SECTION 1.1
Note: “iff” stands for “if and only if”
1.
2.If x ∈ A 
(B 
C), then either x ∈ A or x ∈ B C. In either case, x ∈ (A B) (A C). Conversely, let x ∈ (A B) (A C). If x ∈ A, then x ∈ A (B C), so assume x ∉ A. Then x ∈ B and x ∈ C, so x ∈ A (B C).
3.Amounts to a special case of Problem 4.
4.If 
, then x ∈ at least one Ai. If n is the smallest index such that x ∈ An, then 
. Conversely, if x belongs to one of the sets 
, then, in particular, x ∈ An.
5.(A B) (A C) (A D) (B C) (B D) (C D).
6.(A B Cc Dc) (A Bc C Dc) (A Bc Cc D) (Ac B C Dc) (Ac B Cc D) (Ac Bc C D).
SECTION 1.2
1.Form S by specifying that n ∈ S iff n ∉ Sn, n = 1, 2, . .... If S is on the list, then S = some Sno. Then n0 ∈ S iff n0 ∈ Sn0; but by definition of S, n0 ∈ S iff n0 ∉ Sn0, a contradiction.
2.For each n, there are 2n subsets of {1, 2, ... , n} by Theorem 1.2.1. List them as Sn1, Sn2, ... , Sn2n. Do this for each n, and enumerate all finite subsets as in Theorem 1.2.3.
3.Look at the diagonal where x + y = 3 (positions 6, 7, 8, 9 in the diagram). 
is the sum of the first x + y (= 3) integers, which accounts for all previous diagonals (x + y = 0, 1, 2). Then x (= 2x/2) locates the position within the diagonal; e.g., x = 0 yields position 6, x = 1 position 7, x = 2 position 8, x = 3 position 9. To go backwards, say we are given the integer 11. Since 1 + 2 + 3 + 4 = 10 ≤ 11 < 1 + 2 + 3 + 4 + 5, we are on the diagonal with x + y = 4; x = 0 gives position 10, x =1 gives 11. Therefore x = 1, y =4 – 1 = 3.
4.If the positive rationals are listed as a1, a2, ... , then all rationals will be given by 0, a1, –a1, a2, –a2, ... .
5.There is no guarantee that the number r will be rational.
6.Consider two consecutive entries xn and xn+1. Then 
is a rational number with xn < r < xn+1 a contradiction.
7.n = 1:The rationals are countable.
n = 2:Pairs of rationals are countable, by the same diagonal process that we used to count the rationals.
n = 3:The triple (a, b, c) can be regarded as a pair ((a, b), c), so triples of rationals are countable.
Etc. (We are actually using mathematical induction, to be considered formally in Section 1.4.)
SECTION 1.3
1.You may use either the definitions or the intuitive idea that open means “does not contain any of its boundary points.”
(a)closed
(b)open
(c)neither
(d)
open
(e)closed (note that if xn = n then xn → ∞, but ∞ ∉ R)
2.(a) open
(b)closed
(c)closed
(d)neither
(a)
(b)
(d)
3.If x ∈ R, then any open interval Br(x) is ⊆ R, so R is open. If xn ∈ R and xn → x ∈ R, then xn ∈ R so R is closed.
Note: x = ∞ is not legal here; ∞ is not a real number.
The statements “R does not contain any of its boundary points” and “R contains every one of its boundary points” are both true since there are no boundary points.
4.The empty set 
, by Theorem 1.3.2.
Note: The statements “if 
, there is an open interval 
” and “if 
and xn → x ∈ R, then ” are vacuously true or true by default. For example, if you produce an , I will be happy to find an open interval . We will look at this idea more systematically in Section 1.4.
5.If x ∈ E, there is an open ball Br (x) = {y ∈ Ω : |y – x | < r } ⊆ E }. (Note Br(1) = set of all y in Ω such that |y – 1| < r, so for 0 < r < 1, Br(1) = [1, 1 + r).) If xn ∈ E, xn → x ∈ E, then x ∈ E, so E is closed in Ω (as well as closed in R).
6.Let C = {x1, x2, ... , xn}. If x ∉ C, say xi is the point of C closest to x, with r = d(x, xi) > 0. Then y ∈ Br(x) implies y ∉ C, so Br(x) ⊆ Cc. Thus Cc is open; hence C is closed.
7.Assume a ≠ b. If Br1(a) and Br2(b) are disjoint open balls, then xn ∈ both Br1(a) and Br2(b) for all sufficiently large n, which is impossible.
8.Fix ∈ > 0. For sufficiently large n, an and bn are within ∈ of L. Since an ≤ xn ≤ bn, it follows that |xn – L| < ∈ for all sufficiently large n. Therefore xn → L.
SECTION 1.4
1.There are many, many examples. A simple one is if 1 < x < 2, then 1 < x < 3.
2.
3.
Thus, P ↔ Q is T if and only if P and Q have the same truth value.
4.(a) Thie (for every positive real there is a bigger one).
(b)False (there is no smallest positive real).
5.
means
(here, ∈ is a (positive) real number, and N and n are positive integers).
6.(∃∈ > 0)(∀N)(∃n ≥ N)(d(xn, x) ≥ ∈). There is a positive number e such that for every positive integer N, d (xn, x ) ≥ ∈ for some n ≥ N. In other words, no matter how far out we go in the sequence, we can find an element further along whose distance from x is at least e. This expresses the idea that d(xn, x) does not approach 0.
7.Case 1. P holds.
Case 2. Not P holds.
In either case, we have “not P.”
SECTION 1.5
1.Given any Br(x) we have xn ∈ Br(x) for all sufficiently large n. Since xn ∈ E, xn ≠ x, we can conclude that every open ball about x contains a point of E other than x, so x ∈ E′.
2.E′ = Ē = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
3.
.
4.If x1, x2, ... ∈ E′, xn → x, we must show x ∈ E′. Given Br(x), xn ∈ Br(x) for all sufficiently large n. Pick any xn ∈ Br(x) and let B be a ball about xn entirely contained in Br(x) and with x ∈ B. Since xn ∈ Ε′, Β contains a pointy of E, and y ≠ x since x ∉ B. Thus, y ∈ Br(x), y ∈ E, and y ≠ x, so x ∈ E′.
