Solutions to homework # 9.
1.
By the uniform continuity of
f
, there exists
δ
such that

f
(
p
)

f
(
q
)

<
1 whenever

p

q

< δ
and
p
,
q
∈
E
. Since
E
is bounded, its closure
E
is both bounded and closed,
hence compact, so the following open cover of
E
∪
q
∈
E
U
q
,
where
U
q
:=
{
p
:

p

q

< δ
}
has a finite subcover
∪
n
k
=1
U
p
k
, which can also serve to cover
E
. Taking one point
x
k
∈
E
from each
U
p
k
that contains points of
E
, we see that

f
(
x
)
 ≤
max
k

f
(
x
k
)

+ 1 for all
x
∈
E
.
Thus
f
is bounded on
E
.
The boundedness of
E
is indeed essential, as the example of the unformly continuous
function
f
(
x
) :=
x
on the domain
E
:= IR demonstrates.
2.
If
f
is continuous, then the function
g
:
x
→
f
(
x
)

x
is also continuous. If
f
(0) = 0,
then
f
(0)
>
0 and so
g
(0)
>
0. If
f
(1) = 1, then
f
(1)
<
1 hence
g
(1)
<
0. Thus
g
takes
values of opposite signs at the end of the interval
I
, hence
g
takes the value 0 somewhere
inside
I
, whihc means
f
(
x
) =
x
for some
x
.
3.
Both functions [
x
] and (
x
) are discontinuous at all integer points. Indeed, if
k
∈
ZZ and
x
→
k

(i.e.,
x
tends to
k
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 Spring '09
 Continuity, Continuous function, Metric space, nd, neighborhood Nd

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