Is $BbbQ$ complete as a normed space? [closed]
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I do know that finite dimensional normed vector spaces are complete.
But $mathbbQ^n$ is a finite dimensional normed vector space with (e.g.) the usual norm:
$$ |(q_1, cdots, q_n)| = sqrtq_1^2 + cdots + q_n^2$$
and is not complete. What's wrong with my example $(mathbb Q^n, |cdot|)$?
functional-analysis normed-spaces complete-spaces
edited Aug 13 at 16:15
John Ma
37.5k93669
asked Aug 13 at 4:43
maple
8282922
closed as off-topic by user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce Aug 13 at 14:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce
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up vote
-2
down vote
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I do know that finite dimensional normed vector spaces are complete.
But $mathbbQ^n$ is a finite dimensional normed vector space with (e.g.) the usual norm:
$$ |(q_1, cdots, q_n)| = sqrtq_1^2 + cdots + q_n^2$$
and is not complete. What's wrong with my example $(mathbb Q^n, |cdot|)$?
functional-analysis normed-spaces complete-spaces
edited Aug 13 at 16:15
John Ma
37.5k93669
asked Aug 13 at 4:43
maple
8282922
closed as off-topic by user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce Aug 13 at 14:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce
Could $mathbb Q$ be normed ?
– xbh
Aug 13 at 4:45
@xbh Could you please give more details?
– maple
Aug 13 at 4:55
A vector space comes with its scalar field. Are you taking $mathbbR$ or $mathbbQ$ ?
– ippiki-ookami
Aug 13 at 4:57
So, if you take $mathbbR$ then how can you make sure you have homogeneity ?
– ippiki-ookami
Aug 13 at 4:58
4
Question you linked says "...be a vector space over a complete topological field", which $mathbbQ$ is not.
– Serge Seredenko
Aug 13 at 8:38
 |Â
show 1 more comment
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I do know that finite dimensional normed vector spaces are complete.
But $mathbbQ^n$ is a finite dimensional normed vector space with (e.g.) the usual norm:
$$ |(q_1, cdots, q_n)| = sqrtq_1^2 + cdots + q_n^2$$
and is not complete. What's wrong with my example $(mathbb Q^n, |cdot|)$?
functional-analysis normed-spaces complete-spaces
edited Aug 13 at 16:15
John Ma
37.5k93669
asked Aug 13 at 4:43
maple
8282922
I do know that finite dimensional normed vector spaces are complete.
But $mathbbQ^n$ is a finite dimensional normed vector space with (e.g.) the usual norm:
$$ |(q_1, cdots, q_n)| = sqrtq_1^2 + cdots + q_n^2$$
and is not complete. What's wrong with my example $(mathbb Q^n, |cdot|)$?
functional-analysis normed-spaces complete-spaces
edited Aug 13 at 16:15
John Ma
37.5k93669
asked Aug 13 at 4:43
maple
8282922
edited Aug 13 at 16:15
John Ma
37.5k93669
edited Aug 13 at 16:15
John Ma
37.5k93669
edited Aug 13 at 16:15
John Ma
37.5k93669
37.5k93669
asked Aug 13 at 4:43
maple
8282922
asked Aug 13 at 4:43
maple
8282922
asked Aug 13 at 4:43
maple
8282922
8282922
closed as off-topic by user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce Aug 13 at 14:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce
closed as off-topic by user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce Aug 13 at 14:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, TheSimpliFire, amWhy, Mike Pierce
Could $mathbb Q$ be normed ?
– xbh
Aug 13 at 4:45
@xbh Could you please give more details?
– maple
Aug 13 at 4:55
A vector space comes with its scalar field. Are you taking $mathbbR$ or $mathbbQ$ ?
– ippiki-ookami
Aug 13 at 4:57
So, if you take $mathbbR$ then how can you make sure you have homogeneity ?
– ippiki-ookami
Aug 13 at 4:58
4
Question you linked says "...be a vector space over a complete topological field", which $mathbbQ$ is not.
– Serge Seredenko
Aug 13 at 8:38
 |Â
show 1 more comment
Could $mathbb Q$ be normed ?
– xbh
Aug 13 at 4:45
@xbh Could you please give more details?
– maple
Aug 13 at 4:55
A vector space comes with its scalar field. Are you taking $mathbbR$ or $mathbbQ$ ?
– ippiki-ookami
Aug 13 at 4:57
So, if you take $mathbbR$ then how can you make sure you have homogeneity ?
– ippiki-ookami
Aug 13 at 4:58
4
Question you linked says "...be a vector space over a complete topological field", which $mathbbQ$ is not.
– Serge Seredenko
Aug 13 at 8:38
Could $mathbb Q$ be normed ?
– xbh
Aug 13 at 4:45
Could $mathbb Q$ be normed ?
– xbh
Aug 13 at 4:45
@xbh Could you please give more details?
– maple
Aug 13 at 4:55
@xbh Could you please give more details?
– maple
Aug 13 at 4:55
A vector space comes with its scalar field. Are you taking $mathbbR$ or $mathbbQ$ ?
– ippiki-ookami
Aug 13 at 4:57
A vector space comes with its scalar field. Are you taking $mathbbR$ or $mathbbQ$ ?
– ippiki-ookami
Aug 13 at 4:57
So, if you take $mathbbR$ then how can you make sure you have homogeneity ?
– ippiki-ookami
Aug 13 at 4:58
So, if you take $mathbbR$ then how can you make sure you have homogeneity ?
– ippiki-ookami
Aug 13 at 4:58
4
4
Question you linked says "...be a vector space over a complete topological field", which $mathbbQ$ is not.