5.Assume x is a limit point of E. Every open ball Br(x) contains a point y ∈ E with y ≠ x. Since E ⊆ Ē, we have y ∈ Ē, so x is a limit point of E. Now assume x is a limit point of E. If Br(x) is an open ball about x, then Br(x) contains a point y ∈ Ē with y ≠ x. Since Ē = E E′, we have y ∈ E or y ∈ E′. If y ∈ E, we are finished, so assume y ∈ E′. Let B be a ball about y entirely contained in Br(x), with x ∉ B. Then B will contain a point z ∈ E.
Thus, we have z ∈ Br(x), z ∈ Ε, z ≠ x. We conclude that x is a limit point of E.
6.Method 1. Let x be a limit point of E′. If Br(x) is an open ball about x, then Br(x ) contains a point y ∈ E′ with y ≠ x. Exactly as in Problem 5 (last four sentences), x is a limit point of E.
Method 2. By Theorem 1.5.1 (with E replaced by E′) the set of limit points of E′ is contained in the closure of E′. But by Problem 4, E′ is closed, so its closure is E′ itself. Thus, the set of limit points of E′ is contained in E′; i.e., any limit point of E′ belongs to E′.
7.(a) E′ = [0, 1] {2}.
(b)2 is a limit point of E but not a limit point of E′.
8.Let x ∈ Ē, x not an interior point of E. Then ∃xn ∈ E with xn → x, so each Br(x) contains a point of E Since x is not an interior point, Br(x) 
E so ∃y ∈ Br(x) with y ∉ E; i.e., y ∈ Ec.
Now assume the condition “x ∈ Ē, x not an interior point of E” is false.
Case 1. x ∉ Ē. Then x ∉ E and x ∉ E’, so some open ball Br(x) contains no point of E, and therefore x cannot be a boundary point.
Case 2. x is an interior point of E. Then some Br(x) ⊆ E, so Br(x) contains no point of £c. Again, x cannot be a boundary point of E.
REVIEW PROBLEMS FOR CHAPTER 1
1.There is one–to–one correspondence between co–finite and finite subsets, so by Section 1.2, Problem 2, there are countably many cofinite subsets.
2.True. Any open ball Br(x) contains a point y ∈ Ē with y ≠ x. If y ∈ E we are finished, so assume y ∈ E′. But then, exactly as in Section 1.5, Problem 5, there is an open ball B about y entirely contained in Br(x), with x ∉ B, and B contains a point z ∈ E. Since x ∉ B, we have z ≠ x, so z ∈ Br(x), z ∈ E, z ≠ x, as desired.
3.Let y ∈ Br(x); we show that if s is sufficiently small then Bs(y) ⊆ Br(x). If z ∈ Bs(y), then d(z, y) < s. By the triangle inequality,
Now d(y, x) < r, so let d(y, x) = r – ∈, ∈ > 0. Then d(z, x) < s + r – ∈, which is less than r if s < ∈ = r – d(x, y).
4.(a) (∃∈ > 0)(∀N)(∃n ≥ N)(∃m ≥ N)(d(xn, xm) ≥ ∈).
(b)Take N = 1 to produce n1, m1 with d(xn1, xm1) ≥ ∈. Then take N > max(n1, m1) to produce n2, m2 with d(xn1, xm1) ≥ ∈. Continuing this process yields the desired sequence.
5.If E is a nonempty finite set, then all points of E are isolated.
SECTION 2.1
1.
.
2.If x ∈ V where V is open, then there is an open ball B(x) such that x ∈ B(x) and B(x) ⊆ V. Since each point of V belongs to at least one of the balls, we have
3.Suppose Vx and Vy are not disjoint (see diagram). If z ∈ Vx Vy, then since z ∈ Vx, z belongs to some open interval I with x ∈ I and I ⊆ V. Since Vy I is not empty (it contains z ), Vy I is an open interval.
Since x ∈ I, we have x ∈ Vy I, so Vy I is one of the open intervals involved in the definition of Vx, so Vy I ⊆ V and therefore Vy ⊆ Vx. A symmetrical argument shows that Vx ⊆ Vy; hence Vx = Vy; i.e., Vx and Vy are identical.
4.The set S is formed by choosing a rational number from each Vx. Since the rationals are countable, the result follows.
SECTION 2.2
1.Only (c) is closed and bounded, hence compact ((a) and (e) are unbounded, (a), (b), and (d) are not closed).
2.Let Gn = (1 – δ, n), n = 1, 2, ... , where δ is any positive number.
3.(a) Let Gn = {(x, ƒ) : x2+y2 < 1 – 1/n}, n = 1, 2, 3, ... , so that the Gn form concentric circles whose radii approach 1. Any point in D belongs to Gn for all sufficiently large n.
(b)Since D itself is open, we may take the covering to consist of D alone.
4.For compactness, every open covering must have a finite subcovering.
5.Yes. If 
, Gi, open, then ⊆ any Gi.
SECTION 2.3
1.(a) 0
(b)0
(c)π/2
(d)e–3
(e)no limit
2.If d(x, y) > k, then x ∉ Bk(y). Choose r small enough so that 
. Since xn → x, we have xn ∈ Br(x) for all sufficiently large n. But then d(xn, y) > k, a contradiction. Alternatively d(x, y) ≤ d(x, xn) + d(xn, y)) by the triangle inequality. Since d(x, xn) → 0 and d(xn, y) ≤ k: for all n, the result follows.
3.Let xn = 1/n, n = 1, 2, ... . (Every subsequence converges to 0 ∉ (0, 1).)
4.If ∞ belongs to the set A of the open covering, then for some r, (r, ∞] ⊆ A. Similarly, if –∞ ∈ B, then for some s, [–∞, s) ⊆ B. But [s, r] is compact, so it can be covered by finitely many sets of the open covering. These, along withal and B, give a finite subcovering of 
.
5.For example, take A = R. By Problem 3 of Section 1.3, R is closed in R. But if xn = n, then xn ∈ R, xn → ∞ ∉ R, so R is not closed .
6.No. Assumed A ⊆ R and A closed in . If xn ∈ A, xn → x ∈ R, then 
, so by assumption, x ∈ A.
SECTION 2.4
1.If b is a lower bound of E, then –b is an upper bound of –E, and conversely. By the first part of the proof, –E has a least upper bound –c. Since –c ≤ –b for every upper bound –b of –E, we have b ≤ c for every lower bound b of E, so c = inf E.
2.See the proof of Theorem 2.4.5 (here the convergent subsequence {xnj} is given by hypothesis rather than the Bolzano-Weierstrass theorem).
