Totally bounded and closed implies compact??
Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $B(0,1)$ in $mathbb{R}^n$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact.
Am I right?
general-topology analysis
asked Dec 18 '18 at 14:07
Jiu
435111
add a comment |
Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $B(0,1)$ in $mathbb{R}^n$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact.
Am I right?
general-topology analysis
asked Dec 18 '18 at 14:07
Jiu
435111
1
" closed in itself " ?
– Yadati Kiran
Dec 18 '18 at 14:48
Are you suggesting $B(0,1)$ endowed with subspace topology of $mathbb{R}^n$ ?
– Yadati Kiran
Dec 18 '18 at 14:55
@YadatiKiran yes
– Jiu
Dec 18 '18 at 14:56
it is not closed in $mathbb R^n$ which is presumably what is implied...
– Carmeister
Dec 18 '18 at 19:15
@Carmeister but one does not write simply “metric space” and imply that it is $mathbb{R}^n$.
– Jiu
Dec 19 '18 at 3:01
add a comment |
Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $B(0,1)$ in $mathbb{R}^n$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact.
Am I right?
general-topology analysis
asked Dec 18 '18 at 14:07
Jiu
435111
Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $B(0,1)$ in $mathbb{R}^n$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact.
Am I right?
general-topology analysis
general-topology analysis
asked Dec 18 '18 at 14:07
Jiu
435111
asked Dec 18 '18 at 14:07
Jiu
435111
asked Dec 18 '18 at 14:07
Jiu
435111
asked Dec 18 '18 at 14:07
Jiu
435111
asked Dec 18 '18 at 14:07
Jiu
435111
435111
1
" closed in itself " ?
– Yadati Kiran
Dec 18 '18 at 14:48
Are you suggesting $B(0,1)$ endowed with subspace topology of $mathbb{R}^n$ ?
– Yadati Kiran
Dec 18 '18 at 14:55
@YadatiKiran yes
– Jiu
Dec 18 '18 at 14:56
it is not closed in $mathbb R^n$ which is presumably what is implied...
– Carmeister
Dec 18 '18 at 19:15
@Carmeister but one does not write simply “metric space” and imply that it is $mathbb{R}^n$.
– Jiu
Dec 19 '18 at 3:01
add a comment |
1
" closed in itself " ?
– Yadati Kiran
Dec 18 '18 at 14:48
Are you suggesting $B(0,1)$ endowed with subspace topology of $mathbb{R}^n$ ?
– Yadati Kiran
Dec 18 '18 at 14:55
@YadatiKiran yes
– Jiu
Dec 18 '18 at 14:56
it is not closed in $mathbb R^n$ which is presumably what is implied...
– Carmeister
Dec 18 '18 at 19:15
@Carmeister but one does not write simply “metric space” and imply that it is $mathbb{R}^n$.
– Jiu
Dec 19 '18 at 3:01
1
1
" closed in itself " ?
– Yadati Kiran
Dec 18 '18 at 14:48
" closed in itself " ?
– Yadati Kiran
Dec 18 '18 at 14:48
Are you suggesting $B(0,1)$ endowed with subspace topology of $mathbb{R}^n$ ?
– Yadati Kiran
Dec 18 '18 at 14:55
Are you suggesting $B(0,1)$ endowed with subspace topology of $mathbb{R}^n$ ?
– Yadati Kiran
Dec 18 '18 at 14:55
@YadatiKiran yes
– Jiu
Dec 18 '18 at 14:56
@YadatiKiran yes
– Jiu
Dec 18 '18 at 14:56
it is not closed in $mathbb R^n$ which is presumably what is implied...
– Carmeister
Dec 18 '18 at 19:15
it is not closed in $mathbb R^n$ which is presumably what is implied...
– Carmeister
Dec 18 '18 at 19:15
@Carmeister but one does not write simply “metric space” and imply that it is $mathbb{R}^n$.
– Jiu
Dec 19 '18 at 3:01
@Carmeister but one does not write simply “metric space” and imply that it is $mathbb{R}^n$.
– Jiu
Dec 19 '18 at 3:01
add a comment |
3 Answers
3
active
oldest
votes
You are right.
For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
3
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
add a comment |
$B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $mathbb{R}^n$. But this is true for every $S subset mathbb{R}^n$ if we endow $S$ with the subspace topology.
