Please Answer Exercise 2. Advanced Thankyouu?❤ Definition. A metric space (X, e) is a set X with a distance function or metric p satisfying the followingnmnertieeProperty(4) is called the triangle inequality.Example.1.Let X = R and p(x, y) = |x – yl, the usual absolute value in R. It is easy to verify that (R. p) is a metricspace. The defined p satisfies all the properties.(a) Since the absolute value only gives value of nonnegative real numbers, then it is guaranteed thatfor all x, y eR, we have 0s |x –- yl. Thus, 0s p(x,y).(b) Note that p(x,y) = |x – yl = 0 if and only if x - y = 0 and that is if and only if x = y.(c) For all x, y ER,P(x, y) = |x – yl = l(-1)(y – x)| = |(-1)lly – x| = ly – x| = p(y, x).(d) Recall that the triangle inequality for real numbers states that if a, bER, then la + b| < Ja| + |bl.For all real numbers x, y, z, we have,Study Guide in_ Real AnalysisFM-AA-CIA-15 Rev. 0 03-June-2020Math Elective 2 (Real Analysis)Module 2: Abstract Integration|x - yl = |(x – 2) + (z – y)| s lx – z| + |z –- yl.Thus, p(x, y) < p(x, z) + p(z. y).This is the Euclidean metric on R.2. Consider X to be a non-empty set and definep(x. y) = {6 fx =yWe check if the conditions for p to be a metric are satisfied.(a) Since the possible values of p are only 0 and 1, thus, 0s p(x, y).(b) The definition of p gives us that p(x, y) = 0 if and only if x = y.(c) Note that if x * y then y + x and similarly, if x = y then y = x. Thus, p(x, y) = p(y,x).(d) To prove triangle inequality for p, we have to considerthe cases of equality of any realnumbers x, y. z.(i)Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequalityp(x, y) S p(x,z) + p(z, y) (i)Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequalityp(x, y) < p(x, z) + p(z. y)holds.(ii)Let x * y + z. Then, p(x,y) = p(x, z) = p(z,y) = 1 andp(x, z) + p(z. y) = 1 +1 = 2.Thus, p(x,y) s p(x, z) + p(z, y).(ii)Let x + y and y = z. Then, p(x, y) = p(x, z) = 1, p(z, y) = 0, andp(x, z) + p(z, y) = 1+ 0 = 1.Thus, p(x, y) s p(x, z) + p(z, y).(iv)Let x = y and y # z. Then, p(x, y) = 0, p(x,z) = p(z, y) = 1 andp(x, z) + p(z, y) = 1 +1 = 2.PANGASINAN STATE UNIVERSITY3Study Guide inReal AnalysisFM-AA-CIA-15 Rev. 0 03-June-2020Math Elective 2 (Real Analysis)Module 2: Abstract IntegrationThus, p(x, y) s p(x, z) + p(z, y).Hence, (X, p) is a metric space.Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0,1], and the metric p is thesup-norm, defined asP(f, g) = sup If (x) – g(x)|.Show that (C[0, 1], p) is a metric space.

Name: Please Answer Exercise 2. Advanced Thankyouu?❤ Definition. A metric sp
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Description: Please Answer Exercise 2. Advanced Thankyouu?❤ Definition. A metric space (X, e) is a set X with a distance function or metric p satisfying the followingnmnertieeProperty(4) is called the triangle inequality.Example.1.Let X = R and p(x, y) = |x – yl,

A set (X,d) is called a metric space if these four properties are satisfied for all x, y, z ∈ X: d…

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Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0, 1], and the metric p is the sup-norm, defined as P(f, g) = sup If (x) – g(x)|. Show that (C[0, 1], p) is a metric space.

Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0, 1], and the metric p is the sup-norm, defined as P(f, g) = sup If (x) – g(x)|. Show that (C[0, 1], p) is a metric space.

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Please Answer Exercise 2. Advanced Thankyouu?❤

Definition. A metric space (X, e) is a set X with a distance function or metric p satisfying the following
nmnertiee
Property
(4) is called the triangle inequality.
Example.
1.
Let X = R and p(x, y) = |x – yl, the usual absolute value in R. It is easy to verify that (R. p) is a metric
space. The defined p satisfies all the properties.
(a) Since the absolute value only gives value of nonnegative real numbers, then it is guaranteed that
for all x, y eR, we have 0s |x –- yl. Thus, 0s p(x,y).
(b) Note that p(x,y) = |x – yl = 0 if and only if x - y = 0 and that is if and only if x = y.
(c) For all x, y ER,
P(x, y) = |x – yl = l(-1)(y – x)| = |(-1)lly – x| = ly – x| = p(y, x).
(d) Recall that the triangle inequality for real numbers states that if a, bER, then la + b| < Ja| + |bl.
For all real numbers x, y, z, we have,
Study Guide in_ Real Analysis
FM-AA-CIA-15 Rev. 0 03-June-2020
Math Elective 2 (Real Analysis)
Module 2: Abstract Integration
|x - yl = |(x – 2) + (z – y)| s lx – z| + |z –- yl.
Thus, p(x, y) < p(x, z) + p(z. y).
This is the Euclidean metric on R.
2. Consider X to be a non-empty set and define
p(x. y) = {6 fx =y
We check if the conditions for p to be a metric are satisfied.
(a) Since the possible values of p are only 0 and 1, thus, 0s p(x, y).
(b) The definition of p gives us that p(x, y) = 0 if and only if x = y.
(c) Note that if x * y then y + x and similarly, if x = y then y = x. Thus, p(x, y) = p(y,x).
(d) To prove triangle inequality for p, we have to consider
the cases of equality of any real
numbers x, y. z.
(i)
Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequality
p(x, y) S p(x,z) + p(z, y)

(i)
Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequality
p(x, y) < p(x, z) + p(z. y)
holds.
(ii)
Let x * y + z. Then, p(x,y) = p(x, z) = p(z,y) = 1 and
p(x, z) + p(z. y) = 1 +1 = 2.
Thus, p(x,y) s p(x, z) + p(z, y).
(ii)
Let x + y and y = z. Then, p(x, y) = p(x, z) = 1, p(z, y) = 0, and
p(x, z) + p(z, y) = 1+ 0 = 1.
Thus, p(x, y) s p(x, z) + p(z, y).
(iv)
Let x = y and y # z. Then, p(x, y) = 0, p(x,z) = p(z, y) = 1 and
p(x, z) + p(z, y) = 1 +1 = 2.
PANGASINAN STATE UNIVERSITY
3
Study Guide inReal Analysis
FM-AA-CIA-15 Rev. 0 03-June-2020
Math Elective 2 (Real Analysis)
Module 2: Abstract Integration
Thus, p(x, y) s p(x, z) + p(z, y).
Hence, (X, p) is a metric space.
Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0,1], and the metric p is the
sup-norm, defined as
P(f, g) = sup If (x) – g(x)|.
Show that (C[0, 1], p) is a metric space.

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