3.By Theorem 2.3.1, any sequence in K has a convergent subsequence (with the limit in K). By Problem 2, any Cauchy sequence in K converges to a point of K.
4.Many examples, e.g., Ω1 = {x ∈ R : x ≠ 0}; Ω2 = the set of all rational numbers. Note that if xn = 1/n, n = 1, 2 , ... , then {xn} is a Cauchy sequence of points in Ωι, but does not converge to a point of Ω1.
5.Let xn ∈ K, xn → x; show x ∈ K By hypothesis there is a subsequence xnj. → y ∈ K But since xn → x; , we have xnj; → x ; hence x = y. Thus x ∈ K, proving K closed. If K is unbounded, ∃xn ∈ K with |xn| → ∞. By hypothesis there is a subsequence xnj → y ∈ K. But we also have |xnj| → ∞, a contradiction since a convergent sequence is bounded.
Note; A subsequence of {xn} is still a sequence: y1 = xn1, y2 = xn2 etc.
6.(a) Assume xn < 2. Then 
. Since x1 < 2, we have, by induction, xn < 2 for all n. Assume xn ≥ xn–1. Then 
. Since 
is increasing, again by induction. The result now follows from Theorem 2.4.6.
(b)Let n → ∞ in 
to get 
, which can be solved numerically. (On a calculator, L = 1.8311772.)
7.If ax < ω < bx, then by Theorem 2.4.3 there are numbers y, z such that z < ω < y and (x, y) ⊆ V, (z, x) ⊆ K (hence, (z, y) ⊆ K, because x is assumed in V). But ω; ∈ (z, y), so ω ∈ Vx. Conversely, if ω ∈ Vx, then for some open interval I = (s, r), we have ω ∈ I and x ∈ I ⊆ V (see diagram). Since bx ≥ r and ax ≤ s, we have ax < ω < bx.
8.If x ∈ V (open), then for some open interval I, x ∈ I ⊆ V ; hence x ∈ Vx. Thus 
. But by Section 2.1, Problem 4, there are only countably many distinct Vx, and the result follows.
9.False; e.g., let E be a finite set.
10.False; e.g., let E = [0, 1) (2, 3].
REVIEW PROBLEMS FOR CHAPTER 2
1.[1, ∞) is closed and unbounded, hence not compact.
2.Let I = (x – r, x + r) be an open interval about x. Since x is an upper bound, every y ∈ E is less than or equal to x, hence less than x since x ∉ E. If I contains no point of E, then y ∈ E implies y ≤ x – r, so x – r is an upper bound of E, contradicting the fact that x is the least upper bound of E.
3.Let x = sup E. If I is an open interval containing x, then I contains a point of E (if x ∉ E, use Problem 2), and I also contains points of Ec (since any number greater than x is in Ec).
4.L = 12/(1 + L), so L2 + L – 12 = (L + 4)(L – 3) = 0. By induction, xn > 0 for all n, so L must be 3.
5.By induction, 1 < xn < 2 for all n; also,
so xn + 1 < xn implies xn+2 < xn+1. Thus, {xn} is a bounded, decreasing sequence, so xn → L, where L = 2 – 1/L; i.e., (L – 1)2 = 0. Therefore L = 1.
6.For any x0 we have 
and 
, so
Thus, c is an upper bound of A = {f(x) + g(x) : x ∈ Ω}, and therefore c is at least as big as the least upper bound of A, which is the desired result.
SECTION 3.1
1.a1, b1, c1 d1 a2, b2, c2, d2, a3, b3, c3, d3, ... , where an → –3, bn → 2, cn → 4, dn → 10.
2.Say f(n) ≥ 2 for n ≥ N.
Case 1. xn = 0. Then xn = 0 for all n ≥ N, so xn → 0.
Case 2. xN > 0. Then xN + 1 > 2xN, χN+2 ≥ 4xN, ... , xN + k ≥ 2k xN, so xn → ∞.
Case 3. xn < 0. Then xN + 1 ≤ 2xN, ... , xN+k ≤ 2k xN, so xn → ∞. Thus, xn always converges, and lim sup xn = lim inf xn = 0, ∞, or –∞.
3.If S is the set of subsequential limits of {xn}, the set of subsequential limits of {cxn} is cS = {cx : x ∈ S}. Thus, the largest sub-sequential limit of {cxn} is c lim sup xn, and the smallest subsequential limit of {cxn} is c lim inf xn.
4.Proceed as in Problem 3, noting that since c is negative, the largest subsequential limit of {cxn}; i.e., the largest element of the form cx, x ∈ S, is c times the smallest element of S, i.e., c lim inf xn. Similarly, the smallest subsequential limit of {cxn} is c lim sup xn.
SECTION 3.2
1.If lim inf yn < z, then by Theorem 3.2.2(d), yn < z i.o.; hence, xn < z i.o. Therefore some subsequence of {xn} has a limit ≤ z, so lim inf xn ≤ z. Since z is any number greater than lim inf yn, we have lim inf xn < lim inf yn. If lim sup yn < z, then by Theorem 3.2.2(a), yn < z ev.; hence xn < z ev. Therefore, all subsequential limits of {xn} are ≤ z, so lim sup xn ≤ z. As above, conclude that lim sup xn ≤ lim sup yn.
2.If z > lim sup xn, w > lim sup yn, then by Theorem 3.2.2(a), xn < z and yn < w ev., so xn + yn < z +w ev. Thus, lim sup (xn + yn) ≤ z + w. (If, e.g., lim sup yn = ∞, take w = ∞.) Since z and w are arbitrary, as long as z > lim sup xn w > lim sup yn, we have lim sup(xn + yn) ≤ lim sup xn + lim sup yn. (Here we let z → lim sup xn, w → lim sup yn, and since +∞ and –∞ do not both appear on the right-hand side, z + w → lim sup xn + lim sup yn.) The lim inf case is done exactly as above, with all inequalities reversed.
3.
Therefore, the quadratic polynomial on the right has either no real roots or at worst a real repeated root: so “b2 – 4ac” is ≤ 0; i.e.,
4.
Without Problem 3: If 
, then 
1/n, then 
. Since 
and 
converges by comparison.
5.
By Problem 3,
so |x + y| ≤ |x| + |y|.
Now
which proves the triangle inequality. The other properties of a metric are immediate.
