Measure zero 5.6. Prove that a countable point set has measure zero.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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Section: Chapter Questions
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Question
5.6) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text. I also included the other problem that was previously asked for.
![Measure zero
5.6.
Prove that a countable point set has measure zero.
Let the point set be denoted by x1, X2, X3, X4, . . . and suppose that intervals of lengths less than ɛ/2, ɛ/4,
ɛ/8, ɛ/16, . .., respectively, enclose the points, where ɛ is any positive number. Then the sum of the lengths of
the intervals is less than ɛ/2 + ɛ/4 + ɛ/8 + ...= ɛ [let a = ɛ/2 and r = 1/2 in Problem 2.25(a)], showing that the
set has measure zero.
Infinite series
2.25.
Prove that the infinite series (sometimes called the geometric series)
a + ar + ar2 +
n=1
(a) converges to al(1 – r) if |r| < 1, and (b) diverges if |r|
1.
Let
S, = a + ar + ar² + · . . + ar"-1
Then
rS, = ar + ar² +...+ ar"-1
+ ar"
Subtract
(1 – r)S, = a
- ar"
a(1 – r")
S. =
or
1-r
a(1 – r").
a
by Problem 2.7.
1-r
-
(a) If r<1, lim S, = lim
1-r
n00
n00
(b) If r| > 1, lim S, does not exist (see Problem 2.44).
n00
2.7.
Prove that lim x" = 0 if |x| < 1.
n00
Method 1: We can restrict ourselves to x # 0, since if x = 0, the result is clearly true. Given e > 0, we must
show that there exists N such that |<e for n> N. Now |x" | = |x|"<e when n log,o lx| <log10 €. Divid-
logjo E
ing by log10 |x|, which is negative, yields n>-
= N, proving the required result.
log,0 |x|
Method 2: Let |x| = 1/(1 + p), where p > 0. By Bernoulli's inequality (Problem 1.31), we have |x"| =
|x|" = 1/(1 + p)" < 1/(1 + np) < e for all n> N. Thus, lim x" = 0.
n00](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2F707ec030-aca7-46e3-b5ae-e48d22e37b01%2Ftdpqaij_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Measure zero
5.6.
Prove that a countable point set has measure zero.
Let the point set be denoted by x1, X2, X3, X4, . . . and suppose that intervals of lengths less than ɛ/2, ɛ/4,
ɛ/8, ɛ/16, . .., respectively, enclose the points, where ɛ is any positive number. Then the sum of the lengths of
the intervals is less than ɛ/2 + ɛ/4 + ɛ/8 + ...= ɛ [let a = ɛ/2 and r = 1/2 in Problem 2.25(a)], showing that the
set has measure zero.
Infinite series
2.25.
Prove that the infinite series (sometimes called the geometric series)
a + ar + ar2 +
n=1
(a) converges to al(1 – r) if |r| < 1, and (b) diverges if |r|
1.
Let
S, = a + ar + ar² + · . . + ar"-1
Then
rS, = ar + ar² +...+ ar"-1
+ ar"
Subtract
(1 – r)S, = a
- ar"
a(1 – r")
S. =
or
1-r
a(1 – r").
a
by Problem 2.7.
1-r
-
(a) If r<1, lim S, = lim
1-r
n00
n00
(b) If r| > 1, lim S, does not exist (see Problem 2.44).
n00
2.7.
Prove that lim x" = 0 if |x| < 1.
n00
Method 1: We can restrict ourselves to x # 0, since if x = 0, the result is clearly true. Given e > 0, we must
show that there exists N such that |<e for n> N. Now |x" | = |x|"<e when n log,o lx| <log10 €. Divid-
logjo E
ing by log10 |x|, which is negative, yields n>-
= N, proving the required result.
log,0 |x|
Method 2: Let |x| = 1/(1 + p), where p > 0. By Bernoulli's inequality (Problem 1.31), we have |x"| =
|x|" = 1/(1 + p)" < 1/(1 + np) < e for all n> N. Thus, lim x" = 0.
n00
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