the midpoint of the interval
I
1
, so
c
1
= (
a
1
+
b
1
)
/
2. Then for at least one of the halfintervals
[
a
1
,c
1
] or [
c
1
,b
1
], the following condition holds: there are inﬁnitely many indices
n
such that
s
n
is
an element of that halfinterval. If this condition holds for the left halfinterval [
a
1
,c
1
], let
a
2
=
a
1
and
b
2
=
c
1
. If, instead, this condition does
not
hold for the lefthalf interval [
a
1
,c
1
], so that
it must hold for the righthalf interval [
c
1
,b
1
], let
a
1
=
c
1
and
b
2
=
b
1
. Either way, one gets a
new subinterval, [
a
2
,b
2
], half the length of the original interval [
m,M
], such that
s
n
∈
[
a
2
,b
2
] for
inﬁnitely many
n
. Continuing this process indeﬁnitely, one obtains an inﬁnite
sequence
of intervals
[
a
k
,b
k
], with
k
∈
IN, such that [
a
1
,b
1
] = [
m,M
], and for each
k
∈
IN the interval [
a
k
+1
,b
k
+1
] is one of
the two halves of [
a
k
,b
k
], and for each
k
∈
IN there are inﬁnitely many
n
∈
IN such that
s
n
∈
[
a
k
,b
k
].
Clearly
b
k

a
k
= (
M

m
)
/
2
k

1
for each
k
, so that lim
k
→∞
(
b
k

a
k
) = 0. Thus it follows from
Discussion Exercise (4) from Tuesday, 10/27/2011, that the intersection of these intervals consists
of a single point; call it
d
. Note that
d
= lim
k
→∞
a
k
= lim
k
→∞
b
k
. Now deﬁne a subsequence
(
s
n
1
,s
n
2
, ...s
n
k
, ...
) of
σ
by the following rule:
First, set
n
1
= 1; note that
n
1
is the ﬁrst index such that
s
n
1
∈
[
a
1
,b
1
]. Likewise, deﬁne
n
2
to
be the ﬁrst index larger that
n
1
such that
s
n
2
∈
[
a
2
,b
2
]. Suppose now that indices
n
1
,
n
2
, . . .
n
k
have
already been constructed so that
n
1
< n
2
< ... < n
k
, and
s
n
j
∈
[
a
j
,b
j
] for each
j
= 1
,
2
, ...k
.
Then deﬁne the index
n
k
+1
to be the smallest index such that
n
k
< n
k
+1
and
s
n
k
+1
∈
[
a
k
+1
,b
k
+1
].
(The existence of such an index which is strictly
larger
than
n
k
follows from the fact that we have
constructed these intervals so that for each of them,
s
n
in the interval for inﬁnitely many indices
n
;
in particular,
n
k
cannot be the largest such index.) It follows that
n
1
< n
2
< ... < n
k
< ...
for
all
k
, so that
τ
= (
s
n
1
,s
n
2
, ...s
n
k
...
) is a subsequence of
σ
. It is also clear from the construction
that
s
n
k
∈
[
a
k
,b
k
] for each
k
; that is,
a
k
≤
s
n
k
leq b
k
for each
k
. It now follows from the Squeeze
Theorem, i.e., the result of Exercise 8.5(a), that lim
k
→∞
s
k
=
d
as well, as required.
Remarks
(a) The proof outlined above, of the BolzanoWeierstrass Theorem for Sequences, is
pretty much the standard one, found in most analysis texts.
(b) I’ve appended the phrase ‘for Sequences’ to the text’s name of this theorem because there
are very similar theorems, found in other texts, which are also called ‘the BolzanoWeierstrass
Theorem’, but which apply in other situations. Be careful, when reading other texts or analysis
papers, to know to which version their use of the name ‘BolzanoWeierstrass Theorem’ refers.
Things to Prepare for the Discussions
(These are
not
to be handed in.)
Note: These are problems/examples/topics which the Teaching Assistant (TA) will cover in the
discussion sections. You are expected to work on them
before
you go to the discussions.
(A) (Items for the discussion on Tuesday 11/08/2011. Do not hand in.)
(1) Exercise 11.2
(2) Exercise 12.3 (a, b, d, e) Note: You do not need the results of Section 12 to do this exercise.
It could have been placed it in Section 10 just as well.
(3) Go over Example 4 in Section 11
(B) (Items for the discussion on Thursday 11/10/2011. Do not hand in.) REVISED
(1) Let
σ
= (
s
1
,s
2
, ...
) be a sequence of real numbers. Let
τ
= (
t
1
,t
2
, ...
) be the subsequence
of
σ
given by the oddorder terms of
σ
, and let
ξ
= (
x
1
,x
2
, ...
) be the subsequence of
σ
given by
the evenorder terms of
σ
. That is,
τ
= (
s
1
,s
3
,s
5
, ...s
2
k

1
, ...
) and
ξ
= (
s
2
,s
4
,s
6
, ...s
2
k
, ...
)
,