6.“xn > 3 eventually” can be expressed as
The negation is
i.e., for every n, there is a k > n with xk ≤ 3, which says that xn ≤ 3 infinitely often.
SECTION 3.3
1.Since 
and 
∞, the limit involved in the ratio test does not exist. But
Thus, a = lim sup 
, and the root test gives convergence.
2.Use 1/r = lim sup|cn|1/n.
(a)(nk)1/n → 1 since (k In n)/n → 0, so r = 1.
(b)(3n/n5)1/n → 3, so 
.
3.lim sup |cn|1/n ≥ 3, since at least one subsequential limit is ≥ 3. Thus 
.
4.
5.Since the series converges but not absolutely,
Given any real number r, add enough positive terms to get a partial sum > r, then add enough negative terms to drive the sum < r, then enough positive terms to get the sum > r, etc. Since the nth term → 0, the “overshoot” or “undershoot” → 0, so the rearranged series → r. To achieve + ∞, add positive terms until the sum is > 1, then one negative term, then positive terms until the sum is > 2, then one negative term, etc. Divergence to – ∞ is achieved similarly. For oscillation, add positive terms until the sum is > 1, then negative terms until the sum is < –1, then positive terms until the sum is > 1, etc.
6.(a)
Let m → ∞ to obtain the desired result.
(b)
by part (a) and the identity 
Thus,
(c)Without loss of generality assume r = 1 (consider 
. Given ∈ > 0, choose N as in part (b), and then choose δ > 0 so that
Then x > 1 – δ ⇒ |f(x) – s| < ∈.
REVIEW PROBLEMS FOR CHAPTER 3
1.(a) 1, 0, 1, 0, 1, 0, ...
(b)
2.
.
3.False; e.g., 
.
4.(a) False; e.g., 5, 0, 5, 0, 5, 0, . ...
(b)True (Theorem 3.2.2(b)).
(c)True (Theorem 3.2.2(c)).
(d)False; e.g., xn = 4 + 1/n.
5.1, −1, 2, −2, 3, −3, ....
6.
.
1.(a) |xn + yn – (x + y)| ≤ |xn – x| + |yn → y| → 0.
(b)|xn – yn – (x – y)| ≤ |xn – x| + |y → yn| → 0.
(c)
.
Since a convergent sequence is bounded, for some M > 0 we have |yn| ≤ M for all n; hence, the right-hand side → 0.
(d)This follows from (c) if we prove that 1/yn → 1/y. Now
and since yn → y ≠ 0 we have 
for large enough n. Thus,
2.This is immediate from Problem 1. For example, if xn → x then, by continuity, f(xn) → ƒ (x), g(xn) → g(x). By Problem 1, f(xn) + g(xn) → f(x) + g(x), so f + g continuous at x. The difference, product, and quotient are handled similarly.
3.A polynomial is built from constant functions and the identity function (I(x) = x) using sums and products. By Problem 2, it suffices to show that if h(x) = c and I(x) = x, then h and I are continuous. But if xn → x, then h(xn) = c → c = h(x) and ƒ(xn) = xn → x = I(x), as desired.
4.The statement is not true in general. If 
, then y = f(x) for some 
, so y ∈ f(Ai) for every i, and therefore 
But the inclusion may be proper. For example, let Ω = {1, 2, 3}, Ω′ = {1, 2}. If f(1) = 1, f(2) = 2, f(3) = 1, A1 = {1, 2}, A2 = {2, 3}, then
Note that ƒ(A Β) = f(A) f(B) if f is one-to-one (if z = ƒ(x) = f(y), x ∈ A, y ∈ B, then x = y).
5.If ƒ : Ω → Ω′ is continuous on Ω and A is a closed subset of Ω′, then ƒ –1(A) is closed (see Theorem 4.1.6). In the given case, A = {c}.
1.Yes, with
2.By Theorem 4.2.5, ƒ can be extended to a (uniformly) continuous function on Ē. Since Ē is closed and bounded, it is compact. By Theorem 4.2.1, ƒ (Ē) is compact and therefore closed and bounded. But ƒ(E) ⊆ ƒ(Ē), so ƒ(E) is bounded.
3.Given ∈ > 0 there exists δ > 0 such that if x, y ∈ E and |x – y| < δ, then |ƒ(x) – ƒ(y)| < ∈. Since E is bounded (so that Ē is compact), it can be covered by finitely many (say M) open balls of diameter < δ. But then ƒ(E) is covered by M open balls of diameter < ∈.
4.xn → x means (∀∈ > 0)(∃N)(∀n ≥ N)xn ∈ B∈(x), so xn 
x means (∃∈ > 0)(∀N)(∃n ≥ N)xn ∉ B∈(x). Thus, for some ∈ > 0, 
for infinitely many n. By Theorem 2.3.1, {xn} has a subsequence converging to a limit y ∈ K. Since d(xn, x) ≥ ∈ on the entire subsequence, we have d(y , x) ≥ ∈, so y ≠ x. (If d(y, x) < ∈, then d(xn, x) ≤ d(xn, y) + d(y, x) < ∈ for large n, a contradiction.)
5.Let ƒ (xn) → ƒ(x). By Problem 4, if xn x there is a subsequence xni → y ∈ K with y ≠ x. But by continuity, ƒ(xn) → f(y); hence f(x) = ƒ(y). Since ƒ is one-to-one, x = y, a contradiction. Conclude that xn → x.
6.Take A = R, ƒ (x) = ex; then f(A) = (0, ∞), which is not closed.
SECTION 4.3
1.Look at a picture of the values of ƒ (see diagram), ƒ is increasing, but is discontinuous on the entire positive x and y axes.
2.The only possible discontinuity is at x = 0. Now |x sin(1/x)| = |x||sin(1/x)| ≤ |x| → 0 as x → 0. Thus, f(0+) = f(0–) = 0, and ƒ (0) = 1. Therefore, ƒ has a removable discontinuity at 0.
3.Proceed as in Theorem 4.1.2. If the ∈ — δ condition fails, there is an ∈ > 0 such that for any δ > 0, there is a point t within distance δ of x, with t ≠ x, such that d(f(t), A) > ∈. If we take δ = 1/n, n = 1, 2, ... , and label the corresponding t as tn, then tn ≠ x, tn → x, but ƒ(tn) A. Now if the ∈ – δ condition holds, let tn ≠ x, tn → x. Given ∈ > 0, choose δ > 0 so that t ≠ x, d(t, x) < δ implies d(f(t), A) < ∈. Since tn → x, we have d(tn, x) < δ ev. so d(f(tn), A) < ∈ ev. It follows that f(tn) → A.