The requirement is that the set is closed in $mathbb{R}^n$, not in itself.
Moreover, in a generic tolopogical space X, given $A subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.
answered Dec 18 '18 at 15:56
Hermione
18119
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
add a comment |
Hint: what is the bound on $B(0,1)$ in $B(0,1)$?
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
answered Dec 18 '18 at 14:34
WalterJ
815611
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are right.
For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
3
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
add a comment |
You are right.
For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
3
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
add a comment |
You are right.
For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
You are right.
For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
answered Dec 18 '18 at 15:23
BigbearZzz
7,24921648
7,24921648
3
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
add a comment |
3
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
3
3
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context.
– Paul Sinclair
Dec 18 '18 at 17:10
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check.
– BigbearZzz
Dec 18 '18 at 17:18
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Dec 18 '18 at 23:22
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
@Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is.
– Paul Sinclair
Dec 18 '18 at 23:58
add a comment |
$B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $mathbb{R}^n$. But this is true for every $S subset mathbb{R}^n$ if we endow $S$ with the subspace topology.
The requirement is that the set is closed in $mathbb{R}^n$, not in itself.
Moreover, in a generic tolopogical space X, given $A subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.
answered Dec 18 '18 at 15:56
Hermione
18119
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
add a comment |
$B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $mathbb{R}^n$. But this is true for every $S subset mathbb{R}^n$ if we endow $S$ with the subspace topology.
The requirement is that the set is closed in $mathbb{R}^n$, not in itself.
Moreover, in a generic tolopogical space X, given $A subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.
answered Dec 18 '18 at 15:56
Hermione
18119
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
add a comment |
$B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $mathbb{R}^n$. But this is true for every $S subset mathbb{R}^n$ if we endow $S$ with the subspace topology.
The requirement is that the set is closed in $mathbb{R}^n$, not in itself.
Moreover, in a generic tolopogical space X, given $A subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.
answered Dec 18 '18 at 15:56
Hermione
18119
$B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $mathbb{R}^n$. But this is true for every $S subset mathbb{R}^n$ if we endow $S$ with the subspace topology.
The requirement is that the set is closed in $mathbb{R}^n$, not in itself.
Moreover, in a generic tolopogical space X, given $A subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.
answered Dec 18 '18 at 15:56
Hermione
18119
answered Dec 18 '18 at 15:56
Hermione
18119
answered Dec 18 '18 at 15:56
Hermione
18119
answered Dec 18 '18 at 15:56
Hermione
18119
18119
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
add a comment |
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Dec 19 '18 at 3:04
add a comment |
Hint: what is the bound on $B(0,1)$ in $B(0,1)$?
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
answered Dec 18 '18 at 14:34
WalterJ
815611
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
add a comment |
Hint: what is the bound on $B(0,1)$ in $B(0,1)$?
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
answered Dec 18 '18 at 14:34
WalterJ
815611
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
add a comment |
Hint: what is the bound on $B(0,1)$ in $B(0,1)$?
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
answered Dec 18 '18 at 14:34
WalterJ
815611
Hint: what is the bound on $B(0,1)$ in $B(0,1)$?
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
answered Dec 18 '18 at 14:34
WalterJ
815611
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
edited Dec 18 '18 at 14:57
Davide Giraudo
125k16150259
125k16150259
answered Dec 18 '18 at 14:34
WalterJ
815611
answered Dec 18 '18 at 14:34
WalterJ
815611
answered Dec 18 '18 at 14:34
WalterJ
815611
815611
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
add a comment |
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
Are you suggesting that $B(0,1)$ is not totally bounded?
– Toby Bartels
Dec 18 '18 at 21:47
add a comment |
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1
" closed in itself " ?
– Yadati Kiran
Dec 18 '18 at 14:48
Are you suggesting $B(0,1)$ endowed with subspace topology of $mathbb{R}^n$ ?
– Yadati Kiran
Dec 18 '18 at 14:55
@YadatiKiran yes
– Jiu
Dec 18 '18 at 14:56
it is not closed in $mathbb R^n$ which is presumably what is implied...
– Carmeister
Dec 18 '18 at 19:15
@Carmeister but one does not write simply “metric space” and imply that it is $mathbb{R}^n$.
– Jiu
Dec 19 '18 at 3:01