– Serge Seredenko
Aug 13 at 8:38
Question you linked says "...be a vector space over a complete topological field", which $mathbbQ$ is not.
– Serge Seredenko
Aug 13 at 8:38
 |Â
show 1 more comment
2 Answers
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5
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The proof uses that the field of coefficients (usually $mathbb R $ or $mathbb C $) is complete. And of course if you consider $mathbb Q $ as a $mathbb Q $-vector space, it is not complete; and it is not a real or complex space.
answered Aug 13 at 4:59
Martin Argerami
116k1071165
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
1
What statement?
– Martin Argerami
Aug 13 at 11:04
add a comment |Â
up vote
2
down vote
Typically, a normed linear space is defined to be over one of the scalar fields $mathbbR$ or $mathbbC$, and $mathbbQ$ is not a vector space over either of these fields$^*$. So no, it's not really a counter-example.
$^*$ No countably infinite set could be; a non-zero vector $v$ implies that $lambda v$ is also a vector, and distinct for distinct scalars $lambda$. That is, so long as the vector space isn't trivial, it must contain at least as many elements as the scalar field.
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The proof uses that the field of coefficients (usually $mathbb R $ or $mathbb C $) is complete. And of course if you consider $mathbb Q $ as a $mathbb Q $-vector space, it is not complete; and it is not a real or complex space.
answered Aug 13 at 4:59
Martin Argerami
116k1071165
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
1
What statement?
– Martin Argerami
Aug 13 at 11:04
add a comment |Â
up vote
5
down vote
The proof uses that the field of coefficients (usually $mathbb R $ or $mathbb C $) is complete. And of course if you consider $mathbb Q $ as a $mathbb Q $-vector space, it is not complete; and it is not a real or complex space.
answered Aug 13 at 4:59
Martin Argerami
116k1071165
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
1
What statement?
– Martin Argerami
Aug 13 at 11:04
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The proof uses that the field of coefficients (usually $mathbb R $ or $mathbb C $) is complete. And of course if you consider $mathbb Q $ as a $mathbb Q $-vector space, it is not complete; and it is not a real or complex space.
answered Aug 13 at 4:59
Martin Argerami
116k1071165
The proof uses that the field of coefficients (usually $mathbb R $ or $mathbb C $) is complete. And of course if you consider $mathbb Q $ as a $mathbb Q $-vector space, it is not complete; and it is not a real or complex space.
answered Aug 13 at 4:59
Martin Argerami
116k1071165
answered Aug 13 at 4:59
Martin Argerami
116k1071165
answered Aug 13 at 4:59
Martin Argerami
116k1071165
answered Aug 13 at 4:59
Martin Argerami
116k1071165
116k1071165
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
1
What statement?
– Martin Argerami
Aug 13 at 11:04
add a comment |Â
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
1
What statement?
– Martin Argerami
Aug 13 at 11:04
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
So, could you please tell me what's wrong with the statement?
– maple
Aug 13 at 5:45
1
1
What statement?
– Martin Argerami
Aug 13 at 11:04
What statement?
– Martin Argerami
Aug 13 at 11:04
add a comment |Â
up vote
2
down vote
Typically, a normed linear space is defined to be over one of the scalar fields $mathbbR$ or $mathbbC$, and $mathbbQ$ is not a vector space over either of these fields$^*$. So no, it's not really a counter-example.
$^*$ No countably infinite set could be; a non-zero vector $v$ implies that $lambda v$ is also a vector, and distinct for distinct scalars $lambda$. That is, so long as the vector space isn't trivial, it must contain at least as many elements as the scalar field.
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
add a comment |Â
up vote
2
down vote
Typically, a normed linear space is defined to be over one of the scalar fields $mathbbR$ or $mathbbC$, and $mathbbQ$ is not a vector space over either of these fields$^*$. So no, it's not really a counter-example.
$^*$ No countably infinite set could be; a non-zero vector $v$ implies that $lambda v$ is also a vector, and distinct for distinct scalars $lambda$. That is, so long as the vector space isn't trivial, it must contain at least as many elements as the scalar field.
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Typically, a normed linear space is defined to be over one of the scalar fields $mathbbR$ or $mathbbC$, and $mathbbQ$ is not a vector space over either of these fields$^*$. So no, it's not really a counter-example.
$^*$ No countably infinite set could be; a non-zero vector $v$ implies that $lambda v$ is also a vector, and distinct for distinct scalars $lambda$. That is, so long as the vector space isn't trivial, it must contain at least as many elements as the scalar field.
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
Typically, a normed linear space is defined to be over one of the scalar fields $mathbbR$ or $mathbbC$, and $mathbbQ$ is not a vector space over either of these fields$^*$. So no, it's not really a counter-example.
$^*$ No countably infinite set could be; a non-zero vector $v$ implies that $lambda v$ is also a vector, and distinct for distinct scalars $lambda$. That is, so long as the vector space isn't trivial, it must contain at least as many elements as the scalar field.
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
answered Aug 13 at 6:12
Theo Bendit
12.2k1844
12.2k1844
add a comment |Â
add a comment |Â
Could $mathbb Q$ be normed ?
– xbh
Aug 13 at 4:45
@xbh Could you please give more details?
– maple
Aug 13 at 4:55
A vector space comes with its scalar field. Are you taking $mathbbR$ or $mathbbQ$ ?
– ippiki-ookami
Aug 13 at 4:57
So, if you take $mathbbR$ then how can you make sure you have homogeneity ?
– ippiki-ookami
Aug 13 at 4:58
4
Question you linked says "...be a vector space over a complete topological field", which $mathbbQ$ is not.
– Serge Seredenko
Aug 13 at 8:38