2 ~ O.

(c) Yes, under our definition. By (b), (Xi, Yi) ~ 0 implies G(Xi, Yi) ~ O. The last means that 11/ G(Xi , Yi) I ~ 00, which we accept as 1/ G ~ 00, even though the actual values can go to either 00 or -00.

(d) Yes. ,JG is undefined in quadrants II and IV, but if Xi ~ 0 in the domain of,JG, then ,JG(Xi) ~ 0 by part (b).

4. (a) At 0, limit is 00, because JXi Yi ~ 0+, as in Exercise 3d. At 00, no limit; 11 (X, y) ~ 5 along the curve Y = l/x, ~ 00 along Y = l/x3.

(b) At 0, limit is 1: XiYi ~ 0, so h(xi, Yi) acts like sins/s ~ 1. At 00,

no limit: hex, y) = sin 1 along Y = l/x, but hex, y) ~ 0 along Y =x.

(c) AtO,limitis(O,O): IIf3(xi,Yi)lll = (Xi2_Yi2)2+(2xiYi)2 = (x/+

Yi 2)2 ~ 0, so f3 (Xi, Yi) ~ O. The same equation shows that the limit is 00 at 00.

(d) No limit at either place. Near 0, f4(X, y) ~ (0,0) along X = 0, but Ilf4(x, y) III = 2x2/2x2 along Y = x. Near 00, f4(X, y) = (0, 1) along Y = x, but tends to 00 along Y = O.

5. Let c be a closure point of S. By definition, some sequence (Xi) from S has limit c. Given any neighborhood N of c, by Theorem 2.5, there exists I such that XI, X/+ 1, ... are all in N. Thus, XI is a point from S in N.

6. (a) =} Assume that d is an accumulation point of S. The neighborhood N(d, 1) has some Xl =1= d from S, the neighborhood N(d, 1/2) has some X2 =1= d from S, .... We thereby define (Xi) from S with Xi =1= d

and lid-XiII < l/i, so that (Xi) ~ d.

¢= Assume (Xi) ~ d and d =1= Xi E S. Let N be a neighborhood of d. By Theorem 2.5, XI E N for some I. Thus, N has elements of S different from d. Hence d is an accumulation point of S.

(b) =} Assume that d is an accumulation point of S. The neighborhood N (d, ~) has some Xl =1= d from S, the neighborhood N (d, II d - xIII /2) has some X2 =1= d from S, .... For this sequence, each term is less than half as far from d as the previous. Hence the (Xi) are different from each other as well as from d, plus (Xi) ~ d.

Now let N be a neighborhood of d. As in Exercise 5, we find XI,

X/+l, ... , an infinity of members of Sin N.


{= Trivial; the hypothesis is stronger than the definition of accumulation point.

7. (a) Along y = ax, H(x, y) = a2x 2 /x = a2x. Since (x, y) -+ (0,0) forces x -+ 0 (limit of coordinates), we have H -+ 0 along the line.

(b) Along y = JX, H is constantly 1. Hence H -+ Ion the curve; H has no limit at (0,0).

8. (a) No. If Xi := (_I)i Ii, then Xi E R# and Xi -+ 0, but K(Xi) = (-I)i does not have a limit.

(b) Yes. For every member t of R+, It I = t, so K+(t) = 1. Thus, if ti E R+ and ti -+ 0, then K+(ti) = 1 has a limit.

Understand the difference from (a) to (b): Restricting the domain eliminates some testing sequences, exposing the restricted function to fewer tests, and giving it increased chance to qualify. (See the similar comment in Example 1 of Section 3.4.)

Section 3.3

1. We need the result lims-+o s (ln s) = 0 (from L'Hospital's rule, or the more elementary fact that t /21 -+ 0 as t -+ 00). It implies that for any e > 0, there exists /1 > o for which 0 < s < /1 => Is(lns)1 < e.ByTheorems3.7 and 3.3(b), we see that xy -+ 0 as (x, y) -+ (0,0). Hence (Theorem 3.4) given /1, there is 8 such that 0 < II (x, y) 112 < 8 implies Ixy I < /1. We have shown thateleads to8 with II (x, y)1I2 < 8 => Ixyl < /1 => Ixyln(xy)1 < e. By Theorem 3.4, lim(x,y)-+(O,O) xy In(xy) = O.

2. Compare (1): (x, y) -+ (rr, 0) => x -+ rr, y -+ 0 (Theorem 3.7) => xy -+ 0 (limit of product), sinx/[x - rr] -+ -1 and [cosxy - 1]/x2y2 -+ -.5 (L'Hospital) => f(x, y) -+ (-1, -.5).

3. Let L := limx-+b f(x) < O. Set e := -L/2. By Theorem 3.4, there is a neighborhood N(b, 8) in which b f=. xED implies If(x) - LI < e. For such x, L - e < f(x) < L + e = L/2 = -e.

4. (a) By various theorems, lim f(x) = (lim 4 - [x - 1]2, lim 5 - [y - 2]2_ [z - 3]2) = (4,5); by Theorems 3.3, 3.7, and L'Hospital, lim g(y) = 6 + lim([u - 4] In[u - 4]) + lim([v - 5] In[v - 5]) = 6; but g(f(x» = 6 - (x - 1)2ln( -[x - 1]2) + ... is not defined anywhere, because the two logarithms' arguments are ::s o.

(b) g+ matches g in the vicinity of (4,5), so irrespective of the value of g+ at (4,5), limg+ = limg = 6. The composite g+(f) is undefined, except at the place where f = (4,5), which is (1,2,3). Hence (1,2,3) is an isolated point of the domain of g+(f), and lim g+(f) = g+(f(1, 2,3» = 7.


(c) F is polynomial, so its limit is F(I, 2,3) = (4,5); lim G = 6, its value in the vicinity of (4,5); G(F) is always defined, because G and Fare. But G(F) has no limit: along the line x = 1, F has value (4,5), so G (F) = 8; but along the line x = 1 + (, y = 2 + ( , Z = 3 + (, we have F = (4 + (3, 5 + (3), G(F) = 6 for ( ;;6 O.

(d) Ft has a limit equal to its constant value; G(Ft(x)) is always defined because both functions are; and lim G(Ft(x)) is the constant value 8.

5. Assume that b is an accumulation point of D; as usual, the isolated-point case is trivial.

=} Suppose f has finite limit L at b. Then (Theorem 3.4) for any e > 0, there is 8 > 0 such that IIx - bll2 < 8 =} IIf(x) - LI12 < e/2, lIy -bl12 < 8 =} IIf(y) - Lib < e/2 (for x, y ;;6 b in D). For such x and y, Ilf(x) - f(y) 112 .:::: IIf(x) - Lliz + ilL - f(y)112 < e.

<= Suppose (Xi) is a sequence from D converging to and not reaching b. Let e > O. By hypothesis, there exists 8 such that IIf(x) - f(y) 112 < e for x and y from D with 0 < IIx - bl12 < 8,0 < lIy - bl12 < 8. Because (Xi) must be Cauchy, there exists J (e) such that i, j ~ J (e) implies II Xi - X j 112 < 8. Therefore,i,j ~ J(e) =} IIf(Xi)-f(xj)1I2 < e.Thisshowsthat(f(xi))isa Cauchy sequence. Hence this value-sequence converges, proving that f has a limit at b.

(Why was it essential to assume that f maps into Rm?)

6. Assume lim f(x) = u and lim g(x) = v. If (Xi) -+ b, then (af(xi) + .Bg(Xi)) converges to au +.Bv (Theorem 2.3(b)). Hence af +.Bg has a limit, and the limit is au + .Bv.

7. See Example 2 in Section 3.4.

Section 3.4

1. If lIall < 1 (or lIall > 1) and (Xi) -+ a, then (Theorem 2.4(b)) lim IIXi II = lIall < 1 (respectively, > 1). Hence (property of limits) beginning with some J, II Xi II < (liall + 1)/2 (resp. ». Thus, i ~ J =} IIXili < 1 (resp. » =} f(Xi) = 1 (resp. = 0) = !(a). We have shown that (Xi) -+ a implies !(Xi) -+ !(a);! is continuous at a.

Ifinstead Ilbll = 1, thenYi := (1 + 1/ i)bconverges to b with IIYi II = 1+ 1/ i, so !(Yi) = O. Hence !(Yi) does not approach 1 = !(b); ! is discontinuous at b.

2. Fromllly-bll-lIx-blll.:::: lIy-xll(Theorem1.8),weseethatlly-bll-+ IIx - bll as y -+ x. Thus, the limit equals the value.

3. The domain of! is {x: Ilxll .:::: I} U {O}. On this domain, IIxll is continuous (Exercise 2), IIxll - 1 is a continuous linear combination, and their product


is in the domain of the continuous real-variable function h(s) := ,.;s. By Theorem 3.12, f is continuous. In general, if g(x) = G(lIxlD, where G(t) is continuous for t ~ 0, then g is continuous on its domain.

4. The key is Cauchy's inequality: Iyeb-xebl = I(y-x)ebl ::: lIy-xll21lbll2. so that y -+ x forces y e b -+ x e b. Since only properties of inner product spaces are needed, the answer to (b) is yes.

5. (a) => (b) Because bED, b is trivially a closure point of D, so it is legal to discuss limits at b. Assume that f is continuous at b. This means that for any sequence (Xi) from D converging to b, f(Xi) -+ feb). By definition of function limit (Section 3.2), f has a limit at b, and the limit is feb).

(b) => (c) Assume limx~b f(x) = feb).

Ifb isan isolated point of D, then there is some N(b, 8) in which bis the only point of D. Hence xED with IIx - bll < 8 => x = b => IIf(x) - f(b) II = o < e, no matter what e is.

Suppose instead that b is an accumulation point of D. Theorem 3.4 says that given e > 0, there exists 8 such that Ilf(x) - f(b) II < e for all xED with 0< IIx - bll < 8. Clearly, we may drop the condition 0 < Ilx - bll.

(c) => (d) Part (d) is merely a translation into the language of neighborhoods of what part (c) says in terms of inequalities.

(d) => (a) Assume part (d). Suppose (x;) -+ band N(f(b), e) is a neighborhood of feb). By assumption, there is 8 such that any x E N (b, 8) n D has f(x) E N(f(b), e). By convergence, there exists I such that i ~ I => Xi E N(b,8) (Theorem 2.5). Hence i ~ I => f(Xi) E N(f(b), e). This shows (Theorem 2.5 again) that f(x;) -+ feb); f is continuous at b.

6. (a) Assume that f is continuous at b. Then limx~b f(x) = feb) (Theorem 3.9). By Theorem 3.5(a), there are m and 8 with IIf(x) II < m for x =F b in N(b, 8). Then M := m + IIf(b)II + 1 has IIf(x)1I < M for every x in the neighborhood.

(b) The limit off(x) - y is feb) - y (linear combination) =F O. Theorem 3.5(b) guarantees some e1 and neighborhood N in which IIf(x) -yll > e1, except maybe at b. Then IIf(x) -yll > e := min{e1,lIf(b) -YIl/2} throughout N.

(c) Apply Theorem 3.9(c) to e := f(b)/2 (analogously if feb) is negative). Then in some neighborhood ofb,

If(x) - f(b)1 < e, so f(b)/2 = feb) - e < f(x) < feb) + e.

7. (a) BypropertiesofR,thereisanintegerjwithj::: lOs < j+l,andj = lOs is ruled out by the irrationality of s. From j 110 < s < (j + 1) 110, we see that (j - 1)/10 and (j ± 2)/10, (j ± 3)/10, ... are more than 1110 away from s. Hence one or both of j 110, (j + 1) 1 lOis closest to s, at distance 81 := min{s - j 110, (j + 1)/10 - s}.