4.Assume ƒ continuous at x . Given ∈ > 0, ∃δ > 0 such that d (x, y) < (δ ⇒, d(f(x), f(y)) < ∈. In particular, x < y < x + δ or x – δ < y < x ⇒ |f(x) – f(y)| < ∈. Thus, ƒ(x+)=ƒ (x–) = ƒ(x). Conversely, if ƒ(x+) = ƒ (x–) = ƒ (x), then given ∈ > 0, ∃δl, δ2 > 0 such that
Take δ = mm(δ1, δ2) > 0 and note that when y = x, |f(x) → f(y)| = 0 < ∈. Then |x –y| < δ ⇒ |f(x) → f(y)| < ∈, so ƒ is continuous at x.
1.If A = {y}, then d(x, A) is simply d(x, y). The result follows from the discussion in Section 4.4.2.
2.At step n of the Cantor construction, the set removed consists of the union of 2n–1 intervals, each of length 1/3n (so the length of the set removed is 
, and 
. We may modify the construction so that given α, 0 < α < 1, we remove the union of 2n–1 intervals, each of length 
. Then the length of the set removed is a 
, and 
. The resulting Cantor like set has length 1 – α, and it can be shown that all properties in Theorem 4.4.1 (except (b)) still hold. (At Step 1, remove one interval of length α/2; length 1 – α/2 remains. Step 2: remove two intervals, each of length α/8; OK since α/4 < 1 – α/2; length 1 – α/2 – α/4 remains. Step 3: remove four intervals, each of length α/32; OK since α/8 < 1 – α/2 – a/4; length 1 – α/2 – α/4 – α/8 remains; etc.)
3.Choose y1, y2, ... ∈ A with d(x, yn) → d(x, A). By compactness, there is a subsequence {ynk} converging to a limit y0 ∈ A, so by continuity of distance, d(x, ynk) → d(x, y0). But since d(x, yn) → d (x, A), we have d(x, ynk) → d(x, A) also. By uniqueness of limits, d(x, y0) = d(x, A).
4.Let y be any point of A. Then y ∈ some closed ball Cr(x), and d(x, A) = d(x, A Cr(x)), since points of A outside of Cr(x) will be further away from x than y. Since A Cr(x) is compact, the result follows from Problem 3.
REVIEW PROBLEMS FOR CHAPTER 4
1.
.
2.x = 0, A = (0, ∞).
3.ƒ is continuous at x = 0 (and discontinuous everywhere else). For if xn → 0, then |f(xn)| ≤ |xn|, so ƒ(xn) → 0.
4.By Theorem 4.2.5, ƒ has a continuous extension to the compact set [0, 7]. The result follows from Corollary 4.2.2.
5.This follows from the Intermediate Value Theorem, since f(−1) < 0 and f(0) > 0.
6.No. If so, ƒ would have an extension to a continuous real-valued function g on [0, 1]. But ln x → –∞ as x → 0+, so g(0) = –∞, contradicting the fact that g is real-valued.
7.Let ƒ (x) = ex2 . If V is the complement of the set of integers, then A = ƒ –1(V). Since V is open and ƒ is continuous, A is open.
8.
.
9.Take C = {0}; then f –1(C) = (−∞, 0), which is not closed.
10.There are infinite discontinuities at x = 0 and x = 1 (see diagram).
SECTION 5.1
1.The only possible difficulty is at x = 0. But
since |h sin(1/h)| = |h|| sin(1/h)| ≤ |h|. Thus ƒ′(0) = 0.
2.
Thus, ƒ′ = 0, so ƒ is constant.
3.If p(x) = p(y) = 0 and x < y, then by Rolle’s Theorem, p′(z) = 0 for some z ∈ (x, y). Thus, if p has n distinct real roots, then p′ has n – 1 distinct real roots. Since the degree of p′ is (deg p) –1, there are no other roots of p′.
If p has a repeated root of order k at x0, then write p(x) = (x – x0)k q(x), where q(x0) ≠ 0. But then p′(x) = (x – x0)k q′(x) + k(x – x0)k–1q(x), so p′ has a root of order k – 1 at x0. Thus, k – 1 repetitions of the root of p correspond to a root of p′ of order k – 1. So in any case, if p has n real roots counting multiplicity, then p′ has n – 1 real roots counting multiplicity.
4.Let f(x) =x3, g(x) = x2, on [–1, 1 + δ]. Then
But
Thus, if δ > 0 is sufficiently small, [ƒ (1 + δ) – ƒ(–1)]/[g(1 + δ) – g(–1)] can never equal f′(x)/g′(x).
SECTION 5.2
1.There are many examples; e.g., g(x) = 1 – e x with a = 0, b > 0.
2.Say 
. Choose δ > 0 such that f′(t) ≤ f′(x0) – ∈ / 2 for 0 < |t – x0| < δ (see diagram). By Theorem 5.2.1 there exists y ∈ (x0 – δ, x0 + δ) such that f′(y) = ƒ′(x0) – ∈ / 4, a contradiction.
3.If x ≠ 0, then
by Problem 1 of Section 5.1. Since x sin(1/x) → 0 as x → 0, and cos(1/x) has a nonsimple discontinuity at x = 0 (look at the sequence xn = 1/nπ, for example), it follows that ƒ′ has a nonsimple discontinuity at x = 0.
4.
for some y between 0 and x. But f(n)(y) = ey, so |Rn(x)| ≤ e|x|xn/n! → 0 as n → ∞, so the series converges to ex.
5.Let ƒ be differentiable on R, with |ƒ′(x)| ≤ M for all x. Fix a ∈ R. Then f(b) – f (a) = (b – a)f′(x) for some x between a and b. Given ∈ > 0, it follows that if |b – a|M < ∈, i.e., |b – a| < ∈ / M, then
REVIEW PROBLEMS FOR CHAPTER 5
1.No, although f’(x) = 0 for some x ∈ (–1, 1), by Theorem 5.2.1. For an explicit counterexample, take ƒ (x) = x2.
