Consider the following proposed proof that if f: (0, 1) → R is uniformly continuous, then lim f(x) exists [special case of Proposition 3.4.6]. 2-0 1. Let {n} be a sequence in (0, 1) that converges to 0. In particular, {n} is a Cauchy sequence, so Lemma 3.4.5 implies that the image sequence {f(x)} is a Cauchy sequence, which by Theorem 2.4.5 converges to some limit in R. 2. If {n} is another sequence in (0, 1) that converges to 0, then the same reasoning shows that the sequence {f(yn)} potentially different from the limit of {f(x)}=1. converges to some limit, 3. Define a sequence {n}1 by setting 22k-1 equal to x and %2% equal to y for each natural number k. This sequence converges to 0, so the same reasoning as above implies that {f(zn)}1 converges. 4. By Proposition 2.1.17, all subsequences of a convergent sequence converge to the same limit. Since {f(x)}1 and {f(yn)}1 are subsequences of the convergent sequence {f(zn)}1, the sequence {f(yn)}a_1 converges to the same limit as does {f(x)}x=1. 5. Thus there exists a number L such that for every sequence {n} in (0, 1) converging to 0, the image sequence {f(an)}1 converges to L. So lim f(x) exists (and equals L) by Lemma 3.1.7. x-0 Which one of the following statements best describes this proposed proof? Step 4 is faulty. The proof is valid. Step 1 is faulty. Step 5 is faulty. Step 3 is faulty. Step 2 is faulty.
Consider the following proposed proof that if f: (0, 1) → R is uniformly continuous, then lim f(x) exists [special case of Proposition 3.4.6]. 2-0 1. Let {n} be a sequence in (0, 1) that converges to 0. In particular, {n} is a Cauchy sequence, so Lemma 3.4.5 implies that the image sequence {f(x)} is a Cauchy sequence, which by Theorem 2.4.5 converges to some limit in R. 2. If {n} is another sequence in (0, 1) that converges to 0, then the same reasoning shows that the sequence {f(yn)} potentially different from the limit of {f(x)}=1. converges to some limit, 3. Define a sequence {n}1 by setting 22k-1 equal to x and %2% equal to y for each natural number k. This sequence converges to 0, so the same reasoning as above implies that {f(zn)}1 converges. 4. By Proposition 2.1.17, all subsequences of a convergent sequence converge to the same limit. Since {f(x)}1 and {f(yn)}1 are subsequences of the convergent sequence {f(zn)}1, the sequence {f(yn)}a_1 converges to the same limit as does {f(x)}x=1. 5. Thus there exists a number L such that for every sequence {n} in (0, 1) converging to 0, the image sequence {f(an)}1 converges to L. So lim f(x) exists (and equals L) by Lemma 3.1.7. x-0 Which one of the following statements best describes this proposed proof? Step 4 is faulty. The proof is valid. Step 1 is faulty. Step 5 is faulty. Step 3 is faulty. Step 2 is faulty.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Consider the following proposed proof that if f: (0, 1) → R is uniformly
continuous, then lim f(x) exists [special case of Proposition 3.4.6].
x→0
1. Let {n} be a sequence in (0, 1) that converges to 0. In particular,
{n} is a Cauchy sequence, so Lemma 3.4.5 implies that the image
sequence {f(n)}₁ is a Cauchy sequence, which by Theorem 2.4.5
converges to some limit in R.
2. If {y} is another sequence in (0, 1) that converges to 0, then the same
reasoning shows that the sequence {f(yn)}₁ converges to some limit,
potentially different from the limit of {f(xn)}x=1-
3. Define a sequence {zn}=1 by setting 22k-1 equal to ï and 22k equal
to y for each natural number k. This sequence converges to 0, so the same
reasoning as above implies that {f(zn)}₁ converges.
4. By Proposition 2.1.17, all subsequences of a convergent sequence converge
to the same limit. Since {f(xn)}_1 and {f(yn)} are subsequences of
the convergent sequence {f(zn)}₁, the sequence {f(yn)}-1
n=1
converges to the same limit as does {f(n)}x=1·
5. Thus there exists a number L such that for every sequence {n}=1 in
n=1
(0, 1) converging to 0, the image sequence {f(x)}1 converges to L. So
lim f(x) exists (and equals L) by Lemma 3.1.7.
x→0
Which one of the following statements best describes this proposed proof?
Step 4 is faulty.
The proof is valid.
Step 1 is faulty.
Step 5 is faulty.
Step 3 is faulty.
Step 2 is faulty.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b19493d-2e0b-4edb-977e-29215d3bb0dd%2F42c0a9e9-b7b2-428f-a748-d0830bf92fa8%2F4y0lg8i_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the following proposed proof that if f: (0, 1) → R is uniformly
continuous, then lim f(x) exists [special case of Proposition 3.4.6].
x→0
1. Let {n} be a sequence in (0, 1) that converges to 0. In particular,
{n} is a Cauchy sequence, so Lemma 3.4.5 implies that the image
sequence {f(n)}₁ is a Cauchy sequence, which by Theorem 2.4.5
converges to some limit in R.
2. If {y} is another sequence in (0, 1) that converges to 0, then the same
reasoning shows that the sequence {f(yn)}₁ converges to some limit,
potentially different from the limit of {f(xn)}x=1-
3. Define a sequence {zn}=1 by setting 22k-1 equal to ï and 22k equal
to y for each natural number k. This sequence converges to 0, so the same
reasoning as above implies that {f(zn)}₁ converges.
4. By Proposition 2.1.17, all subsequences of a convergent sequence converge
to the same limit. Since {f(xn)}_1 and {f(yn)} are subsequences of
the convergent sequence {f(zn)}₁, the sequence {f(yn)}-1
n=1
converges to the same limit as does {f(n)}x=1·
5. Thus there exists a number L such that for every sequence {n}=1 in
n=1
(0, 1) converging to 0, the image sequence {f(x)}1 converges to L. So
lim f(x) exists (and equals L) by Lemma 3.1.7.
x→0
Which one of the following statements best describes this proposed proof?
Step 4 is faulty.
The proof is valid.
Step 1 is faulty.
Step 5 is faulty.
Step 3 is faulty.
Step 2 is faulty.
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