(b) and (c) Same arguments as (a), beginning from j .::: 100s < j + 1, etc., or j .::: J s < j + 1.

8. (a) If x = (x, y) is on the unit circle, then IIx - (3,0) 112 = ([x - 3f + y2)1/2 > 3 - x unless y = 0, and 3 -x > 2 unless x = 1. Hence x := (1,0) gives the minimum IIx - (3,0)112 = 2, and this is necessarily the infimum.

(b) As in (a), IIx - (3,0) 112 2: 2, but no point of the neighborhood gives equality.However,xi := (1-I/i, O)isinN(O, I),and IIxi-(3, 0)112 = 2 + 1/ i. Hence inf Ilx - (3,0) 112 = 2.

(c) No and yes. First, let b := (1,0). From lib - (1 - I/i, 0)112 = I/i, we judge that deb, N(O, 1» = O. Hence deb, N) = 0 does not imply bEN. Second, let e E S. Then trivially dee, S) := inf{lIe - xII: x is in S} = lie - ell = O.

(d) Assume S = {Xl, ... , xd. Then inf{lIb - xII: XES} = min{lIbxIII, ... , lib - xkll} = lib - xiII for some i. This is 0 iffb = Xi E S.

(e) In part (d), let b := .j2. The distance from b to T := {ijj E Q: 0< j < 101O0} is the same as the distance to the subset S := {i/j E

T: 0 < j < 10100 and j .::: i .::: 2j} (the ones between 1 and 2), because the rest of T is more than 0.4 away from b. Since S is clearly finite and b E S, part (d) tells us that ~ := deb, S) > O. Therefore, i/j E T ::::} li/j - bl 2: deb, S) > .9~.

Section 3.5

1. Because IIG(Xi) - G(b)lIl = (GI(Xi) - GI(b»2 + ... + (Gm(Xi) -Gm (b»2 .::: (Gj(Xi) - Gj(b»2 for each j, we see that G(Xi) -+ G(b) as i -+ 00 iff G j (Xi) -+ G j (b) for each j. It follows that G is continuous iff every G j is continuous. (Compare the argument in Theorem 3.13.)

2. We have seen that additive is subtractive: <l>(x - y) + <l>(y) = <l>(x) implies <l>(x - y) = <l>(x) - <l>(y). If <l> is also of bounded magnification, then II <l>(x) - <l>(y) II = II <l> (x - y)11 .::: Mllx - yll, and y -+ X forces <l>(y) -+ <l> (x); <l> is continuous.

3. K is a linear combination of functions that are separately continuous: Kl (f) := -31 + 1 is of first degree; H(f) := 12 was proved continuous in Co[O, 1] just below Theorem 3.14; and K2(f) := 13 is continuous

because II/i 3 - 1 3 110:= sup lii- Illf?+ l;f+ 1 21.::: llii- 1110(11 Ii 1102+ llii 110111110 + 11/1102) -+ 0 as Ii -+ I·

4. AhomogeneousmaphastomapOtoO:U(20) = 2U(0)forcesU(0) = O. If U has unbounded magnification, then to each i there corresponds Xi with IIU(xi)1I > ilixill·ThenYi :=xi/(iIiXill) (Why is Ilxili :j=.O?)hasYi -+ 0, but IIU(Yi)11 = IIU(xi/[ilixillDll = IIU(xi)/[ilixiIIJII > 1, so that (U(Yi» does not converge to U(O).


5. In any vector space, let b be a fixed nonzero vector. Then rex) := x + b is of first degree, but not additive.

6. We assume that L is linear. (a) => (b) is trivial.

(b) => (c) Assume that L is continuous at b. Then Xi --+ 0 => Xi + b --+ b => L(xi) + L(b) = L(xi + b) --+ L(b) => L(xi) --+ 0 = L(O). That makes L continuous at O.

(c) => (a) Assume that L is continuous at O. For any e, Xi --+ e => Xi - e --+ o => L(xi) - L(e) = L(xi - e) --+ L(O). By Exercise 4, L(O) = O. Thus, Xi --+ e => L(xi) - L(e) --+ 0 => L(xi) --+ L(e).

7. (a) Yes.Infaet,Tiseontraetive:Letmax/(x) = I(a),maxg(x) =g(b); then T(f) - T(g) = I(a) - g(b) .:::: I(a) - g(a) .:::: III - gllo and T(f)-T(g) ~ I(b)-g(b) ~ -1/(b)-g(b)1 ~ -1I/-gllo·Henee IT(f) - T(g)1 .:::: III - gllo·

(b) No. For the functions hk discussed just above Theorem 3.15, hk --+ 0, but T(hk) = hk (~) = 1.

8. No to (c), yes to the others. U is clearly linear, so we look at its magnification. We have IU(f(x»1 = I/(x)llgo(x)1 for every x. Taking sups, we find that IIU(f)lIo .:::: IIgoliollfllo, so that (a) U has bounded magnification on Co. Doing integrals instead, we obtain either

( I ) 1/2

IIU(f) 112:= fo [f(x)go(x)]2 dx

( I )IP .:::: Ilgoili fo l(x)2 dx = II gO 110 II f112,

meaning that (b) U has bounded magnification on C2; or

( I ) 1/2

IIU(f)112':::: 11/110210 g(x)2 dx = IIgo1l211/110,

which says that (d) U is bounded as a mapping from Co to C2. To get a counterexample for (c), look at the functions hk defined above Theorem 3.15. They have IIhkll2 ~ ./3/2k/2 and IIhkllo = 1. Setting go := 1, we obtain IIU(hk) 110 ~ 2k/2/./3l1hkIl2' and U is unbounded from C2 to Co.


1. (a) Yes. I is a composite of continuous functions, so it is continuous on the box. By Theorem 4.3, it attains a maximum.

(b) No. g is not even bounded above: Along the line y = x, the value g(1/ i, 1/ i) is 2 exp(i 2) / i2, which tends to 00 as i --+ 00.


(c) Yes. h is undefined on the two axes, but its limit is 0 at each place there: Near (a, 0), Ixy sin(l/xy)1 :::: Ixyl :::: (Ial + 1)Iyl -+ 0 as (x, y) -+ (a, 0); and similarly near (0, b). Hence h can be extended to a continuous h+, defined throughout the box. This h+ must have a maximum h + (b) in the box. Point b is not on the axes, because h + = 0 on the axes, whereas h+ has such positive values as h+(1, 1) = sin 1. Therefore, b is one of the places where h is defined. Then clearly, h (b) is maximal.

2. (a) The biggest (x2 + y2) 1/2 -such a maximum must occur--comes when Ixl and Iyl are maximal. The box is defined by -3 :::: x :::: 1,3 :::: y :::: 4. Therefore, Ixl < 3, except where x = -3, and Iyl < 4, except where y = 4. The farthest place is (-3,4).

(b) Generalizing (a) in R2: In [a, b], we have al :::: x :::: bl. If 0 :::: aI, then Ix I < bl, except where x = bl; if bl :::: 0, then Ix I < -aI, except wherex = -al; and if al < 0 < hI, then Ixl < max{ -aI, hI}, except where x is the one of aI, bl farther from O. Similarly for y. Thus, the outlying place has coordinates (al or bl, a2 or b2). Those name the four "corners." The argument extends to the 2n corners in Rn.

3. On [0, 00) (the first octant plus its edges):

(a) x 2 + y2 + Z2 is unbounded;

(b) tan- I (xy) sin Z is between -rr /2 and rr /2, but cannot reach either;

(c) x 2 is nonuniformly continuous. Set c := 1. No matter what 8 you name,o := (1/8,0,0) and v := (1/8 + 8/2,0,0) have 110 - vl12 < 8, but (1/8 + 8/2)2 - (1/8)2 = 1 + 82/4> c.

4. (a) For i ~ 2, Xi := a + (b - a)/ i is in (a, b), but Xi -+ a, which is not.

(b) I(x) := 1/lix - al12 is a continuous composite defined in (a, b) and tending to 00 as X -+ a.

(c) g(x) := Ilx - allz can approach 0 but not reach it; and, because the diagonal is the longest segment in the closed box, g can get near to, but not as large as, lib - a1l2.

(d) The function in (b) is nonuniformly continuous, because for the sequence (Xi) in (a), the distances Ilxi - Xj II become small, but the value differences I (Xi) - I (X j) get large.

5. (a) trivial.

(b) If each a j :::: b j and b j :::: a j, then a j = b j for every j, and a = b.

(c) If each a j :::: b j and b j :::: C j, then a j :::: C j for every j, and a :::: c.

6. "a is not to the left of b" denies that every a j :::: b j . Hence it means that some ak exceeds the corresponding bk. Sinceak > bkonoxcanhaveak :::: Xk :::: bk. This last is required for a :::: X :::: b.


Section 4.2

1. (a) Closed and unbounded. Closed because distinct members ofZ2 are.fi or more apart. Hence any Cauchy sequence from Z2 must eventually stabilize: (Xi) -+ X forces X = XI = X/+ I = ... E Z2. Unbounded because (i, i) -+ 00; apply Theorem 4.6(b).

(b) Not closed, because (Iii, Iii) -+ 0 I/. I/Z2. Bounded, because 1I(1lx, Ily)1I2 ::: vItTI.

(c) Closed, because (Xi, 0) -+ (X, y) forces y = 0 (Theorem 2.8). Unbounded, because (i, 0) -+ 00.

(d) This part is deceptive, because the graph "approaches" an asymptote. It is nevertheless closed: If (Xi, exp(Xi» -+ (X, y), then by Theorem 2.8 Xi -+ X and exp(Xi) -+ y; since !(s) := eS is continuous, the last forces y = exp(x), which says that (x, y) is on the graph. The graph is unbounded, because (i, ei ) -+ 00.

(e) Not closed, because (Iii, 1) -+ (0,1), which is not on the graph. Unbounded, because (i, 1) -+ 00.

2. (a) This set T + T is bounded iff T is. If T is bounded, then Ix + yl ::: Ix 1+ Iy I ::: 2 sup Ix I, making T + T bounded. If T is unbounded, then some (Xi) from T tends to 00, and so does (Xi + XI) from T + T.

(b) Same argument, with Ix - yl and sequence (Xi - x[).

(c) Same argument, with xy and sequence (XiXI), where XI is any nonzero term in (Xi). Such terms must exist when (Xi) -+ 00. Note that we are unworried about the sign of the product.

(d) New argument is essential: TIT is bounded iff T is bounded and does not have 0 as accumulation point.

=} If either T is unbounded (some (Xi) -+ (0) or 0 is an accumulation point (some (Yi) -+ 0 with Yi f=. 0), then xi/xI -+ 00 for appropriate lor yt!Yi -+ 00.

<= If there are £ and M such that £ < I Y I < M for those Y f=. 0 in T, then Ix I Y I < M I £for xl YET IT.

3. Xi E (-00, b] means thatJl"[ (Xi) ::: bl , ... , Jrn (Xi) ::: bn . If (Xi) -+ C, then Jrk(Xi) -+ Ck for each k (Theorem 2.8), forcing Ck ::: bk, or C E (-00, b]. Hence (-00, b] is closed. Similarly for [a, (0).

4. (a) Suppose (Xi) is a sequence from S := {ai, ... , ad. Necessarily some member, say ai, is repeated infinitely often. Thus some subsequence (Xj(i) has Xj(i) = ai, and converges to al.lfnow (Xi) -+ X, then X = al E S (Theorem 2.4(a»; S is closed. The converse is false, since Rn is closed and infinite.

