This preview shows pages 1–2. Sign up to view the full content.
Homework # 10
p. 193, # 33
Consider the characteristic function
f
of the dyadic rational numbers
Z
[1
/
2].
(a) What is the set of discontinuities of
f
?
The discontinuities of
f
will be
R
. Let
x
∈
R

Z
[1
/
2], then for any
± <
1, and for any
δ >
0, there exists
y
∈
Z
[1
/
2], such that

x

y

< δ
. Let
n
be such that 1
/
2
n
< δ
. By
the Archimedean principle, for some
k
,
k/
2
n
< x <
(
k
+ 1)
/
2
n
, and therefore

x

k/
2
n

<
1
/
2
n
< δ
. Then

f
(
x
)

f
(
y
)

= 1
> ±
, so
f
is not continuous at
x
.
If
x
∈
Z
[1
/
2], then again there exists
y
∈
R

Z
[1
/
2] such that

x

y

< δ
for any
δ
,
and therefore

f
(
x
)

f
(
y
)

= 1
> ±
. Applying the previous case to
x

√
2, we see there is
y
0
∈
Z
[1
/
2] such that

x

√
2

y
0

< δ
. Then let
y
=
√
2 +
y
0
∈
R

Z
[1
/
2], so

x

y

< δ
.
(b) At which points is its oscillation
≥
κ
?
Since
f
takes on only values of 0
,
1,
osc
x
f
≤
1, for all
x
. The above argument shows that
osc
x
f
= 1, for all
x
∈
R
, since for any
δ
,
diam
(
f
(
x

δ,x
+
δ
)) = 1. Therefore, if
κ >
1, then
osc
x
f < κ
for all
x
, and if
κ
≤
1, then
osc
x
f
≥
κ
for all
x
.
(c) Is
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 RIEMAN
 Math

Click to edit the document details