1. Exercise 6.2.16. For each
n
∈
N
, let
f
n
be a function deﬁned on [0
,
1]. If (
f
n
) is bounded on
[0
,
1]—that is, there exists an
M >
0 such that

f
n
(
x
)
 ≤
M
for all
n
∈
N
and
x
∈
[0
,
1]—and
if the collection of functions (
f
n
) is equicontinuous (Exercise 6.2.15), follow these steps to
show that (
f
n
) contains a uniformly convergent subsequence.
(a) Use Exercise 6.2.14 to produce a subsequence (
f
n
k
) that converges at every rational point
in [0
,
1]. To simplify the notation, set
g
k
=
f
n
k
. It remains to show that (
g
k
) converges
uniformly on all of [0
,
1].
Proof.
Since (
f
n
) is bounded and
Q
∩
[0
,
1] is countable, Exercise 6.2.14 implies that
there is a subsequence (
f
n
k
) that converges at all points in
Q
∩
[0
,
1].
(b) Let
± >
0. By equicontinuity, there exists a
δ >
0 such that
g
k
(
x
)

g
k
(
y
)

<
±
3
for all

x

y

< δ
and
k
∈
N
. Using this
δ
, let
r
1
, r
2
, . . . , r
m
be a
ﬁnite
collection of
rational points with the property that the union of the neighborhoods
V
δ
(
r
i
) contains
[0
,
1].
Explain why there must exist an
N
∈
N
such that

g
s
(
r
i
)

g
t
(
r
i
)

<
±
3
for all
s, t
≥
N
and
r
i
in the ﬁnite subset of [0
,
1] just described. Why does having the
set
{
r
1
, r
2
, . . . , r
m
}
be ﬁnite matter?
Proof.
Let
± >
0. Since
r
1
, . . . , r
m
∈
Q
∩
[0
,
1], we know that each sequence (
g
k
(
r
i
))
converges and, hence, is Cauchy. Therefore, for each
i
= 1
, . . . , m
, there exists
N
i
∈
N
such that
r, s
≥
N
i
implies that

g
s
(
r
i
)

g
t
(
r
i
)

<
±
3
.
Let
N
= max
{
N
1
, . . . , N
m
}
(this is where it’s essential that the collection
{
r
1
, . . . , r
m
}
is ﬁnite; if it were inﬁnite, then there might well not be a maximum
N
i
). Then, for any
r, s
≥
N
and for any
i
∈ {
1
, . . . , m
}
, we have that
s, t
≥
N
i
and so

g
s
(
r
i
)

g
t
(
r
i
)

<
±
3
,
as desired.
(c) Finish the argument by show that, for an arbitrary
x
∈
[0
,
1],

g
s
(
x
)

g
t
(
x
)

< ±
for all
s, t
≥
N
.
1