(b) If (Xi) comes from B(b, 8) and (Xi) -+ X, then IIx - bll = lim IIXi - bll (Theorem 2.4(b» ::: 8, so that X E B(b, 8). The ball is closed.


(c) Similar argument with lim IIXi - bll = 8.

(d) Suppose that S and T are closed. If(Xi) comes from snT and (Xi) ~ X, then Xi E S forces XES and Xi E T forces X E T, so XES n T; the intersection is closed. If instead (Xi) comes from S U T, then it cannot be that each of Sand T contributes finitely many terms; that is, a subsequence (Xj(i) must come from just one of Sand T. Since (Xj(i)

must converge to X also, we have X in the same one, and XES U T. The union is closed.

5. (a) We have seen (Exercise 2 in Section 3.4) that f(x) := IIx - bll2 is a continuous function on S. By Theorem 4.8, it has a minimum value f(c). Then deb, S) := inf{lIx - bib: XES} = IIc - bll2.

(b) Yes, there must be a maximum fed) = lid - bll2.

(c) G, ~). Using vectors: Suppose x is on the circle. Let e be the angle between x and (3, 4). By the law of cosines, IIx- (3,4) Iii = IIxlli+ II (3, 4)lIi-211xII211 (3,4) 112 cos e = 26-lOcos e. Least value occurs when e = o. Necessarily (Theorem 1.4(d» x = t(3, 4) for a positive t, which then has to be ~.

Using calculus instead: d2 := (3 - cos u)2 + (4 - sin u)2 is minimal when tan u = 1, which describes the closest point (~, ~) and the

farthest (-~, -~).

(d) Your sketch will make the answer obvious; here is an analytical proof. For any x in the ball, let e be the angle between x - a and b - a. By the law of cosines (compare (c», IIx-blii = IIx-alli+llb-alli-211x - all211b - a 112 cose = (lib - all2 -lix - a1l2)2 +211x - a11211baIl2(1- cos e). Clearly, e = 0 is necessary for a minimum, so we need x - a = t(b - a), or x := a + t(b - a).

If b is outside the ball, then lib - all2 > 8 :::: IIx - alb for every x, so the minimum comes with IIx - all2 = 8, forcing t = 8/lib - a1l2. That describes the place a + 8(b - a)/lib - all2 where the segment ab crosses the sphere.

Ifb is in the ball, then obviously we should make IIx-all2 = Ilb-aI12. That gives t = 1 and x = b, which makes sense: b is closest to b.

(e) Pick a fixed c E S. Let T := B(b, lib - C1l2) n S, which includes c. Clearly, every point of S outside T is more than Ilb-cll2 from b. Hence deb, S) := inf{llx - b112: XES} = inf{lly - b112: YET} = deb, T). Since T is closed (Exercises 4b, d) and bounded (contained in a ball), part (a) applies. Thus, there exists dE T with deb, S) = lib - dll2-

There need not be a farthest point: in R2, no point of the x-axis is farthest from (0, 1).


6. {oJ is closed (Exercise 4a), but its set of accumulation points is 0.

7. (a)::::} (b) Suppose every XES satisfies IIxll ::: M. Then clearly, every term of (Xi) from S satisfies the same inequality.

(b) ::::} (c) Assume that no neighborhood of the origin contains S. Then for each i, there exists Xi E S outside of N (0, i). Since IIXi II > i, the sequence (Xi) is unbounded.

(c) ::::} (d) trivial.

(d)::::} (a) Assume S ~ N(b, r). Since N(b, r) ~ N(O, IIbll +r) (Theorem 2.2(c», we have S ~ N(O, IIbll + r). Consequently, every XES satisfies

IIxll < IIbll + r. (a) ::::} (e) Assume IIxlll := x l2 + ... + xn2 ::: M2 for all XES. Then -M ::: Xk ::: M for each k, and X E [(-M, ... , -M), (M, ... , M)]. Hence S is a subset of a box.

(e) ::::} (f) Let S ~[a, b). Then ak ::: Xk = 1Tk(X) ::: bk for each k and every XES. Hence 1Tk(S) := (1Tk(X)} is bounded.

(f) ::::} (a) Assume that 1Tk(S) is bounded, say l1Tk (x) I ::: Mk for all XES. Then x E S::::} x/ + ... +xn2 ::: M/ + ... + Mn2, and S is bounded.

8. part (b) Let M := sup{f(x): XES}, which is finite by part (a). By definition of sup, there is a sequence (f(Xi» ~ M. Because S is bounded, BWT applies, and there is a subsequence (x j (i) converging to some b. S is closed, so b E S. Since (f(Xj(i») must converge to the parent limit, we have M = lim f(xj(i», which is f(b) by continuity. Hence the sup of f is a value f(b), making f(b) the maximum. Similarly for min.

(c) Suppose there were an E > 0 for which no closeness 8 guaranteed function differences less than E. Thus, for each i, there would be Xi

and Yi with IIxi - yilI2 < Iii, yet having IIf(Xi) - f(Yi)ll2 2: E.

From (Xi), find a subsequence (Xj(i» ~ b. Necessarily b E Sand (Yj(i» ~ b. By continuity, f(xj(i» ~ f(b) and f(Yj(i» ~ f(b). But then IIf(xj(i» - f(Yj(i»1I2 ~ 0, contradicting the functiondifference inequality.

9. (a)::::} (b) Assume (a), and let (Xi) be a bounded sequence. If {Xl, X2, ... } is finite, then at least one Xk is repeated infinitely many times in the sequence. That Xk qualifies as a sublimit. If instead {Xl, X2, ... } is an infinite set, then it fits the hypothesis of (a), so it has an accumulation point b. N (b, 1) must have infinitely many members of the set (Exercise 6b in Section 3.2). Call one of them Xj(1), and write rl := lib - Xj(l) 11/2. N(b, rl) must have infinitely many members; among them, there must be Xj(2) with j(2) > j(I). Set r2 := lib - Xj(2) 11/2, and continue the selection process. Then (Xj(i» ~ b, and b is a sub limit.


(b) =} (a) Assume (b), and suppose S is infinite and bounded. By the stated definition, it is possible to find a sequence (Xi) from S of unequal vectors. By Theorem 4.6(b), (Xi) is bounded. By hypothesis, it must have a sublimit e = lim Xk(i). For each neighborhood N(e, 8), there is I such that the terms Xk(l), Xk(l+l),··· are in N(e, 8). These terms being unequal, they constitute an infinity of members of Sin N(e, 8). We conclude thate is an accumulation point of S.

Section 4.3

1. Each function in Example 1 is a composite of xy (polynomial in R2), absolute value, exponential, and inverse tangent (all continuous in R); that makes them continuous.

The rangeofxy is (-00,00). Therefore, in (a), tan-1 xy ranges over (-n /2, n/2), without extremes, and I tan- 1 xyl ranges over [0, n/2). Similarly in (b), eXY has range (0, 00), never hitting 0, and e1xyl has range [1, 00), value = 1 at O.

2. Same words as in the solution to Exercise 8c in Section 4.2, with the conclusions in the third and fourth sentences justified by sequential compactness.

3. (a) Continuity is automatic at isolated points.

(b) You have to have arbitrarily close points. Let T := {(1,0), (~,O), G, 0), ... } ~ R 2 . (1/ i, 0) is isolated, because no other point of T is within 1/ i -1/ (i + 1) of it. f(x, y) := I/x is continuous on T, but not uniformly: Set E := 1, and see that no matter what 8 is, there are points with II(I/i, 0) - (I/j, 0)112 < 8 but If(I/i, 0) - f(I/j, 0)1 >. E.

4. Since IIx - bll is a continuous function ofx E S, it must achieve greatest and least values (Theorem 4.11). Those values occur at the farthest and closest points.

5. Fix X and y. For Z E S, Ilx - zil s IIx - yll + Ily - zll. Hence d(x, S) := inf{lIx - z: z E S} s IIx - yll + inf{lly - zll: z E S} = Ilx - yll + dey, S). Therefore, d(x, S) - dey, S) s IIx - YII, and the absolute value inequality follows by symmetry.

6. By definition ofinf, there are sequences (Xi) from Sand (Yi) from T such that IIXi -Yi II -+ deS, T). Let Sand T be sequentially compact. Then there is a subsequence (Xj(i») -+ a E S and corresponding (Yk(j(i») -+ bET. From IIXk{j(i» - Yk{j(i» II -+ lIa - bll, we judge that deS, T) = Iia - bll (mUltiple uses of Theorem 2.4).

Alternative proof: By Exercise 5, d(x, T) is contractive on S, and therefore continuous. Hence there is a minimum value dCa, T).


By Exercise 4, there is bET with dCa, T) = lIa - bll. Now take any XES, YET. Then IIx - yll ~ d(x, T) ~ dCa, T) = lIa - bll. Of necessity, deS, T) := inf IIx - yll = Iia - bll·

7. (a) If x, Y E B(b, r), then Ilx - yll ~ IIx - bll + lib - yll ~ 2r. Since the points b ± ru, for any unit vector u, are in the ball 2r apart, we conclude that sup Ilx - YII = 2r.

(b) The diameter d is finite, because S has to be bounded (Theorems 4.11 and 4.10). Let IIxi - Yi II -* d. As in Exercise 6, take subsequences (Xj(i)) -* a, (Yk{j(i))) -* b. Then IIxk{j(i)) - Yk{j(i)) II tends to both d and lIa - bll, so these last two must be equal.

8. One example fits all: S := quadrant I in R2, b := 0, T := {OJ.

9. (a) Since d(x, T) is contractive (Exercise 5) on S, it has a minimum value dCa, T). That makes a as close to T as a member of S can get. Also, XES, YET => IIx - yll ~ d(x, T) ~ dCa, T), giving deS, T) ~ dCa, T); while dCa, T) := inf lIa - YII ~ inf IIx - YII (inf of a smaller set is larger) = deS, T).

(b) In (a), we found a with deS, T) = dCa, T). Were this distance 0, there would be (Yi) from T such that lIa - Yi II -* 0; that would make a a closure point of T, forcing a E T, violating S n T = 0. Hence dCa, T) > O.

(c) S:= x-axis and T := {(x, y): y = eX} are both closed (Exercise Ic,d in Section 4.2), but have points in arbitrarily close proximity: (-i, e-i )

on the graph is II ei from (-i, 0) on the x-axis.

(d) S:= {OJ and T := quadrant I in R2.

(e) By (a), there is a with deS, T) = dCa, T). By definition, there is a sequence (Yi) in T with lIa - Yi II -* dCa, T). (Yi) has to be bounded, because IIYi 112 ~ lIall2 + IIYi - a1l2, and the sequence on the right converges. Since we are in Rn, some subsequence (y j(i)) -* bET, and deS, T) = dCa, T) = lib - a1l2. This says that lib - all2 inf IIx - YII2 ~ inf Ilx - bib and so b is closest to S.

Section 4.4

1. (a) Open: If e := (a, b) is in quadrant I, so that a and b are positive, let r := min{a, b}; then N(e, r) s;: quadrant I, because (x, y) E N(e, r) => Ix - ai, Iy - bl both ~ (x - a)2 + (y - b)2 < r2 => x > 0, y > O. Not closed: (11 i, II i) -* 0, which is not in quadrant I.

(b) Not open: (1, 1) is on the graph, but N((l, 1), r) has the point (1,1 + rI2), which is not. Closed: Suppose Xi := (Xi, I/xi) is a sequence from the graph converging to (a, b); then Xi -* a and Ilxi -* b; a cannot be 0, because then 11 I Xi I would approach 00; by continuity


I/Xi -+ l/a; necessarily b = l/a, which says that (a, b) is on the graph.

