2. Is the proof correct? Either state that it is, or circle the first error and explain what is incorrect about it. If the proof is not correct, can it be fixed to prove the claim true? Claim: If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and S f(x) for r€ (0,1] l9(x) for r€ [1,2) f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) = is uniformly continuous on (0,2).
2. Is the proof correct? Either state that it is, or circle the first error and explain what is incorrect about it. If the proof is not correct, can it be fixed to prove the claim true? Claim: If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and S f(x) for r€ (0,1] l9(x) for r€ [1,2) f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) = is uniformly continuous on (0,2).
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.4: Ordered Integral Domains
Problem 7E: For an element x of an ordered integral domain D, the absolute value | x | is defined by | x |={...
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![2. Is the proof correct? Either state that it is, or circle the first error and explain what is
incorrect about it. If the proof is not correct, can it be fixed to prove the claim true?
Claim:
If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and
Sf(x) for r€ (0, 1]
[9(x) for r € [1, 2) '
f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) =
is uniformly continuous on (0,2).
Proof:
Let e >0.
Since f is uniformly continuous on (0, 1], there exists d > 0 such that if r, y € (0,1] and
|r – y| < d1, then |f(x) – f (y)| < e/2.
Since g is uniformly continuous on [1,2), there exists d2 > 0 such that if x, y E [1,2) and
|r – y| < d2, then [g(x) – g(y)| < €/2.
Let 8 = min{d1, 82}.
Now suppose r, y E (0, 2) with r < y and |x – y| < 8.
If r, y E (0, 1], then |r – y| < ô < 8i and so |h(x) – h (y)| = |f(x) – f (y)| < €/2 < e.
If x, y € [1,2), then |r – y| < d < dz and so h(x) – h(y)| = [g(x) – g(y)| < €/2 < e.
If r € (0, 1) and y E (1,2), then |r – 1|< |r – y| < 8 < ôị and |1 – y| < |r – y| < 8 < 82
so that
|h(x) – h(y)| = |f (1) – f(1) + g(1) – g(y)| < |f(x) – f(1)| + \g(1) – g(y)| < e/2+ €/2 = e.
So in all cases, h(x) – h(y)|< €, as needed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f41cb94-d8e0-4009-bc66-df82df6fba5b%2F5d67f22b-b71e-4f9b-a296-76c3758ee26d%2Fbolk5dj_processed.png&w=3840&q=75)
Transcribed Image Text:2. Is the proof correct? Either state that it is, or circle the first error and explain what is
incorrect about it. If the proof is not correct, can it be fixed to prove the claim true?
Claim:
If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and
Sf(x) for r€ (0, 1]
[9(x) for r € [1, 2) '
f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) =
is uniformly continuous on (0,2).
Proof:
Let e >0.
Since f is uniformly continuous on (0, 1], there exists d > 0 such that if r, y € (0,1] and
|r – y| < d1, then |f(x) – f (y)| < e/2.
Since g is uniformly continuous on [1,2), there exists d2 > 0 such that if x, y E [1,2) and
|r – y| < d2, then [g(x) – g(y)| < €/2.
Let 8 = min{d1, 82}.
Now suppose r, y E (0, 2) with r < y and |x – y| < 8.
If r, y E (0, 1], then |r – y| < ô < 8i and so |h(x) – h (y)| = |f(x) – f (y)| < €/2 < e.
If x, y € [1,2), then |r – y| < d < dz and so h(x) – h(y)| = [g(x) – g(y)| < €/2 < e.
If r € (0, 1) and y E (1,2), then |r – 1|< |r – y| < 8 < ôị and |1 – y| < |r – y| < 8 < 82
so that
|h(x) – h(y)| = |f (1) – f(1) + g(1) – g(y)| < |f(x) – f(1)| + \g(1) – g(y)| < e/2+ €/2 = e.
So in all cases, h(x) – h(y)|< €, as needed.
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