is an increasing sequence of real numbers, and
(sn) is bounded [by K + 1, in fact]. So by
Theorem 10.2, (sn) converges to 10.
Monotone Sequences and Cauchy Sequences
59 a real number we traditionally write as
K.d1d2d3d4 . For example, 3.3333
represents l
Note: If sup S is in S, its sufficient to define sn
= sup S for all n. 12 lim sups and lim infs Let
(sn) be any sequence of real numbers, and let
S be the set of subsequential limits of (sn).
Recall lim sup sn = lim N sup cfw_sn : n>N =
sup S (*) and lim
Define subsequences as in (3) of Definition
11.1. 11.7 Let (rn) be an enumeration of the
set Q of all rational numbers. Show there
exists a subsequence (rnk ) such that limk
rnk = +. 11.8 8 Use Definition 10.6 and
Exercise 5.4 to prove lim inf sn = lim su
exists a number N such that n>N implies |an|
< . Thus lim an = 0. The converse of Corollary
14.5 does not hold as the example 1/n = +
shows. We next give several tests to assist us
in determining whether a series converges.
The first test is elementary bu
. (1) Since lim n/(n + 1) = 1, neither the
Ratio Test nor the Root Test gives any
information. Since 1 n diverges, we will not
be able to use the Comparison Test 14.6(i) to
show (1) converges. Since the terms of the
series (1) are not all nonnegative, we
nonnegative numbers. Prove lim sup sntn
(lim sup sn)(lim sup tn). 12.9 (a) Prove that if
lim sn = + and lim inf tn > 0, then lim sntn =
+. (b) Prove that if lim sup sn = + and lim
inf tn > 0, then lim sup sntn = +. 13. * Some
Topological Concepts in Metr
discuss. Many of the notions introduced in this
chapter make equally good sense in more
general settings. For example, the ideas of
convergent sequence, Cauchy sequence and
bounded sequence all make sense for a
sequence (sn) where each sn belongs to the
p
preceding discussion also holds. That is, every
nonnegative real number x has at least one
decimal expansion. This will be proved, along
with some related results, in 16. Unbounded
monotone sequences also have limits. 10.4
Theorem. (i) If (sn) is an unbou
need to prove lim sn = s. Let > 0. Since s = lim
vN there exists a positive integer N0 such that
|s supcfw_sn : n>N0| < . Thus supcfw_sn : n>N0 < s
+ , so sn < s + for all n>N0. (1) Similarly, there
exists N1 such that |s infcfw_sn : n>N1| < ,
hence infcf
Topological Concepts in Metric Spaces 89 sets
is cfw_x : d(x,x0) = r. In the plane R2, the sets cfw_(x1,
x2) : x1 > 0 and cfw_(x1, x2) : x1 > 0 and x2 > 0
are open. If > is replaced by , we obtain
closed sets. Many sets are neither open nor
closed. For ex
denoted lim inf sn and lim sup sn, respectively.
10.6 Definition. Let (sn) be a sequence in R. We
define lim sup sn = lim N sup cfw_sn : n>N (1)
and lim inf sn = lim N inf cfw_sn : n>N. (2)
Note that in this definition we do not restrict
(sn) to be bounde
monotonic sequences converging to t.
Henceforth, we assume cfw_n N : sn = t is
finite. Then cfw_n N : 0 < |sn t| < is infinite
for all > 0. Since these sets equal cfw_n N : t <
tcfw_n N : t< t + , and these sets get smaller
as 0, we have cfw_n N : t < t
(a) Show that if L < 1, then lim sn = 0. Hint:
Select a so that L<a< 1 and obtain N so that |
sn+1| < a|sn| for n N. Then show |sn| <
anN |sN | for n>N. (b) Show that if L > 1, then
lim |sn| = +. Hint: Apply (a) to the sequence
tn = 1 |sn| ; see Theorem 9
1r . This proves n=0 arn = a 1 r if |r| < 1.
(2) If a = 0 and |r| 1, then the sequence (arn)
does not converge to 0, so the series arn
diverges by Corollary 14.5 below. Example 2
Formula (2) of Example 1 and the next result
are very important and both sho
in terms of open sets [see Definitions 13.8,
13.11, and 22.1] are called topological, hence
the title of this section. 88 2. Sequences 13.8
Definition. Let (S, d) be a metric space. A
subset E of S is closed if its complement S \ E is
an open set. In othe
= , then for every is infinite. If L =
and >L, the set cfw_n : sn > is finite; otherwise
supcfw_sn : n>N > for all N and hence L = lim
sup sn >L, a contradiction; see Definition
10.6. If L = and is infinite; otherwise
there exists a positive integer N
number N such that m, n > N implies |sn sm|
< . (1) Nothing is lost in this definition if we
impose the restriction n>m. Moreover, it is
only a notational matter to work with m 1
where m n instead of m where m 0 there
exists a number N such that n m>N imp
> 0 and let = 1 M . Then 54 2. Sequences > 0, so
there exists N such that n>N implies | 1 sn 0|
< = 1 M . Since sn > 0, we can write n>N
implies 0 < 1 sn < 1 M and hence n>N implies
MN0. (a) Prove that if lim sn = +, then lim tn
= +. (b) Prove that if lim
belongs to U. Hence (1) holds. Exercises 13.1
For points x, y in Rk, let d1(x, y) = maxcfw_|xj yj
| : j = 1, 2,.,k and d2(x, y) = k j=1 |xj yj |.