(c) Not open: (1, 0) (rectangular or polar) is on the graph, but (1 + r, 0) is not for 0 < r < e2:n: - 1. Not closed: Xi := polar (e- i , -i) -+ 0, which is not on the graph.

(d) Not open: (1/2n,0) is on, but (1/2n, r/2) is not. Not closed: (1/ in, 0) -+ (0,0), and no point with x = 0 is on the graph.

2. (Compare the solution to Exercise la. Also, note that (a, 00) is not the complement of (-00, a].)

Assume C E (a, 00), so that a < c. Let r be the least of C! - aI, ... , en - an. Then X E N(c, r) ::::} each IXk - ql < r ::::} each Xk > q - r ~ ak ::::}

x E (a, 00). Hence N (c, r) ~ (a, 00), and the latter is open. Similarly with c E (-00, b).

3. (Compare the solution to Exercise 6 in Section 4.1.) a < b means that each ak < bk; to deny it is to say that some a j ~ b j. In that case, no x can satisfy a j < x j < b j, so no x has a < x < b.

4. For every x E V, N (x, 1) ~ V trivially, so V is open. At the other extreme, "x E 0 ::::} N (x, 1) S; 0" is true vacuously.

5. Only the empty set. If 0 is open and bE 0, then some N(b, 8) ~ 0, and o has all the vectors b + (8/i)b/llbll, i ~ 2.

6. (a) Assume T S; W. In V, x E r- I (T*) {} rex) E T* {} rex) fi T {} x fi r-1 (T) {} x E r- I (T)*. Hence r- I (T*) = r- I (T)*.

(b) r- I (open) = open comes from Theorem 4.18 (whose proof, to be sure, depends on Exercise 9 below). Suppose instead T is closed. Then T* is open (Theorem 4.14, whose proof is Exercise 7). By Theorem 4.18, r- I (T*) is open. By (a), r- I (T*) = r- I (T)*. Hence r- I (T) is closed.

7. ::::} Assume that 0 is open. Suppose (Xi) is a sequence from 0* converging to b. If b were in 0, there would exist N (b, 8) ~ 0, so that (Theorem 2.5) the Xi would eventually get into N (b, 8) and thereby into O. Hence b must be in 0*. We have shown that 0* is closed.

{=: Assume that 0 * is closed. Let b EO. By Theorem 4.7, b is not a closure point of 0*. By definition, there exists N (b, 8) containing no points from 0*. Necessarily N (b, 8) ~ 0, proving that 0 is open.

8. Let bED, c > O. The set 0 := N(r(b), c) is open, so by hypothesis there exists an open P such that D n P is r- I (0). Since b E r- I (0) ~ P, there is N(b, 8) ~ P. If now X E N(b,8) n D, then rex) E O. By Theorem 3.9(d), r is continuous at b.


9. (a) Ifx E OnP,bothopen,thenthereareN(x,r) ~ OandN(x,s) ~ P. Let t := min{r, s}. Clearly, N(x, t) is contained in each of 0 and P, so this neighborhood of x is a subset of 0 n P. Hence 0 n P is open.

(b) Same argument with t := min{Tj, ... , Tk}.

10. By one of DeMorgan's laws, (CI U··· U Ck)* = Cj n··· n Ck. If each Cj is closed, then each Cj is open (Theorem 4.14 = Exercise 7), so their intersection is open (Exercise 9b). From (CI U ... U Ck)* being open, we conclude that the union is closed.

Section 4.5

1. (a) Not connected, therefore not arc-connected. (0,0) is in the set, but not (0, 1), even though y = sin x/x -+ 1 as x -+ O. Accordingly, 01 := N (0, IT /6) and 02 := {x: IIxll2 > IT /6} disconnect the set.

(b) Arc-connected, therefore connected. You would think it impossible to get from (1, 0) (polar () = 0, r = 1) to the origin (polar () = -00, r = 0) along the spiral, since the distance appears to be infinite. Without deciding whether that appearance is reality, we merely remark that you can cover infinite distance in finite time by walking fast. In this case, set() = tant,r = elant (rectangular x = rcos() = elantcos(tant), y = r sin () = elant sin (tan t)) for -IT /2 < t < IT /2. Clearly, the mapping t r-+ (x, y) is continuous, stays on the spiral, covers the whole spiral, and has x -+ 0, y -+ 0 as t -+ -IT /2. Therefore, we can extend it by setting x := 0, y := 0 when t := -IT /2. The resulting extension maps [ -IT /2, IT /2) continuously onto the set, so that we can use it to define an arc from any point to another.

2. (a) Qn does not have the IVP, so it is not connected (Theorem 4.21). For example, Ilxll} is continuous on Qn, has values IIOII} = 0 and

11(2,0, ... , O)II} = 4, but never reaches,J2.

(b) In R I, the irrationals are disconnected by 01 := {x > O} and 02 := {x < OJ. In Rn, the not-all-rationals (complement of Qn) are arcconnected. To illustrate, you travel from (,J2, y, z) to (x, ,J3, u) along

the segments from (,J2, y, z) to (,J2, ,J3, z) to (x, ,J3, z) to (x, ,J3, u), staying in (Qn)*; and from (,J2, y, z) to (J5, v, w) along the seg

ments from (,J2, y, z) to (,J2,,J3, z) to (J5,,J3, z) to (v'S, v, z) to

(v'S, v, w).

3. Projections are continuous (Theorem 3.8, among other places). By Theorem 4.23, the projected image of a connected set is connected. The converse is false: In R 2, the union of the two lines y = x and y = x + 1 is a disconnected set whose projections are connected.


4. 0 is trivially connected, because you cannot nontrivially partition it. V is connected for many reasons; for example, it is convex: If a, b E V, then the segment ab ~ V.

5. => Let 01 and 02 disconnect S. Write Cl := OJ, C2 := O~; these are closed, by Theorem 4.14. Since 01 n Sand 02 n S partition S, we have x E Cl n S {} XES and x ~ 01 {} X E 02 n S. Thus, Cl n S = 02 n S. Similarly C2 n S = 01 n S. Hence {Cl n S, C2 n S} is a partition of S.

{= Same argument with C's interchanged with O's.

6. Suppose S is neither 0 nor V. Then Sand S* are both nonempty, so {S, S*} is a partition of V. Since V is connected (Exercise 4), Sand S* cannot both be open. Hence either S is not open, or S* is not open, in which case S is not closed.

7. Let S be arc-connected. According to Theorem 4.19, we need to show that S has the IVP. Suppose f is continuous on S, a and b E S, and f(a) < Y < f(b). By the arc-connectedness, there exists a continuous function g mapping some interval [r, s] to S with g(r) = a, g(s) = b. Then f(g(t» is continuous on [r, s] and has f(g(r» < y < f(g(s». By the IVT for real functions, there is a place u E [r, s] where f(g(u» = y. Set c := g(u), and we have c E S with f(c) = y. Hence S has the IVP.

8. (a) f(t) := b, 0 S t S 1, defines an arc from b to b within S.

(b) If S is open and bE S, then there is N(b, 8) ~ S. If now x E N(b, 8), then g(t) := (1 - t)x + tb, 0 S t S 1, defines an arc from x to b within N(b, 8) (Theorem 2.1(b», and therefore within S.

(c) See the proof of Theorem 5.5(b).

9. (a) Assume that S is star-shaped and x, YES. By hypothesis, the segments bx and by are subsets of S. As in Exercise 8b, these are arcs from b to x and from b to y. As in 8c, xb is an arc from x to b. Hence we have an arc from x to b and one from b to y. See the argument in Theorem 5.5(b) to establish an arc from x to y.

(b) No; the unit circle in R2 is itself an arc, but contains no line segments.

(c) No; draw a "star." Alternatively, look at the union of the two axes in R2.

10. (a) Let x := alXl + ... + akXk and y := {31Yl + ... + {3jYj E CH(S). The segment xy has vectors (1 - a)x + ay = (1 - a)alxl + ... + (1 - a)akxk + a{31Yl + ... + a{3jYj· In this last combination, the vectors are from S and the coefficients are nonnegative and sum to (1-a)(al + .. ·+ak)+a({31 + .. '+{3j) = 1. Hence (1-a)x+ay E CH(S), proving that CH(S) is convex.


(b) Ifx E S, then Ix E CH(S). Hence S ~ CH(S); the latter is one convex (part (a» superset of S. Assume that T is any convex superset of S. Then T has the convex combinations IXI of a single vector from S. Suppose T includes all combinations of k - 1 or fewer vectors. Then

aIXI + ... + ak-IXk-I x := aIXI + ... + akXk = (1 - ak) + akXk

(1- ak)

is on the segment from one vector in T to another, unless 1 - ak = 0, in which case x = Xk E S. Either way x E T, since T is convex. By induction, T possesses all the members of CH( S). Hence CH( S) ~ T, and the hull is the smallest convex superset of S.

(c) {:::: If S = CH(S), then S is convex by part (a).

:::} If S is convex, then necessarily S is the smallest convex superset of S; by (b), S = CH(S).

(d) The combination IXI consists of both endpoints of a degenerate segment. The combination aIXI + a2x2lies on a segment with one endpoint in S and the other also in S. The combination aIXI + a2X2 + a3x3 = (1 - (3)(al [1 - a3]-I xi + a2[1 - a3]-I x2) + a3x3 lies on a segment with one endpoint X3 in S and the other (al [1 - a3]-I xi + a2[1- a3]-I x2) on a segment with one endpointx2 in S and the other Xl also in S ....

(e) No, no, yes. The convex hull of a neighborhood or ball is the neighborhood or ball, by part (c) and Example 3(a). A hull has to be connected; being convex, it is arc-connected.

Section 4.6

1. (a) Positive-definiteness: Since IIxlli ~ each Ix} I, IIxlli = 0 forces x} = 0 for all j.

Radial homogeneity: lIax 111 := laxil + ... + laxn I = la I (ixil + ... + Ixnl) = lalllxlII.

Subadditivity: IIx+ylli := IXI + YII + ... + IXn + Ynl ~ IXII + IYII + ... + IXnl + IYnl = IIxlli + lIyllI·

(b) Clearly, Ix}1 ~ (x? + ... + Xn2) 1/2 ~ (lIxlI?)1/2. Hence IIxllI ~ nllxII2 ~ nllxlIl. Set m := 1, M := n.

2. (a) Recall (Exercise 1 in Section 1.6) that in C2[0, 1], these sines and cosines are orthogonal and have length I/.J2. Hence II f 112 = (a2 /2 +

p2 /2)1/2. We can find the extremes of f by calculus or trigonometry:

f(x) = (a2 +p2)I/2 sin(2JTx+e),wheree:= ±cos-I (a/[a2 +

p2]1/2) (same sign as (3), yielding sup If I = (a2 + (32)1/2. Accord

ingly, IIfllo = .J211f112.


(b) If (fi) converges uniformly to /, then Ii -+ / in Co[O, 1]. Since V, being two-dimensional, must be a closed subset of Co[O, 1] (Theorem 4.30), we must have / E V.

3. The basic principle here is that the cubic polynomials form a 4-dimensional subspace.

(a) By hypothesis, lllilil -+ 28. Necessarily (II Ii 112) is a bounded sequence of rea Is, so (Theorem 4.29(a» (II/illo) has to be bounded. If, say, II Ii 110 < M, then IIi(x)1 < M for each i and every x.