2.The only difficulty occurs at x = 0. We have
(let h = 1/y and note that ye–y2 → 0 as y → ± ∞). Thus, ƒ′(x) = (2/x3)e–1/x2, x ≠ 0; ƒ′(0) = 0. A similar argument shows that
and by repeating this procedure we obtain ƒ (n)(0) = 0 for all n. The key point is that e–1/h2 will approach zero as h → 0 faster than any polynomial can approach infinity.
4.Let f(x) = 1 + 2x, g(x) = 1 + x, or f(x) = x + ex, g(x) = ex. L’Hospital’s rule does not apply since f(x) and g(x) do not approach 0 as x → 0.
5.
.
6.The remainder is of the form f(n)(y)xn/n!. Since |f(n)| ≤ 1 for all n and xn /n! → 0 as n → ∞ for any fixed x, the result follows.
7.(a) We have ƒ(x)/g(x) = f′(y)/g′(z) and ƒ′(x)/g′(x) → L, but it does not follow (even if we ignore the problem of division by 0) that f′(y)/g′(z) → L. For example, consider the two identical sequences
Then xn/yn is 1 for all n, but xn/yn+1 is 2 for all n.
(b)If f′ and g′ are continuous at a, and g’(a) ≠ 0, then
SECTION 6.1
1.Let
If xk – 1 ≤ 0 ≤ xk, then Mk Δαk = 1, mk Δαk = 0, so if P is any partition of [a, b] where a < 0 < b, then U(P, f, α) = 1, L(P, ƒ, a) = 0. Thus, 
does not exist.
2.Since any interval of positive length contains both rationals and irrationals, U(P, ƒ, α) is always 1 and L(P, ƒ, α) is always 0. Thus, 
does not exist.
3.Let ƒ (x) = 1, x rational; f(x) = –1, x irrational. As in Problem 2, ƒ is not Riemann integrable on [0, 1], but |ƒ(x)| = 1 for all x, so |f| is Riemann integrable.
4.Given ∈ > 0 and a partition P of [a, b] with m subintervals, by definition of sup and inf (specifically by Theorem 2.4.3) we may choose 
such that (Mk – ƒk) Δαk < ∈/2m and if 
such that (ƒk – mk) Δαk < ∈ / 2m. With the 
we get
and with the 
we get
But since 
, it follows that S(P(1), ƒ, α) and S(P(2), ƒ, α) will differ by less than ∈ / 2 for sufficiently small |P|. But then
By (1), (2), and (3), U and L both → 
.
SECTION 6.2
1.If f ∈ R(α), then 
; hence U(P) – L(P) → 0. If f ∉ R(α), then for some sequence of partitions Pn with |Pn| → 0, we do not have U(Pn), L(Pn) → the same finite limit. By passing to convergent subsequences, we may assume that U (Pn) → U, L(Pn) → L, with U ≠ L. But then U(Pn) – L(Pn) 0. (There is one other case to dispose of, namely, U (Pn), L(Pn) both approach a limit, but for some other sequence 
and 
both approach a different limit. A common refinement 
of Pn and 
will decrease upper sums and increase lower sums (see the discussion before (2) of Section 6.1.1), which yields a contradiction. To spell this out, say U(Pn) and L(Pn) → Lx and 
and 
. Then 
will have a subsequence on which 
will approach a limit < L1, and 
will approach a limit ≥ L2, contradicting 
.)
2.Since U (P) – L(P) on [a, c] and on [c, b] are less than or equal to U(P) – L(P) on [a, b], the result follows from Problem 1.
3.
4.α has jumps of size 1 at all integers, so
SECTION 6.3
1.By uniform continuity of ƒ, there is a δ > 0 such that |x – y| < δ implies |f(x) – f(y)| < ∈/2. By definition of variation, there is a partition P: c = x0 < x1 < … < xn = b such that 
. Adding more points to P can only increase 
so we can assume 0 < x1 – x0 < δ; hence |f(x1) – ƒ(x0)| < ∈/2.
2.F(b) – F(c) = V(f; [c, b]) by Theorem 6.3.2(e), so by Problem 1,
Therefore F(x1) – F(c) < ∈. If c < c′ ≤ x1, then, since F is increasing,
proving that F is right-continuous at c.
3.Choose δ as in Problem 1. Again by definition of variation, there is a partition P: a = x0 < x1 < … < xn = c such that
As in Problem 1 we may add points to P so that xn – xn–1 < δ; hence |ƒ (xn) – f(xn–1)| < ∈ / 2.
4.F(c) –F(a) = V(f; [a, c]) by Theorem 6.3.2(e). By Problem 3,
Thus, F (c) <F(xn–1) + ∈. If xn–1 ≤ c′ < c, then F (c) – F(c′) ≤ F(c) – F(xn–1),since F is increasing; hence 0 < F(c) – F(c′) < ∈, proving F left-continuous at c.
SECTION 6.4
1.By the Cauchy-Schwarz inequality for sums,
Let |P| → 0 to obtain the desired result. Alternatively,
As in Section 3.2, Problem 3, the discriminant must be ≤ 0, and the result follows.
2.We have (see diagram)
sum over n to get
Since 
iff 
the series converges if and only if the integral is finite.
3.Let f(x) = (–1)n–1/n, n – 1 ≤ x < n, n = 1, 2, 3, ... (see diagram); we have 
, but 
(finite).
4.(a) Take 
.
(b)Take f(x) = 1/x.
1.
2.(a) False; ƒ and α can be discontinuous at the same point.
(b)Thie; see the discussion before (2) of Section 6.1.1.
(c)True by Theorem 6.4.1, since 
exists.
3.Let F be an antiderivative of f; e.g., 
. Then
Differentiation yields, by the chain rule,
4.Let f(x) = 1 if x is rational; f(x) = –1 if x is irrational.
5.
6.By the fundamental theorem of calculus, F′ = ƒ on [a, b]. Therefore F has a bounded derivative on [a, b], and the result follows from Theorem 6.3.2(c).
7.Our Change of Variable Theorem 6.4.2 breaks down when the substitution y = h(x) is made and h is not one-to-one.
1.
as before, but 
.
2.fn → f pointwise, where 
. Thus,
Since 
as x → 1, and |fn(x) – f(x)| increases with 
. Thus 
uniformly.
3.sup0≤x≤1 |xn /(n + xn)| = l/(n + 1) → 0, so fn → 0 uniformly.
4.
and
Now supx>1 n/(n + xn) = n/(n + 1) → 1, so 
uniformly on [1, ∞). But supx ≥ 1 + δ n/(n + xn) = n/[n + (1 + δ)n] → 0, so fn → f uniformly on [1 + δ, ∞).
