THEOREM 5.7. (1) If a set A is relatively compact, it is also totally bounded, in the metric space (X, d). (2) If (X, d) is complete and A is totally bounded, then A is relatively compact. ® How does one of the spheres have infinite no of xn Ⓒ How do diagonal elements yield couchy seq 7. © Request explain this step (1) Suppose A is relatively compact. We can apply the preceding theorem to assert that, from any sequence in A, a convergent subsequence can be extracted. We need only apply Theorem 5.3 now to conclude that A is totally bounded. (2) Let (X, d) be a complete metric space and let A be a totally bounded subset of X. By Theorem 5.6 we must show that it is possible to extract a convergent subsequence from every sequence of points in A. To this end, let {x} be a sequence of points from A. Since A is assumed to be totally bounded, for each &= 1/k, where k is a positive integer, there must be a finite e-net: {y(), y(), . Considering the collection of spheres of radius 1 about each point in the k = 1 net, we see that this collection must cover all of A. In particular, the x, (n = 1, 2, ...) must be contained in a finite number of these spheres. Further, one of these spheres must contain an infinite number of the x,. Hence we have some sphere, S, say, containing the subsequence of {x}: x₁(1), x₂(1), xy(1), ..., where x₁(¹) € {x₁, x₂,...} (for i=1, 2, ...). Reasoning along the same lines for the collection of neighborhoods of radius about the points in the k= 2 net, we can get some sphere of radius from this collection containing infinitely many of the x¹). Thus, calling this sphere S₂ and denoting the points it contains by x, (2), x₂(2), ..., we have S₂: x₁(2), x₂(²), x₂(2), ..., where x₁ (²) € (X₁, X₂, ...) Proceeding in this fashion, we arrive at Fig. 5.2. (i=1,2,...). S₁: x₁,x₂,... S₂: x₂ FIG. 5.2. It is now clear that selecting the “diagonal" terms (see arrows in Fig. 5.2, the sequence {x}) yields a Cauchy sequence which, since the space is complete, must have a limit in X. Since each of the sequences {x} (n is the index here, while i is fixed) is a subsequence of {x-1)} it is clear that {x} is a subsequence (3 of the original sequence {x}. Hence we need only apply the preceding theorem to complete the proof.
THEOREM 5.7. (1) If a set A is relatively compact, it is also totally bounded, in the metric space (X, d). (2) If (X, d) is complete and A is totally bounded, then A is relatively compact. ® How does one of the spheres have infinite no of xn Ⓒ How do diagonal elements yield couchy seq 7. © Request explain this step (1) Suppose A is relatively compact. We can apply the preceding theorem to assert that, from any sequence in A, a convergent subsequence can be extracted. We need only apply Theorem 5.3 now to conclude that A is totally bounded. (2) Let (X, d) be a complete metric space and let A be a totally bounded subset of X. By Theorem 5.6 we must show that it is possible to extract a convergent subsequence from every sequence of points in A. To this end, let {x} be a sequence of points from A. Since A is assumed to be totally bounded, for each &= 1/k, where k is a positive integer, there must be a finite e-net: {y(), y(), . Considering the collection of spheres of radius 1 about each point in the k = 1 net, we see that this collection must cover all of A. In particular, the x, (n = 1, 2, ...) must be contained in a finite number of these spheres. Further, one of these spheres must contain an infinite number of the x,. Hence we have some sphere, S, say, containing the subsequence of {x}: x₁(1), x₂(1), xy(1), ..., where x₁(¹) € {x₁, x₂,...} (for i=1, 2, ...). Reasoning along the same lines for the collection of neighborhoods of radius about the points in the k= 2 net, we can get some sphere of radius from this collection containing infinitely many of the x¹). Thus, calling this sphere S₂ and denoting the points it contains by x, (2), x₂(2), ..., we have S₂: x₁(2), x₂(²), x₂(2), ..., where x₁ (²) € (X₁, X₂, ...) Proceeding in this fashion, we arrive at Fig. 5.2. (i=1,2,...). S₁: x₁,x₂,... S₂: x₂ FIG. 5.2. It is now clear that selecting the “diagonal" terms (see arrows in Fig. 5.2, the sequence {x}) yields a Cauchy sequence which, since the space is complete, must have a limit in X. Since each of the sequences {x} (n is the index here, while i is fixed) is a subsequence of {x-1)} it is clear that {x} is a subsequence (3 of the original sequence {x}. Hence we need only apply the preceding theorem to complete the proof.
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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Question
Request explain how does a sphere contain infinite xn , how do diagonal elements give a cauchy sequence and step 3
![THEOREM 5.7.
(1) If a set A is relatively compact, it is also totally bounded, in the metric space
(X, d).
(2) If (X, d) is complete and A is totally bounded, then A is relatively compact.
□ How does one of the spheres have
infinite no of xn
a
☺ How do diagonal elements yield
couchy seq, 7.