(b) We just saw that (fi) is a bounded sequence in a 4-dimensional subspace of Co[O, 1]. Hence some subsequence converges (Theorem 4.27(c» in Co[O, 1]; that is, some subsequence converges uniformly.

4. From the inequality, we conclude that (Xi) -+ X relative to II lIa <=> IIXi -xlla -+ 0 (definition) <=> IIXi - Xllb -+ 0 <=> (Xi) -+ X relative to II lib.

5. We use the characterization in Theorem 4.25.

Reflexivity: ll1xlla S IIxlia S ll1xlla, so II lIa is equivalent to itself.

Symmetry: If mllxlla < IIxlib S Mllxlla, then M-Illxllb S IIxlia < m-Illxllb'

Transitivity: If mllxlla < IIxlib S Mllxlla and kllxllb S IIxlic S Kllxllb, then mkllxlla S IIxlic S M Kllxlla.

6. Write IIxll for the norm in V and IIxliB for the "Pythagorean norm," as defined in the proof of Theorem 4.26. By Theorems 4.25 and 4.26, there are m and Mwithmllxll S IIxliB S Mllxll for all x.

(a) :::::} If (Xi) converges to y := alVI + ... + anvn, then for each k, lak-TIk(xi)f S (al-TII(xi»2+. +(an -TIn(Xi»2 = IIY-Xi IIi S M 211y - Xi II~ -+ O. This says that each sequence (TIk(Xi» converges to ak = TIk(Y).

{::: If, conversely, each coordinate sequence (TIk(Xi» converges to 13k. then z := 131 VI + ... + f3nvn has liz -Xi II = II (131 - TIl (Xi»VI + ... + (f3n- TI n(xi»)vn ll S If3l- TI I(Xi)llIvIII+·· ·+lf3n- TIn(Xi)llIvn ll-+ 0, and (Xi) converges to z.

(c) FromITIk(Xi)1 S IIxillB S Mllxill,weconcludethatif(xi)isbounded, then so is each sequence (TIdXi». As in the proof ofBWT (Theorem 2.10), we extract a subsequence (Xj(i» for which all the coordinate sequences (TIk(Xj(i)) converge, and their limits are the coordinates of a finite sublimit for (Xi).


7. Write II lIa and II lib for two norms in V. We now know there are m and M with mllxlla S IIxlib S Mllxlla.

(a) If (Xi) -+ X relative to lilia, meaning that IIx - Xi lIa -+ 0, then IIx - Xi lib S Mllx - Xi lIa -+ 0, and (Xi) -+ X relative to II lib· (Conversely, by symmetry.)

(b) Case 1: b is isolated relative to II Ila. Thus, no member of D has IIx-blia < o.ThennomemberofDhasllx-blib < mo. Consequently, b is isolated relative to II lib, and under both norms, the limit is feb).

Case 2: b is an accumulation point of D relative to II lIa. First, there is a sequence (Xi) from D with 0 < II b - Xi II a -+ 0, and the limit of f is c:= limf(Xi). By part (a), 0 < IIb-xillb -+ 0; that is, some sequence from D converges to b, without reaching b, relative to II lib. Next, let (Yi) be another such sequence. Then also lib - Yi lIa -+ O. Because f has an a-limit, we have f(Yi) -+ c. We have established that for every (Yi), the value sequence f(Yi) has limit c. It follows (Section 3.2) that f has a b-limit matching its a-limit c.

(c) The conclusion says that Xi E D and lib - Xi lib -+ 0 imply IIf(b) -f (Xi) II w -+ O. Its proof is immediate, because lib - Xi lib -+ 0 forces lib - Xi lIa -+ O.

8. (a) Assume that S is a-closed, meaning that S possesses the a-limit of any sequence from S. Thus, Xi E Sand IIx - Xi lIa -+ 0 =} XES. Since IIx-xi lib -+ o forces IIx-xi Iia -+ o (Theorems 4.26 and 4.24), we see that S possesses the b-limit of any sequence from S, and S is b-closed. The conclusion about boundedness follows from IIxllb S Mllxlla.

(b) =} If S ~ V has the EVP, then it is closed and bounded by Theorem 4.10.

{= Assume that S is closed and bounded. Let (Xi) be a sequence from S. It has to be bounded, so by Theorem 4.26(c), there is a convergent subsequence. Because S is closed, the limit of the subsequence is in S. Hence S is sequentially compact. By Theorem 4.11, S has the EVP.

(c) Suppose S is closed and bounded and f is continuous there. By (b), S is sequentially compact. By Exercise 2 in Section 4.3, f is uniformly continuous.

9. Let L: V -+ W, with norms II Ilv and II Ilw. By theorems from linear algebra, the range L(V) is a finite-dimensional subspace of W. Let B := {VI, ... , vn } and C := {WI, ... , wm } be bases for V and L(V). There are scalars CXjk with L(vk) = CXlkWI + ... + CXmkWm for each k. For any X =


th VI + ... + f3n vn , we then have

IIL(x)lI~ :::: M2I1L(x)IIE (by equivalence of norms in L(V))

= M 211f3I(allwl + ... +amIWm)

+ ... + f3n(al nwl + ... + amnWm) liE

:= M2 ([f3!a!! + ... + f3na!n]2 + ... + [f3!aml + ... + f3namnf)

:::: M2([f3? + ... + f3n2] [ad + ... + al;]

+ ... + [f3? + ... + f3n2] [ad + ... + al;])

by Cauchy's inequality. Writing K2 for the sum of all the a /f, we have

IIL(x)lI~ :::: M2[f312 + ... + f3n2]K2 = M2K211xll§ :::: M2K2m2I1xll~,

the last by equivalence of norms in V. This says that L is a bounded linear map from V to W. By Theorem 3.16, L is continuous.

10. (a) =::} (b) If the norms are equivalent, then there are M and m with m IIxlia :::: IIxlib :::: Mllxll a, so that IIxlla/llxllb :::: 11m and Ilxllblllxlla :::: M are both bounded.

(b) =::} (c) If Ilxlla/llxllb :::: m and IIxllblllxlla :::: M, then II M :::: IIxlla/llxllb :::: m says that the fraction is bounded and bounded away from zero.

(c) =::} (d) Assume 0 < 8 :::: IIxlla/llxllb :::: B. Then IlIxlial :::: Bllxllb says that the a-norm is a function of bounded magnification relative to the b-norm; and analogously IlIxllbl :::: (1/8)lIxlla ·

(d) =::} (e) Assume that the a-norm is a function of bounded magnification relative to the b-norm, so that Ilxlia :::: Mllxllb.1f (Xi) -+ X in the b-norm,

then IlIxli a -llxi lIal :::: IIX-Xi lIa (difference of the norms) :::: Mllx-Xi lib -+ O. This says that IIxli a is a continuous function under the b-norm. Similarly for the opposite direction.

(e) =::} (a) Assume that each norm is continuous relative to the other. Since IIxlia is b-continuous at the origin, there exists 8 such that IIx - Ollb < 8 => Illxlia - 1I01la I < 1. Therefore, for every X :I 0, since the vector y := 8x/(2I1xllb) satisfies lIyllb < 8, we know that lIylla < 1, or Ilxli a < (218)lIxlib' Clearly, also 1I01la :::: (218)1I0Iib. Similarly, we find B such that Ilxlib :::: (2IB)lIxll a , and the norms are equivalent (Theorem 4.25).


1. (a) This is just a consequence of continuity. Since the x-value 1TI (f(a))

is 1011T, there is (Theorem 3.1O(c)) a neighborhood N(a, 8) such that 1T! (f(t)) > 5 11T for tEN (a, 8) n [a, b) = [a, a + 8). All these points f(t) are therefore on the curved part.


(b) Considering (a), let c := a +sup{8: rrj (f(t)) > 0 for all t E [a, a +8)}. We show that this c is the first point of the arc on the straight part. First, if aSs < c, then by definition of sup there exists 8 such that s < a + 8 S c and f(t) is on the curved part for f E [a, a + 8); in particular, f(s) is on the curved part. Second, rrj (f(c)) cannot be positive. If it were, then we would have c < b, plus there would be a neighborhood (c - /)., c + /).) in which rrj (f(f)) would stay positive, contradicting the definition of c. Hence rrj (f(c)) = 0; f(c) is on the y-axis.

Alternatively: Put c := inf{t: rrj (f(f)) = OJ. The indicated set is the inverse image of {OJ under rrj (f), so it is closed. Therefore, its inf is its least element. That makes rrj (f(c)) = 0 and rrl (f(t)) > 0 for f < c, answering (a) along with (b).

(c) Let (s, sin lOis), 0 < s S IO/rr, be a point of the curved part up to f(a). Since the x-value rrl(f(t)) is a continuous function of f, with values 10/rr when f = a and 0 when f = c, it must achieve the intermediate value s at some f = c. Since rrl (f(c)) = sand fCc) is on the curve, necessarily fCc) = (s, sin lOis).

(d) Observe that the order has to be a < f5 < f9 < ... < c. That is, rrl (f(t9)) = 20/9rr < 20/2rr = rrl (f(a)), so the intermediate value 2015rr must have been achieved between f = a and f = f9;

that proves a < f5 < f9, and similarly for the other fj+4i. Thus, (tl+4i) is an increasing sequence, converging to some d S c. Since rrj(f(d)) = lim rrl (f(tj+4i)) = lim20/(rr +4rri) = 0, and c is the first place where x = 0, we have d 2: c. We see that (tj +4i) -+ c.

(e) Same argument as (d).

(f) We now have rr2(f(c)) = limi-HlO rr2(f(fl+4i)) = 1 and rr2(f(c)) = limi->oo rr2(f(t3+4i)) = -1.

2. The interval (00, -00) is empty, and by convention inf 0 = 00, sup 0 = -00.

3. (a) See the proof of Theorem 5.5(b).

(b) See the solution to Exercise 8b in Section 4.5.

4. (a) Exercise 8a in Section 4.5 shows that JOINS is reflexive, and Exercise 8c shows that it is symmetric. Exercise 3a in this section shows it to be transitive.

(b) Let la be an equivalence class. Any two x, y E la are related to a, so there are arcs within 0 from a to x and from a to y. Clearly, any point along these arcs is joined to a by part of that arc, so these points are in la. Then we put the arcs together (Exercises 8c in Section 4.5 and 3 in Section 5.1) to give us one arc within la from x to y. We conclude that la is arc-connected.


(c) Assume fa U {b} S; T with b ~ fa. By definition of equivalence class, b is not related to a. That is, no arc within S joins a to b. Hence T is not arc-connected, and fa is maximal.

Section 5.2

1. If t E (0, 1], then 0 < t ~ 1 < 1 + 1/ i for each i, so t is in the intersection. Conversely, if t is in the intersection, then 0 < t < 1 + 1/ i for each i; passing to the limit in the second inequality, we have t ~ 1, and t E (0, 1].

2. (a) Assume x < O. Then (1 - a)( -1,0) + a(x, y) = (a - 1 + ax, ay) has (strictly) negative x-coordinate for every 0 ~ a ~ 1.

(b) Assume x > 0, and suppose g : [r, s] --+ R2 is a continuous function with g(r) = (-1,0), g(s) = (x, y). Then 1T1 (g) is continuous on [r, s], with 1T1 (g(r» = -1, 1T1 (g(s» = x > O. By the intermediate value theorem, there exists C E (r, s) with 1T1 (g(c» = O. Then g(c) is a point of the arc on the y-axis.

3. For x > 0, f = 0 at x = 1,~, t, .... In each of the intervening inter

vals, f must be of one sign. The intervals of positivity are (1, 00), (t, ~), (~, !) , .... Since f is even, we add the corresponding ones on the x < 0 side.