5.fn(x) = (xe–x)n, and the maximum value of xe–x occurs at x = 1. Thus, supx ≥ 0 fn(x) = e–n → 0, so fn → 0 uniformly on [0, ∞).
SECTION 7.2
1.Take fn(x) = n for all x. Then 
, so 
converges uniformly, although fn does not.
2.Let fn → f, gn → g uniformly. Then |fn + gn – (f + g)| ≤ |fn – f| + |gn – g|, so fn + gn → f + g uniformly.
3.Let fn(x) = 1/n → 0 uniformly on R; gn(x) = x → x uniformly on R. But fn(x)gn(x) = x/n → 0 point wise but not uniformly on R, since supx∊R |x/n| = ∞.
4.Say fn → f uniformly. Fix ∊ > 0; eventually, |fn – f| ≤ ∊, say for n ≥ N. Then |f(x)| ≤ |fN(x)| + ∊ ≤ MN + ∊ < ∞, so f is bounded. But then |fn| ≤ Mn for n < N, and |fn| ≤ |f| + ∊ ≤ MN +2∊ for n ≥ N. Thus, |fn(x)| ≤ max(M1,..., MN – 1, MN + 2∊)for all n and all x.
5.Let fn(x) = x for all n = 1, 2,... and all x ∊ R; f(x) = x. Then fn → f uniformly on R, but the fn are not uniformly bounded.
SECTION 7.3
1.The Weierstrass M–test applies, since e−nx ≤ e−na, and ∑ne−na < ∞ if a > 0.
2.The nth partial sum is
Thus, |sn(x) − s(x)| = e−(n+1)x / (l − e−x), and since 1 − e−x → 0 as x → 0, we have supx>0 |sn(x) − s(x)| = ∞. Therefore, 
uniformly on (0, ∞).
3.This follows immediately from Theorem 3.3.2, since the radius of convergence is at least |r|.
4.On [−a, a], |anxn| ≤ |an|an, and 
by Problem 3. The result follows from the Weierstrass M -test.
5.This follows from Theorems 7.2.2 and 7.2.3. To apply Theorem 7.2.3, we need only check that the differentiated series 
has the same radius of convergence as the original series. But this follows from Theorem 3.3.2, since lim sup |nan|1/n = lim sup |an|1/n (note that n1/n → 1).
6.
by Theorem 6.2.1(a). Let n → ∞; by Theorem 7.2.2, 
; i.e., the series 
converges to 
.
SECTION 7.4
1.Take fn(x1) = 1 for all n, fn(x2) = 2 for all n, fn(x3) = 3 for all n, etc.
2.(a) See diagram.
(b){fn} converges pointwise on the dyadic rationale k/2n, k = 0, 1, ..., 2n, n = 1, 2, ...; {fn} is uniformly bounded since 0 ≤ fn ≤ 1 for all n. If the second and third digits of the binary expansion of x are both 1, then 
; if the digits are both 0, then 
. Similarly, if the (n + l)st and (n + 2)nd digits of the binary expansion of x are both 1, then 
; if the digits are both 0, then 
. Thus, given any subsequence {fnk}, we may assume without loss of generality that nj+1 > 2 + nj for all j, and we can select the digits in the binary expansion of x to produce oscillation in the values of fnk(x), so {fnk} cannot converge pointwise.
3.Given ∊ > 0, choose δ > 0 such that d(x, y) < δ implies d(fn(x), fn(y)) < ∊/3 for all n. If x, y ∈ K and d(x, y) < δ, then d(f(x), f(y)) ≤ d(f(x), fn(x)) + d(fn(x), fn(y)) + d(fn(y), f(y)) < ∊ for sufficiently large n, by pointwise convergence. Thus f is uniformly continuous on K. Now adjust the choice of δ so that if d (x, y) < δ, then, in addition to d(fn(x), fn(y)) < ∊/3 for all n, we have d(f(x), f(y)) < ∊/3. By compactness, there are points x1, ..., xm ∈ K such that 
Then
If x ∈ Bδ(xi), then d(x, xi) < δ, so the first and third terms on the right are < ∊/3. The second term is < ∊/3 for large n, say n ≥ N, by pointwise convergence. It follows that d(f(x), fn(x)) < ∊ for n ≥ N and all x, proving uniform convergence.
4.(a)
Given ∊ > 0, there is a positive integer N such that the first and third terms on the right are < ∊/3 for n ≥ N and all x, y ∈ K (by uniform convergence). For some δ > 0, d(x, y) < δ implies d(f(x), f(y)) < ∊/3 , by uniform continuity of f. (f is a uniform limit of continuous functions, so is continuous. Since K is compact, f is uniformly continuous.) Since f1, ..., fN−1 are uniformly continuous on K (compactness is also used here), there is a δ > 0 such that d(x, y) < δ implies d(fn(x), fn(y)) < ∊ for all n, proving equicontinuity.
(b)
Given ∊ > 0, the first term on the right is < ∊/2 for large n, by uniform convergence. The second term is < ∊/2 for large n since f, a uniform limit of continuous functions, is continuous. Thus, d(fn(xn), f(x)) < ∊ for large n.
(c)The argument of (a) uses compactness, but (b) does not. For an explicit counterexample to (a), take fn(x) = f(x) = 1/x on (0, ∞) for all n. Then {fn} is not equicontinuous because x → 1/x is not uniformly continuous on (0, ∞). (See the discussion before Definition 4.2.3.)
1.Let g(y) = f(a + (b − a)y), 0 ≤ y ≤ 1. If p is a polynomial such that | g(y) − p(y)| < ∊ on [0, 1], then 
= (with x = a + (b − a)y) | g(y) − p(y) | < ∊ for a ≤ x ≤ b.
2.One example: let ann = n, an,n+1 = −n, n = 1, 2, ...; anj = 0 for all other pairs (n, j) of positive integers. Then
3.For every 
. Since 
by hypothesis, the result follows.
4.Since the xk, k ≥ 1, are isolated points, the fn are automatically continuous there (if xj converges to the isolated point y, then xj = y eventually). So all that needs to be verified is that fn(xk) → fn(x0) as k → ∞. But, by hypothesis, 
for each n; in particular, 
converges. But then 
.
5.
for any m. Since
and
we may let m → ∞ to get the desired result.