( Request explain this atefe
step
Proof.
(1) Suppose A is relatively compact. We can apply the preceding theorem to
assert that, from any sequence in A, a convergent subsequence can be extracted.
We need only apply Theorem 5.3 now to conclude that A is totally bounded.
(2) Let (X, d) be a complete metric space and let A be a totally bounded subset
of X. By Theorem 5.6 we must show that it is possible to extract a convergent
subsequence from every sequence of points in A. To this end, let {x} be a sequence
of points from A. Since A is assumed to be totally bounded, for each & = 1/k,
where k is a positive integer, there must be a finite &-net:
{y₁(k), y₂), ..., y()}.
Considering the collection of spheres of radius 1 about each point in the k = 1
net, we see that this collection must cover all of A. In particular, the x, (n = 1, 2, ...)
must be contained in a finite number of these spheres. Further, one of these spheres
must contain an infinite number of the x,. Hence we have some sphere, S₁ say,
containing the subsequence of {x,}:
x₁(¹), x₂(¹), x3(¹), ...,
where
x (¹) € {x₁, x₂,...}
(for i = 1, 2, ...).
Reasoning along the same lines for the collection of neighborhoods of radius
about the points in the k = 2 net, we can get some sphere of radius from this
collection containing infinitely many of the x¹). Thus, calling this sphere S₂ and
denoting the points it contains by x₁ (2), x₂(2), ...,
we have
S₂: x₁(²), x₂(²), x₂(²), ...,
where
x₁ (²) € {x₁, x₂,...}
Proceeding in this fashion, we arrive at Fig. 5.2.
S₁: x₁(¹), x₂ (¹)
2
S₂: x₂(2)x²₂ (2)
(i=1,2,...).
S₁: x₂ (n), x₂ (n)...
FIG. 5.2.
2
It is now clear that selecting the "diagonal” terms (see arrows in Fig. 5.2, the
sequence {x}) yields a Cauchy sequence which, since the space is complete,
must have a limit in X. Since each of the sequences {x} (n is the index here,
while i is fixed) is a subsequence of {x(i-¹)} it is clear that {x} is a subsequence
of the original sequence {x}. Hence we need only apply the preceding theorem to
complete the proof.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F0144b026-fef2-4c1f-a75f-91db7354b88c%2F7dk6mut_processed.png&w=3840&q=75)
Transcribed Image Text:THEOREM 5.7.
(1) If a set A is relatively compact, it is also totally bounded, in the metric space
(X, d).
(2) If (X, d) is complete and A is totally bounded, then A is relatively compact.
□ How does one of the spheres have
infinite no of xn
a
☺ How do diagonal elements yield
couchy seq, 7.
( Request explain this atefe
step
Proof.
(1) Suppose A is relatively compact. We can apply the preceding theorem to
assert that, from any sequence in A, a convergent subsequence can be extracted.
We need only apply Theorem 5.3 now to conclude that A is totally bounded.
(2) Let (X, d) be a complete metric space and let A be a totally bounded subset
of X. By Theorem 5.6 we must show that it is possible to extract a convergent
subsequence from every sequence of points in A. To this end, let {x} be a sequence
of points from A. Since A is assumed to be totally bounded, for each & = 1/k,
where k is a positive integer, there must be a finite &-net:
{y₁(k), y₂), ..., y()}.
Considering the collection of spheres of radius 1 about each point in the k = 1
net, we see that this collection must cover all of A. In particular, the x, (n = 1, 2, ...)
must be contained in a finite number of these spheres. Further, one of these spheres
must contain an infinite number of the x,. Hence we have some sphere, S₁ say,
containing the subsequence of {x,}:
x₁(¹), x₂(¹), x3(¹), ...,
where
x (¹) € {x₁, x₂,...}
(for i = 1, 2, ...).
Reasoning along the same lines for the collection of neighborhoods of radius
about the points in the k = 2 net, we can get some sphere of radius from this
collection containing infinitely many of the x¹). Thus, calling this sphere S₂ and
denoting the points it contains by x₁ (2), x₂(2), ...,
we have
S₂: x₁(²), x₂(²), x₂(²), ...,
where
x₁ (²) € {x₁, x₂,...}
Proceeding in this fashion, we arrive at Fig. 5.2.
S₁: x₁(¹), x₂ (¹)
2
S₂: x₂(2)x²₂ (2)
(i=1,2,...).
S₁: x₂ (n), x₂ (n)...
FIG. 5.2.
2
It is now clear that selecting the "diagonal” terms (see arrows in Fig. 5.2, the
sequence {x}) yields a Cauchy sequence which, since the space is complete,
must have a limit in X. Since each of the sequences {x} (n is the index here,
while i is fixed) is a subsequence of {x(i-¹)} it is clear that {x} is a subsequence
of the original sequence {x}. Hence we need only apply the preceding theorem to
complete the proof.
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