4. (a) The sum of the lengths is t + ~ + ~ + ... = (1) / (1 - ~) = 1.

(b) Let C := [0,1] - O. Since C = [0,1] n 0*, C is closed. C is also infinite, since the components of 0 do not abut (do not share endpoints), and those endpoints are in C. On the other hand, (a) says that in some sense C occupies 0% of the space in [0, 1]. So C is weird, and deserves more study.

5. (a) and (b) In the figure, the shaded portion shows Q2 U Q3 U Q4 near the unit square. It makes clear that each Qi crosses the "main canal" Q2. Accordingly, 0 is arc-connected, and has just one component.

(c) By DeMorgan's law, S - 0 = S n Q~ n Q3 n· ... For each power-oftwo group (2, then 3, 4, then 5, 6, 7,8), its intersection cuts the square(s) from the group before into four times as many, each one-third as big. In symbols, S n Q~ has four squares of size 1 x 1, S n Qi n Q3 n Q:

has sixteen ~ x ~,and so on. Hence for each i, S - 0 is subset of a union of squares with total area 4i /9 i ; 0 occupies 100% of the unit area of S.

6. (a) In the figure, the shaded portion is PI U ... U PI +8+64. The picture suggests that the gasket is a square with an infinity of square holes.

(b) Yes. Since the Pi are disjoint convex open sets, they must be the components of O.


1 Exercise 5 Exercise 6

(c) PI has area (~) 2, P2 through PI +8 all have area (~) 2, and so on. Thus:

1 square of area (~) 2, 8 squares of area (1/32) 2, 64 squares of area

(1/33)2, ... ; total area 1/32 + 8/34 + 64/36 + ... = (1/32)/(1 -8/32) = 1. The holes have all the area.

(d) No. All these constructions behave alike: If some neighborhood N (b, 8) ~ S were free of points of 0, then the area occupied by 0 would be 1 - Jr82 or less.

7. (a) This is trivial: The way we defined "component" (elements joined to a) matches the definition of "equivalence class" (elements related to a).

(b) A set harboring an equivalence relation is always the disjoint union of the equivalence classes, and by (a) the classes are the components.

(c) See the solution to Exercise 4b in Section 5.1.

8. (a) T:= {OJ has only one equivalence class, and it is not open.

(b) Remember S from Example 1 in Section 5.1 and Exercise 1 in that section. Clearly, you can join any point of the straight part to (0,0), and you can join any two points in the curved part. We-decided that you cannot join from the curved part to (0,0). Hence the MACS (equivalence classes) are the curved part and the straight part. But neither is alone a maximal connected subset, because S is connected.

(c) Let U := {(x, y): x is irrational}. Then each vertical line contained in U is an equivalence class under JOINS, and the lines are uncountably numerous.

Section 5.3

1. (a) Let ~ be a class of closed sets and S := nCEl: C their intersection. Then each complement C* is open (Theorem 4.14), the union T :=


UCE'E C* of the complements must be open (Theorem 5.3(a», and T = S* (DeMorgan). Thus, S has open complement, so S is closed.

(b) Same argument with 1: := {Cj, ... , Cd and Theorem 5.3(b).

2. (a) B(O, 1) U B (0,1) U B (0, ~) u··· = B(O, 1), which is closed (Exercise 4b in Section 4.2) but not open (any neighborhood N«I, 0), 8) of (1, 0) has the point (1 + 812, 0) from outside B(O, 1».

(b) B (0, ~) U B (0, ~) U B (0, ~) U ... = N(O, 1) (Explain!), which is open but not closed.

(c) The rings {(x, y): I/(i + 1) ~ x 2 + y2 ~ IIi}, i = 1,2, ... , add up to {(x, y): 0 < x 2 + y2 ~ 1} (Explain!), which is neither open nor closed (Explain, too!).

(d) B(O, 1) U B(O, 2) U B(O, 3) U· .. = R2. By Exercise 6 in Section 4.5, the only open and closed candidates are R2 and 0, and it is impossible to express 0 as the union of unequal sets.

3. (a) Tis not open, because every neighborhood N (0,8) has points (0,812) from outside T. It is not closed, because Xi := ([1 + Ili]i, 0) -+

(e,O) rf. T.

(b) Every set is the union of its MACS (Exercise 7b in Section 5.2). lithe MACS are closed and finitely numerous, then the set is the finite union of closed sets, and is therefore closed.

(c) See the solution to Exercise 8c in Section 5.2. (Why is each line there, like {(J2, Y): y E R}, closed?)

(d) No. MACS are disjoint, so a collection of open MACS is a collection of disjoint open sets. The argument in Theorem 5.7 applies: Every such collection has to be countable.

4. By Example 2(b), if b < c are in C, then there exists z rf. C with b < z < c. Then {C n (-00, z), C n (z, oo)} is a disconnection of C.

5. See it geometrically and analytically. (Consult Example 2.)

For the geometric look, let c E C. Since no two component subintervals from 0 abut, c cannot be the endpoint of subintervals on both sides. Thus, on one side-say the right-every interval (c, c + £) has a point x E C. The interval (c, x) has a point y from 0, because C contains no intervals (Example 2(b». The component of 0 to which y belongs must be a subset of (c, x), because otherwise (if it stretched left of c or right of x) either c or x would be in it. That gives evidence for (c), which implies the others.

Analytically, let c E C have the ternamal #aja2 ... , with each ai = 0 or 2. Then Xi := #aja2 ... ai 111 ... , which is not in C, is less than 3- i from c. Hence Xi E N(c, £) for big enough i, proving (a). The same neighborhood has Yi := #aj a2 ... ai 1000 ... and Zi := #aj a2 ... ai 1222 .... Those are


in C, because the l's are removable. The interval (Yi, Zi) is necessarily a component of 0, because anything in it, having an irremovable 1 in temamal place i + 1, is in O. That proves (b). It also proves (c), because no two Yi match.

6. (a) Map the endpoints Xi, Yi of component Pi (defined in Example 3 in Section 5.2) into 2i - 1, 2i, respectively. The mapping is well-defined, because no component of 0 shares an endpoint with any of its predecessors, and is clearly one-to-one.

(b) Expansions have two properties: Each real r E [0, 1] has a "binimal" r = &hb2 ... := bj/2 + b2/4 + b3/8 + ... , where every bi is 0 or 1; and the binimal is unique if we require that the sequence (bi) not end in 0, 0, .... With that principle in mind, we see that fer) :=

(2bj)/3 + (2b2)/9 + (2b3)/27 + ... = #(2bj)(2b2) ... maps r into a temamal with only O's and 2's. Since two such temamals are equal only if produced by identical sequences (2bi), f maps [0, 1] into C one-to-one.

(c) If C were countable, there would be a one-to-one g: C ~ N. But then g(f) (using f from part (b» would map [0, 1] one-to-one into N. That would contradict the uncountability of [0, 1].

Section 5.4

1. (a) The y-axis has empty interior, because the neighborhood N«O, b), 8) of point (0, b) has the point (8/2, b) noton the axis. Hence all its points are boundary points. There are no other boundary points, because if a =I 0, then N«a, b), lal/2) has no points ofthe axis. Hence bd(y-axis) = y-axis.

(b) Quadrant I is an open set (Exercise 1 in Section 4.4), so it is its own interior. Its boundary is the union ofthe nonnegative axes: Near (0, b), with b 2: 0, we have (8/2, b) from quadrant I and (-8/2, b) from outside it; similarly near (a, 0), with a 2: 0; and if either e < 0 or d < 0, then the smaller of N«c, d), leI) and N«c, d), Idl) is devoid of quadrant I points.

(c) We have seen (Example 2 in Section 3.4, among others) that every neighborhood has points from Q2 and from its complement. That tells us that bd(Q2) is all of R2. Then nothing is left for int(Q2), so this set must be empty.

2. (a) Every ball is closed (Exercise 4b in Section 4.2); by Theorem 5.11(b), a ball equals its closure. This means that its complement is open, so its exterior := interior of complement = complement.

(b) cl(N (b, 8» = B(b, 8) and ext(N (b, 8» is the rest of R2 (outside the ball). The ball B(b, 8) is a closed superset of N(b, 8), so B(b, 8) ~


cl(N(b, 8» (Theorem 5.11(d». By the argument in Example 2(a), the points outside the ball are not closure points of the ball, let alone of N (b, 8), socl(N(b, 8» £ R(b, 8). We conclude both thattheclosureis the ball and that the complement of the ball, being open, is ext( N (b, 8».

(c) Closure = line, exterior = rest. Lines are closed sets: If (Xi, Yi) is a sequence from the line given by ax + by = c and (Xi, Yi) -+ (s, t), then as + bt = lim(axi + bYi) = c, so (s, t) is on the line. Hence the closure is the line, and exterior = interior of (open) complement = complement.

(d) Adapt the argument from the solution to Exercise l(b): The points on the nonnegative axes {(x, 0): X ~ O} and {CO, y): Y ~ O} are boundary points, and the points outside [0,00) are exterior. Consequently cl(quadrant I) = [0,00), and the rest is the exterior.

( e) Since every point is a boundary point of Q2, the closure is R 2 and the exterior is 0.

3. (a) int(int(S» = int(S). The interior is open (Theorem 5.9(c», so it is its own interior (Theorem 5.9(b».

(b) int(ext(S» = ext(S). The exterior is by definition int(S*), which by (a) is its own interior.

(c) ext(bd(S» = int(S) Uext(S). The boundary is closed (Theorem 5.10). Hence its complement is open, is therefore its own interior, is therefore ext(bd(S». As our figures have shown, the complement bd(S)* is int(S) U ext(S).

4. Exercises 4 and 5 are intended to show that the six combinations not covered in Exercise 3 are as we would expect for a neighborhood, but can be strange in general. Let N := N (b, 8) and S := S(b, 8). Then:

(a) int(bd(N» = 0. The boundary is the sphere: Points in N are interior (N is open), points outside the ball are exterior (Exercise 2b), and YES has nearby points Y ± .58 (y - b) /8 from inside and outside N. Those same points are off the sphere, so they tell us that the sphere has no interior points. Hence int(bd(N» = int(S) = 0.

(b) bd(int(N» = bd(N). The interior of N is N, so its boundary is that of N (which is S by the previous paragraph).

(c) bd(bd(N» = bd(N). First, bd(N) = S. Second, we have seen that S has no interior, so every point of S is a boundary point. Last, there are no other boundary points of S, because points of N are surrounded by N and points outside the ball are surrounded by outsiders. Hence bd(S) = S.

(d) bd(ext(N» = bd(N). Points in N are surrounded by N, because N is open. Similarly, points in ext(N) are surrounded by ext(N). That


leaves S, whose points have nearby points from ext(N) (Example 2(a) or part (a) here). Hence bd(ext(N)) = S.

(e) ext(int(N)) = ext(N). int(N) is N, so its exterior is ext(N) (= complement of ball, by Exercise 2a).

(f) ext(ext(N)) = int(N). We know ext(N)* = (complement ofball)* = ball. Therefore, ext(ext(N)) := int(ext(N)*) = int(ball) = N = int(N).

5. (a) and (c) In Example 2(b), wesawthatbd(T) is the annulus consisting of the unit circle, the radius-2 circle, and the points in between. From the arguments in Exercise 4, it is clear that the in-betweens are interior to this set and the two circles form its boundary. Hence (a) int(bd(T)) = {x: 1 < IIxII2 < 2}, which is not empty; and (c) bd(bd(T)) = {x: Ilxti2 = lor 2}, which is only a subset of bd(T).

