(a) Directly use the Bolzano-Weierstrass theorem (Theorem 2.3.8) to prove that every Cauchy sequence of real numbers is convergent. That is, only make use of the fact that every bounded sequence of real numbers has some convergent subsequence (not necessarily converging to either lim sup/lim inf). (Remark: Proving Cauchy-completeness from Bolzano-Weierstrass without lim sup/inf in this way generalizes more easily to R".) (b) One interesting thing about the Cauchy sequence definition is that it can be stated without reference to real numbers at all: We say a sequence of rational numbers {rn} is Cauchy (in the absolute value metric on the rational numbers) if for every & EQ satisfying > 0, there exists some MEN such that for all n, k≥ M, we have rn - Tk < e. Prove that is not Cauchy complete, that is, show that there exists a Cauchy sequence
(a) Directly use the Bolzano-Weierstrass theorem (Theorem 2.3.8) to prove that every Cauchy sequence of real numbers is convergent. That is, only make use of the fact that every bounded sequence of real numbers has some convergent subsequence (not necessarily converging to either lim sup/lim inf). (Remark: Proving Cauchy-completeness from Bolzano-Weierstrass without lim sup/inf in this way generalizes more easily to R".) (b) One interesting thing about the Cauchy sequence definition is that it can be stated without reference to real numbers at all: We say a sequence of rational numbers {rn} is Cauchy (in the absolute value metric on the rational numbers) if for every & EQ satisfying > 0, there exists some MEN such that for all n, k≥ M, we have rn - Tk < e. Prove that is not Cauchy complete, that is, show that there exists a Cauchy sequence
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![This question looks at the relationship between Bolzano- Weierstrass and the
"Cauchy completeness" property of R.
(a) Directly use the Bolzano-Weierstrass theorem (Theorem 2.3.8) to prove that every
Cauchy sequence of real numbers is convergent. That is, only make use of the fact
that every bounded sequence of real numbers has some convergent subsequence (not
necessarily converging to either lim sup/lim inf).
(Remark: Proving Cauchy-completeness from Bolzano-Weierstrass without lim sup/inf
in this way generalizes more easily to Rn.)
(b) One interesting thing about the Cauchy sequence definition is that it can be stated
without reference to real numbers at all:
We say a sequence of rational numbers {rn} is Cauchy (in the absolute value metric
on the rational numbers) if for every & EQ satisfying e > 0, there exists some MEN
such that for all n, k≥ M, we have rn - Tk < E.
Prove that is not Cauchy complete, that is, show that there exists a Cauchy sequence
{n} which does not converge to some limit r € Q.
(Hint: Look at problem 4b)
Let {dn} be a sequence of digits, that is dn = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for all n € N.
Let us consider the decimal
0.d₁d₂d3... := Σ
dn
10n
Show that the series is absolutely convergent to a number x = [0, 1].
(Hint: Use the comparison test.)
201](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F97e612ef-1556-436b-b62c-352b280e9e69%2Fb6bd003b-4ed5-4fc8-a7e9-10223694d865%2F33ntsc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:This question looks at the relationship between Bolzano- Weierstrass and the
"Cauchy completeness" property of R.
(a) Directly use the Bolzano-Weierstrass theorem (Theorem 2.3.8) to prove that every
Cauchy sequence of real numbers is convergent. That is, only make use of the fact
that every bounded sequence of real numbers has some convergent subsequence (not
necessarily converging to either lim sup/lim inf).
(Remark: Proving Cauchy-completeness from Bolzano-Weierstrass without lim sup/inf
in this way generalizes more easily to Rn.)
(b) One interesting thing about the Cauchy sequence definition is that it can be stated
without reference to real numbers at all:
We say a sequence of rational numbers {rn} is Cauchy (in the absolute value metric
on the rational numbers) if for every & EQ satisfying e > 0, there exists some MEN
such that for all n, k≥ M, we have rn - Tk < E.
Prove that is not Cauchy complete, that is, show that there exists a Cauchy sequence
{n} which does not converge to some limit r € Q.
(Hint: Look at problem 4b)
Let {dn} be a sequence of digits, that is dn = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for all n € N.
Let us consider the decimal
0.d₁d₂d3... := Σ
dn
10n
Show that the series is absolutely convergent to a number x = [0, 1].
(Hint: Use the comparison test.)
201
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