6.
By Problem 4, f(xk) → f(x0) as k → ∞, as desired.
7.
The result is finite because all partial sums are bounded by 
.
8.Choose N and k so that sNk will be “close” to s. If s is finite, this means |s − sNk| < ∊; if s = ∞, take sNk > M, where M is an arbitrary positive number. Since anj ≥ 0, sNk increases with N and k, and it follows that the double sum (in either order) is arbitrarily close to s.
9.
since the power series Σanxn and Σbnxn converge absolutely with their interval of convergence. Now
1.Since xn → 0 for 
pointwise. Now
(note xn/(3 + xn) = 1 − 3/(3 + xn), which increases with x), so {fn} does not converge uniformly on [0, 1).
2.Take fn(x) = (−1)n/n for all real x.
3.For any fixed x, fn(x) is eventually 0, so fn → 0 pointwise. But 
, so 
uniformly.
4.Power series converge uniformly on any closed subinterval of the interval of convergence.
SECTION 8.1
1.Let x ∈ A ∩ F. Since A is open in E, for some r > 0 we have {y ∈ E : d(x, y) < r} ⊆ A. But F ⊆ E, so {y ∈ F : d(x, y) < r} ⊆ A also. But then {y ∊ F : d(x, y) < r} ⊆ A ∩ F, proving A ∩ F open in F.
2.Let xn ∈ A ∩ F, xn → x ∈ F. Then xn ∈ A, xn → x ∈ E, so (because A is closed in E) x ∈ A. But we already have x ∈ F, so x ∈ A ∩ F, proving A ∩ F closed in F.
3.If h(x) = arctan x, then h is a homeomorphism of 
and [−π/2, π/2]. Thus, h ∘ f is a bounded, continuous function on E, so by Theorem 8.1.3, there is a continuous function g0 on Ω such that g0 = h ∘ f on E and sup{|g0(x)| : x ∈ Ω} = sup{|h(f(x))| : x ∈ E}. If g = h−1 ∘ g0, then g is continuous on Ω and g = f on E. (Note that if g0(x) = π/2, then g(x) = h−1(g0(x)) = ∞, and similarly, if g0(x) = −π/2, then g(x) = −∞, so we cannot replace the codomain of g by R.)
SECTION 8.2
1.Let 
, so that we have points yn ∈ Br(x) with yn → y. But then d(x, yn) → d(x, y). (See Section 4.4, Problem 1.) Since d(x, yn) < r for all n, we have d(x, y) ≤ r.
2. Let Ω be any set with at least two points, and put the following metric on Ω :
Then 
.
3.The result follows from the definition of convergence:
(Note that if for every positive integer m, |fn(x) − f(x)| is eventually less than 1/m, then for every ∊ > 0, |fn(x) − f(x)| is eventually less than ∊.)
SECTION 8.3
1.Assume 
. Since 
, the sets A and B are disjoint. If xn ∈ A, xn → x ∈ A ∪ B, then x ∈ Ā; hence x ∉ B (since 
). But then x must be in A, proving A closed in A ∪ B. An identical argument proves B closed in A ∪ B. Now assume A, B separated. If x ∈ Ā ∩ B, let xn ∈ A, xn → x. Since x ∈ B ⊆ A ∪ B and A is closed in A ∪ B, we must have x ∈ A, so 
, a contradiction. Therefore 
, and similarly 
.
2.Assumed 
. If 
and 
, then since 
we have 
, and since 
we have 
. Now 
, and 
. Now assume there exist open sets G1, G2 with the specified properties. Then (since 
(closed); hence 
. Since B ⊆ G2, we have, 
. Similarly, 
(since 
), so 
. Since A ⊆ G1, we have 
.
3.If x ∉ E, then E = [E ∩ (−∞, x)] ∪ [E ∪ (x, ∞)]. Since (−∞, x) and (x, ∞) are open in R, E ∩ (−∞, x) and E ∩ (x, ∞) are open in E (Section 8.1, Problem 1). Since E is connected, either 
or 
. In the first case, y ∈ E implies y ≥ x, so x is a lower bound of E, contradicting a = inf E. ln the second case, x is an upper bound of E, contradicting b = sup E.
4.If f(E) = A ∪ B, where A and B are disjoint and both open (and closed) in f(E), then E = f−1(A) ∪ f−1(B), where f−1(A) and f−1(B) are disjoint and (by continuity of f) both open (and closed) in E. Since E is connected, 
or 
. But f maps onto f(E), so either A or B is empty (e.g., if y ∈ A, ∃x ∈ E with f(x) = y, so x ∈ f−1(A)). Therefore f(E) is connected.
5.Let f be a continuous, real-valued function on [a, b], and assume f(a) < c < f(b). By Problem 4, f[a, b] is connected, so by Problem 3, f[a, b] is an interval I. But f(a), f(b) ∈ I implies c ∈ I, so c = f(x) for some x ∈ [a, b]. Since f(a) ≠ c, f(b) ≠ c, we must have x ∈ (a, b).
SECTION 8.4
1.Let fn(x) = 1 − n|x|, −1/n ≤ x ≤ 1/n; fn(x) = 0 elsewhere. Each fn is continuous, but
which is not LSC (it is USC).
2.Let fn(x) = nx, −1/n ≤ x ≤ 1/n; fn(x) = −1, x ≤ −1/n; fn(x) = 1, x ≥ 1/n. Then
and f is neither USC nor LSC.
3.Let f(x) = 1/x, 0 < x ≤ l; f(0) = 0.
4.f is LSC at x iff given ∊ > 0∃δ > 0 such that d(x, y) < δ ⇒ f(y) > f(x) − ∊; f is USC at x iff given ∊ > 0∃δ > 0 such that d(x, y) < δ ⇒ f(y) < f(x)+∊. The proofs are done as in Theorem 4.1.2, with d(f(x), f(y)) < ∊ replaced by f(y) > f(x) − ∊ or f(y) < f(x) + ∊.
5.If f is nonnegative and finite-valued, then (see the proof of Theorem 8.4.6) so are the fn. If f ≥ 0 but can assume the value +∞, map the nonnegative extended reals one-to-one onto [0, π/2] via h(x) = arctan x and proceed exactly as in Theorem 8.4.6. The functions fn will automatically be nonnegative.
1.(a) Always
2.(b) Sometimes (see Section 8.1)
3.(a) 
(b)Cantor set