(b) and (e) Taking from T the part in the boundary annulus, we see that int(T) = N(O, 1). Hence (b) bd(int(T)) = S(O, 1), a subset ofbd(T); and (e) ext(int(T)) = complement of unit ball, a superset of ext(T).

(d) and (f) We have identified int(T) and bd(T), so what is left is exterior: ext(T) = {x: IIxII2 > 2}. Hence (d) bd(ext(T» = radius-2 circle, a subset ofbd(T); and (f) ext(ext(T» = N(O, 2), a superset of int(T).

(In these last three paragraphs, those extra remarks about "subset" and "superset" are not coincidences; check that they are always true.)

6. For (c): We know that C is closed, so cl(C) = C. For the others, in view of Exercise 5 in Section 5.3:

(a) int(C) is empty.

(b) bd(C) = C. Since there are no interior points, C ~ bd(C). Since the definition of boundary implies that bd(C) ~ cl(C), we have bd(C) ~ C.

(d) Set is empty. Every C E C has nearby intervals from C*, an infinity of these intervals ending in Cantor points 1= c. Hence no point of Cis isolated.

(e) Set is C. By (d), every C E C is an accumulation point of C.

7. (c) Suppose (Xi) is a sequencefromcl(S) convergingtox. For any N(x, s), there is I such that Xl , Xl + 1, ... are in N (x, s). Since Xl is in the open set N(x, s), there is a neighborhood N(XI, 8) ~ N(x, s). Because Xl is a closure point of S, some YES must be in N (Xl, 8) (Exercise 5 of Section 3.2). This Y is a member of S inside N(x, s). We have shown that X E cl(S), leading to the conclusion that cl(S) is closed.

(d) By part (a), S ~ cl(S), and by (c), cl(S) is closed. Thus, cl(S) is a closed set containing S.


Suppose T is another such set. If x E cl(S), then by definition there exists (Xi) from S converging to x. Since S S; T, (Xi) comes from T. The latter being closed, (Xi) ~ X forces X E T. We have proved that cl(S) is the smallest closed superset.

8. The key is Theorem 3.1. Using that result, we have X E cl(S)* ¢> X rt cl(S) ¢> there exists N (x, E) devoid of points of S ¢> there exists N (x, E) S; S* ¢> x E int(S*).

9. (a) Apply Theorem 3.1 again: x E cl(S) ¢> every neighborhood ofx has points from S ¢> (every neighborhood has points from both Sand S*) or (every neighborhood has points from S and some neighborhoods have points from only S). The first clause in parentheses says that x E bd(S), the second says that x E int(S), and their disjunction ("or") amounts to x E bd(S) U int(S).

(b) By (a), cl(S) = bd(S) U int(S). We know that int(S) is contained in S. Therefore, cl(S) S; S iff bd(S) S; S. By parts (a) and (b) of Theorem 5.11, bd(S) S; S iff S is closed.

(c) By Theorem 5.9(b), S is open iff every point of S is an interior point. Since every set partitions into interior points and boundary points, we conclude that S is open iff no point of S is in bd(S).

10. (a) Let (Xi) be a sequence of accumulation points of S converging to b. Any neighborhood N (b, E) possesses some XI, X/+ 1, .... If XI = b, then b is already an accumulation point of S. If not, let 8 := min{lIxI - bll, E - IIxI - bJIl. Then (draw the picture) N(XI, 8) S; N(b, E) and does not reach b. Within N (XI, 8), there must exist y =1= XI from S. This y cannot be b, so we have found y =1= b from S in N(b, E). Hence b is an accumulation point; the set of such points is closed.

(b) Partition the space into int(S), bd(S), and ext(S). The points of int(S) are accumulation points of S but not S*; those in ext(S) are accumulators of S* but not of S. In the boundary, there are points of S surrounded by S* , meaning isolated points of S, which are accumulators of S* but not S; points of S* surrounded by S, which are accumulators of S but not S*; and points with neighbors from both Sand S*, which are accumulation points for both, irrespective of which they belong to.

(c) No. S := {( i, 0) , (~, 0) , ... } has only isolated points, and is not closed. (Why?)

11. (a) Yes to all. The closure of the interior of a ball is the ball (equal); cl(int(line» is empty (smaller); cl(int(neighborhood» = ball (bigger); S := N(O, 1) U {(2, O)} S; R2 has int(S) = N(O, 1), cl(int(S» = B(O, 1) (not subset of S, not superset).

(b) Yes again: int( cl(neighborhood» = neighborhood; int( cl(line» = empty; int(cl(Q2» = R2; S .- {x: 0 < IIxli ~ I} has int(cl(S» = {x: Os IIxJJ < I}.


(c) First question: ext(cl(S)) = ext(S) by Theorem 5.12 and Exercise 3b. Second question: cl( ext( S)) is int( ext( S)) U bd( ext( S)) by Exercise 9a. Of the last two, int( ext( S)) is predictable, being ext( S) because ext( S) is open; the other is not, as we see by Exercise 5d. It has to be a subset of bd(S), because it cannot have any of int(S) or ext(S), and it would include any isolated points of S.

12. (a) If S is bounded, say S ~ N (0, M), then B(O, M) is a closed superset of S, so that cl(S) ~ B(O, M) (Theorem 5.11(d)). The closure is therefore bounded.

(b) Assume that S is convex and x, Y E cl(S). By definition of closure, there exist sequences from S with (Xi) ~ X and (Yi) ~ y. If (X and {3 are nonnegative reals summing to 1, then «(xXi + {3Yi) is a sequence from S (because S is convex) with lim «(xXi + {3Yi) = (Xx + {3y. This shows that (Xx + {3y E cl(S), so that the closure is convex.

(c) The result is trivial if S = 0, so assume that S is connected and nonempty. Let {R, T} be a partition of cl(S). Examine R n Sand Tn S. If one of them is empty, say R n S, then S ~ T; that says that R, which must have closure points of S, has closure points of T. If instead both are nonempty, then they partition S. Because S is connected, Theorem 5.13 tells us that one, say R n S, has closure points of the other. This says that R has closure points of T. We have shown that if {R, T} partitions cl(S), then one of R, T holds closure points of the other. By Theorem 5.13, cl(S) is connected.

Section 5.5

1. (a) The hypotheses make S closed and bounded. By the Heine-Borel theorem, a closed bounded subset ofRn is compact. Its continuous image is compact (Theorem 5.19), and this compact image is closed (Theorem 5.15).

(b) No. Any constant function maps open sets into nonopen sets.

(c) First, r-I(T) = B(O, 1) - r-I(T*): X E r-I(T) #- rex) E T #

rex) ~ T* #- X is in the part of B(O, 1) that does not map to T*. Second, r- I (T*) is the intersection of B(O, 1) with some open set 0: T is closed, so T* is open, so r- I (T*) is B(O, 1) n 0 (Theorem 4.17). Therefore, r-I(T) = B(O, 1) - 0 = B(O, 1) n 0*. This last is an intersection of closed sets, so it is closed. Being also a bounded set in Rn , it is compact.

You should see that the compactness of B(O, 1) is essential. Any constant function will have r-I(range) = domain, with "range" compact but not necessarily "domain."

2. (a) By Theorem 5.20, r can be extended continuously to cl(S). The closure of a bounded set is bounded (Exercise 12 in Section 5.4), so cl(S) is


compact. Therefore, r( cl(S)) is compact (Theorem 5.19), and its subset reS) must be bounded.

(b) No. Rn is closed and f(x) := tan-I IIxll2 is uniformly continuous on it (by the mean value theorem), but feRn) = (-T( /2, T( /2).

(c) Yes to both. If S is closed and bounded in Rn , then it is compact, and so must reS) be.

(d) No, yes. See Exercises 1b and 2a, respectively.

(e) Yes. f(x) := 1/(1- Ilxll2) maps N(O, 1) continuously onto [1, 00).

3. Assume that S is compact and T ~ S is closed. Let (Xi) be a sequence from T. The Xi belong also to S, and S is sequentially compact (Theorem 5.15(b)). Hence there is XES with Xi ~ x. We now have a sequence from T converging to x, forcing X E T. We have proved that T is sequentially compact, and by the remarks following Theorem 5.16, T is compact.

4. We have here the hypothesis of Theorem 5.18, plus diameter ~ O. Consequently, we know that TI n T2 n ... is not empty. The only question is whether it can hold two different members. Suppose x, yare in the intersection. Then x, Y E Ii for each i, so that IIx - Y II ::: diam(1i) ~ O. We conclude that X = y.

5. (a) Assume that S is compact. Let n be a family of closed subsets of S, and suppose that n has empty intersection: ncen C = O. By DeMorgan, Ucen C* is the whole space. In particular, S is covered by {C*}, and these sets are open. Hence some finite subcollection Cr, ... , C; suffices to cover S. That is, each XES is in q U ... u C;, so that X E CI n ... n Ck is impossible. Thus, from the assumption that ncen C = 0, we constructed a finite subfamily {CI, ... , Ck} having empty intersection. By contraposition, if n has the finite intersection property, then it has nonempty intersection.

(b) N(O, 1) ~ Rn is not compact, but if n is a family of closed subsets of N (0, 1) with the finite intersection property, then the members of n are also subsets of B(O, 1) and must therefore (part (a)) have nonempty intersection. (If this appears to contradict a theorem from topology, bear in mind that "closed subsets of N (0, I)" means "closed subsets ofRn that happen to be contained in N(O, 1).")

6. (a) LetybeanymemberofS.Recallthatifoy := IIx-yJJ/2,thentheneighborhoods N(x, Oy) and N(y, Oy) are disjoint. The family {N(y, Oy): YES} is clearly an open cover for S. There is therefore a finite subcover; label it N(YI, 01), ... , N(Yk, Ok). If we set 0 := least of 01, ... ,Ok. then 01 := N(x, 0) does not intersect any of the N(Yj, OJ). Consequently, 02 := N (YI, 01) u ... U N (Yk. Ok) is an open set with S ~ 02, 01 n 02 = 0, and X E 01.


(b) Let Z be any member of S. By (a), there exist disjoint open sets 01 (z) and 02(Z) with Z E 01 (z), T ~ 02(Z). The collection {Ol (z)} covers S, so there exists a subcover 01 (ZI), ... , 01 (zm). Each corresponding 02(Zj) contains T, so T ~ 02 := 02(Zj) n··· n 02(Zm). 02, being a subset of every 02(Zj), is disjoint from every 01 (Zj)' Hence 02 is disjoint from their union: 01 := 01 (ZI) U· .. U 01 (zm) does not meet 02, and S ~ 01.

(c) In R2, x := 0, T := {OJ, S := quadrant I offer counterexamples for (a) and (b).

7. We can look back to Exercise 9b in Section 4.3. The hypotheses here give us S sequentially compact, T closed and disjoint from S. Hence 8 :=

deS, T)/2 > O. For any XES and YET, we must have IIx - yll ~ 28. Therefore, N(x, 8) does not reach N(y, 8), and 01 := UXES N(x, 8) does not intersect 02 := UYET N(y, 8). (Why are 01, 02 open?)

8. The distance function d(x, S) is a good start, since it is constantly 0 on S and is at least c := deS, T) > 0 on T (Exercise 9b in Section 4.3). It is also continuous, in fact, contractive (Exercise 5 in Section 4.3). Set f(x) := d(x, S)/c if d(x, S) < c, := 1 if d(x, S) ~ c. Clearly, f = 0 on S, f = 1 on T. The reason f is continuous is that f is everywhere the smaller of d(x, S)/c and 1. Whenever you have continuous functions g and h, the function min{g(x), hex)} is necessarily continuous: min{g(x), hex)} = (g(x) + h(x))/2 -Ig(x) - h(x)I/2, and the expression on the right is a continuous composite.