(a) Show d1 and d2 are metrics for Rk. (b)
Show d1 and d2 are complete metrics on Rk.
13.2 (a) Prove (1) in L
be a compact nonempty subset of Rk, and let
= supcfw_d(x, y) : x, y E. Show E contains points
x0, y0 such that d(x0, y0) = . 13.15 Let (B, d)
be as in Exercise 13.3, and let F consist of all x
B such that supcfw_|xj| : j = 1, 2,. 1. (a)
Show F is closed
this case, applying the Root Test is not much
more difficult provided we recall lim n1/n = 1.
It is also possible to show (1) converges by
comparing it with a suitable geometric series.
Example 7 Consider the series an where an = 2
(1)n 3
n . (1) The form
may as well try the Ratio Test first. Example 3
Consider the series n=2 1 3 n = 1 9 1 27 +
1 81 1 243 + . (1) This is a geometric series
and has the form n=0 arn if we write it as
(1/9) n=0(1/3)n. Here a = 1/9 and r = 1/3,
so by (2) of Example 1 the sum i
subsequence (snk ) of (sn) will refer to the
subsequence defined by (1) and (2) or by (3),
depending upon your point of view. Example 1
Let (sn) be the sequence defined by sn =
n2(1)n. The positive terms of this sequence
comprise a subsequence. In this ca
There exists N0 so that supcfw_sn : n>N < t + for N
N0. In particular, sn < t + for all n>N0. We
now claim cfw_n N : t < t + is infinite. (1)
Otherwise, there exists N1 > N0 so that sn t
for n>N1. Then supcfw_sn : n>N t for N N1,
so that lim sup sn < t,
bound property in 4.4 as a theorem. We did
not do so because the concept of least upper
bound in R seems to us more fundamental
than the concept of Cauchy sequence. We will
prove Rk is complete. But we have a notational
problem, since we like subscripts f
this case. We have now established (1) in all
cases. For the reversed inequality, we resort to
a little trick. First note that we may ignore the
first few terms of (sn) and assume all sn = 0.
Then we can write lim 1 sn = 1 s by Lemma 9.5.
Now we apply (1)
By Theorem 10.7 we need only show lim inf sn
= lim sup sn. (1) Let > 0. Since (sn) is a Cauchy
sequence, there exists N so that m, n > N
implies |sn sm| < . In particular, sn < sm + for
all m, n > N. This shows sm + is an upper
bound for cfw_sn : n>N, so
and show how to select nk. This will give us an
infinite increasing sequence (nk)kN and
hence a subsequence (snk ) of (sn) satisfying
(3). Since we will have snk1 snk for all k,
this subsequence will be monotonically
increasing. Since (3) also will imply
real number S, in which case we define n=m
an = S. Thus n=m an = S means lim sn = S or
limn n k=m ak = S. A series that does not
converge is said to diverge. We say that
n=m an diverges to + and we write n=m
an = + provided lim sn = +; a similar remark
a
above, F is a union of 2k k-cells having
diameter 2 . At least one of these 2k k-cells,
which we denote by F1, cannot be covered by
finitely many sets from U. Likewise, F1
contains a k-cell F2 of diameter 4 which
cannot be covered by finitely many sets fr
Um0 . It follows that E is bounded. To show
E is closed, consider any point x0 in Rk \ E. For
m N, let Vm = x Rk : d(x,x0) > 1 m . 13.
* Some Topological Concepts in Metric Spaces
91 Then each Vm is open in Rk and V = cfw_Vm : m
N covers E since m=1Vm =
first show lim sup sntn s lim sup tn. (1) We
have three cases. Let = lim sup tn. Case 1.
Suppose is finite. By Theorem 11.7, there
exists a subsequence (tnk ) of (tn) such that
limk tnk = . We also have limk snk = s
[by Theorem 11.3], so limk snk tnk = s.
has the following form. There exist closed
intervals [a1, b1], [a2, b2], . . . , [ak, bk] so that
F = cfw_x Rk : xj [aj , bj ] for j = 1, 2,.,k,
which is sometimes written as F = [a1, b1]
[a2, b2] [ak, bk], so it is a k-dimensional
box in Rk. Thus a 2-ce
induction step, assume (2) holds for some n
2. To show sn+2 < sn+1, we need s2 n+1 + 5
2sn+1 < sn+1 or s2 n+1 + 5 < 2s2 n+1 or 5 < s2
n+1, but this holds because sn+1 > 5 by the
assumption (2) for n. To show sn+2 > 5, we
need s2 n+1 + 5 2sn+1 > 5 or s2 n