Section 5.6

1. {:= Suppose V has finite dimension. We know, from considerable experience, that R(O, 1) is closed and bounded. By Theorem 4.29(b), R(O, 1) has the EVP. By Theorem 5.16, it is compact.

=} Suppose V has infinite dimension. By Theorem 5.23, V has a sequence (Ui) of unit vectors separated by distance ~ 1. That separation means that (Ui) has no Cauchy subsequences (same argument used in Theorem 5.24(a)). Therefore (Ui) has no convergent subsequences. Thus, R(O, 1) has a sequence with no sublimits; it is not sequentially compact.

2. Compare the solution to Exercise 3 in Section 4.6.

(a) Let P2 be the set of all quadratic polynomials over the domain [0, 1]. P2 is a vector space of dimension 3. Hence all norms on P2 are equivalent. In particular, IIp(x)11! := Ibl + lei + Idl and IIp(x) 110 := sUP-19:o1 Ip(x)1 are equivalent, so there exists M with IIplll :::: Mllp 110 independent of p. The subset (not subspace) P is precisely the II 110-unit ball: P = {p: IIpllo :::: I}. Hence for any PEP, we have Ibl :::: IIplll :::: Mllpllo:::: M.


(b) Set Pi (x) := (1 - x)i. Clearly, 0 ~ Pi (x) ~ 1 for every i and x, but the second-degree coefficient in Pi (x) is i(i - 1)/2.

3. This is a standard vector-space result. The expression p in brackets is the projection of x onto the subspace spanned by U1, ... , Uk. (See the remarks between Theorems 5.21 and 5.22.)

For any j, (x - p) • Uj = x. Uj - p. Uj = x. Uj - ([x. u11u1 • Uj + ... + [x • Uk]Uk • U j). In the parentheses, each of U1 • U j, ... , Uk • U j is 0, save for Uj • Uj, which is 1. Hence (x - p). Uj = x. Uj - x. Uj = O. Thus, x - P is orthogonal to every U j. By linearity of the dot product, x - p is orthogonal to any combination a1 U1 + ... + akuk.

4. Refer to Section 1.6, especially Example 1.

(a) As we saw, the functions S j := sin 2j JT X are mutually orthogonal

and of length -J2 /2. Therefore {-J2s j } is an orthonormal set. Clearly,

a j = f • -J2s j and !k := a1-J2s1 + ... + ak-J2sk is the projection of f onto the span of -J2s1, ... , -J2sk. By the principles cited between Theorems 5.21 and 5.22, fk is the combination of -J2s1, ... , -J2sk closest to f, making it likewise the closest combination of Sl, ... , Sk.

(b) and (c) This is just the Pythagorean theorem. We have f = (f - fk) + fk. By Exercise 3, f - fk is orthogonal to fk. Hence IIflll = IIf - fklll + II !k III ~ II fkll t The inequality is strict, unless fk actually is f·

5. Write C := B(U2, 1/2) U B(U3, 1/3) U .... Since each Ui has unit norm, the biggest possible norm in C is 1.5, so C is bounded. Also, since the centers are 1 or more apart, the closest two of the component balls can get is ! - t = ~. Let (Xi) be a sequence from C converging to x. Then (Xi) is Cauchy, so there is! such that II Xi - X j II < ~ for any i, j ~ I. Of necessity, all of Xl, X/+1, ... come from a single ball B(uk, 1/ k). Since this ball is closed and (Xi) -+ x, we conclude that x E B(uk, 1/ k) S; C; the latter is closed.

Now define f(y) := 2klly-ukll fory E B(uk, 1/ k),k = 2, 3, .... Then f is continuous in each ball separately, because distance from Uk is continuous. That makes f continuous on C, because the test sequences (those with Yi -+ y) must, as described above, come eventually from a single ball. However, f is not uniformly continuous, because the points [1 + .25/ k]Uk and [1- .75/ k]Uk are "close together" (distance = 1/ k) in C, butthe function difference f([1 - .75/k]uk) - f([1 + .25/k]uk) = 1 is not "small."

6. Let S := {OJ and T := {(4/3)U2, (5/4)U3, (6/5)U4, ... }. S obviously fits the bill. T is closed, because its members are far apart: Each member comes from a different one of the balls in Exercise 5, and those balls are at least ~ from one another; therefore, you cannot make Cauchy sequences from T, except by repetition. But the members of T are at distances 1, i, ... from S; there is no minimum.

References

We have made references to just four books. Here we list them, together with comments on why they are our favorites.

Morris Kline, Mathematical ThoughtfromAncient to Modern Times, Oxford University Press, New York, 1972.

Kline's book is a monumental achievement among histories of the sciences and mathematics. Although our focus on continuous functions restricts our interest in history to Europeans during (roughly) 1810-1870, Kline is a wonderful source of information about the development of mathematical ideas worldwide and over thousands of years.

David C. Lay, Linear Algebra and Its Applications, 2nd edition, Addison Wesley Longman, Reading MA, 1997.

Lay is a good source for the material we have assumed from elementary linear algebra.

Kenneth A. Ross, Elementary Analysis: The Theory of Calculus , Springer-Verlag, New York, 1980.

This is the best introduction to advanced calculus we know. The subtitle suggests that Ross's mission is an axiomatization of elementary calculus. The book carries out that mission with an admirable combination of mathematical rigor and attention to pedagogy, the latter absolutely essential in a student's entry to this level of theory.

204 References

H. L. Royden, Real Analysis, 3rd edition, Macmillan, New York, 1988.

This graduate-level text is best described as "elegant." It may therefore seem like cheating to refer to it in a treatment that claims to be elementary. Royden's first two chapters, however, are introductory, and happen to cover the properties of continuous functions of one real variable.

Index

accumulation point, 62 additive function, 30 angle between vectors, 11 arc (from point to point), 108 arc-connected set, 109

ball, 33 Bolzano-Weierstrass theorem (BWT),

45,47,48,95 boundary of a set, 135 bounded

function, 87 linear mapping, 82 sequence, 41 set, 90

box, 86 infinite box, 89,94 open box, 90

Cantor set, 131 Cantor, Georg, 133 Cauchy's criterion, 44, 48, 49, 72 Cauchy's inequality, 11 closed

interval, 86, 101, 122, 141 set, 91

closure of a set, 136 closure point, 60 compact set, 140 complete normed space, 50 component

of a vector function, 56 of a vector, in a direction, 10 of an open set, 124

composite function (continuity of), 76

connected set, 107, 138 continuous function, 73

continuity of a composite, 76 continuity of a contractive func

tion,79 continuity of a linear combina

tion, product, quotient, 76 continuity on a set, 73

contractive mapping, 44 convergence, 37

to infinity, 39 uniform convergence, 52

convex combination, 110 hull,110

206

convex (continued) set, 108

coordinate projections; See also projection mappings, 42

countable set, 133 Cremonini, Sera, 155

diameter of a set, 100 Dirichlet's function, 74 disconnected set, 107 disconnection, 107 distance

between sets, 100 from point to set, 77

distance, See also metric, 19 dot product, 8

equivalence relation, 118, 123 Euclidean space, 15 exterior of a set, 136, 139 extreme value property (EVP), 96 extreme value theorem, 87

Fernandez, Ted,45 finite intersection property, 148 finite subcover(ing), 141 finite-dimensional subspace, 116 first-degree real or vector function,

56 Fourier coefficients, 153

Gram-Schmidt process, 49,149

Heine-Borel property, 140 Heine-Borel theorem, 143

image of a set, 99, 103 inverse image, 103

indexed family, 141 infinite set, 95, 133 inner product, 8 inner product space, 9, 10 interior of a set, 134 intermediate val ue property (IVP), 106 intermediate value theorem, 88, 107

interval, 122 closed interval, 101, 122 open interval, 101, 122

inverse image, preimage, 103 isolated point, 62, 138

law of cosines, 12 left (a is to the left of b), 86 length,9

limit Pythagorean length, 14

of a composite, 70 of a function, 64

Index

of a linear combination or a product, 65

of a quotient, 69 of a sequence, 38

limit point; See closure point, 60 line, 4

determined by points, 5 line segment from a to b, 6 linear mapping, 82, 84

bounded linear mapping, 82 linear space, 4 linear transformation; See linear map

ping, 82

maximal arc-connected subsets, 123, 129,130

maxnorm, 16,26 metric (as adjective), 119 metric; See also distance, 19 midpoint formula, 6

n-tuple, 1,2 neighborhood, 33

of infinity, 41 nested intervals theorem, 148 nested sets theorem, 145 norm, 13

equivalent norms, 111, 112 mean-norm, 25 standard norm, 15 uniform norm, 52

normed linear space, normed vector space, 13

Index

open ball, 101 box or interval, 101, 122 cover(ing), 141 set, 100

orthogonal vectors, 9

pair, ordered pair, 1 parallelogram identity, 28 partial order, 90 partition, nontrivial partition, 107 perpendicular vectors, 9 persistence of sign, 68, 75 points, 4 polynomial real or vector function,

57 projection, 11

mappings (same as projections), 42

of a vector onto (line of) a vector, 9

of a vector onto a subspace, 149 Pythagorean theorem, 10

Rn,l

rational real or vector function, 57 real function, real-valued function,

55 rectangle, 86 relation, 1 rescaling, 111 restriction (and continuity), 73

207

scalar function, scalar-valued function; See real function, 55

scalar product, 8 sequence, 38

bounded sequence, 41 Cauchy sequence, 44 convergent sequence, 38

sequentially compact set, 97 Siwanowicz, Jan, 98 sphere, 33 star-shaped set, 110 subcover(ing), 141 sublimit, 45 subsequence, 38

ternary (representation of reals), 131 topology, 119 totall y disconnected set, 131 translate, 4 triangle inequality, 14 triple, ordered triple, 1

uncountable set, 133 uniform continuity, 88 uniform convergence, 52

vector function, vector-valued function, 55, 58

vector space, 4

Forthcoming by Alberto Guzman

Derivatives and Integrals of Multivariable Functions ISBN: 0-8176-4274-9

This work examines derivatives and integrals of functions of several real variables. Topics from advanced calculus are covered, including: differentiability and its relation to partial derivatives, directional derivatives and the gradient, surfaces, inverse and implicit functions, integrability and properties of integrals, and the theorems of Fubini, Stokes, and Gauss. The order of topics reflects the order of development in calculus: limits, continuity, derivatives, integrals---a sequencing that allows generalizations from and analogies to elementary results, such as the second-derivative test and the Chain Rule.

Derivatives and Integrals of Multivariable Functions has a definition-theoremproof format, together with a conversational style, including historical comments, an abundance of questions, and discussions about strategy, difficulties, and alternative paths. It is aimed at advanced undergraduate pure mathematics majors whose next course will be real analysis with measure theory. Required background includes theoretical work in linear algebra, one-variable calculus, properties of continuous functions, and related topological material. The last two are used in the context of Euclidean space, but a strong grounding in the corresponding realline topics will suffice.

Table of Contents:

1. Differentiability of Multivariable Functions

2. Derivatives of Scalar Functions

3. Derivatives of Vector Functions

4. Integrability of Multivariable Functions

5. Integrals of Scalar Functions

6. Vector Integrals and the Field Theorems

Bibliography

Index

solutions to exercises - springer978-1-4612-0083... · 2017. 8. 25. · solutions to exercises 157...

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