norm of element in equivalent class in quotient space
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If we have a quotient space $Ebackslash L_0$ where $E$ is a linear normed space and $L_0$ it's subspace the norm of an element $L$ in $Ebackslash L_0$ is defined as $$lVert LrVert = inf_{x in L}{lVert xrVert}$$ What I'm trying to understand better is why we can find an element $xin L$ s.t. $lVert x rVert < lVert L rVert + epsilon$. I understand that such element is not going to be an infimum, but how are we sure that this element is going to be in $L$? Does is have something to do with the fact that $L$ is a closed set?
Thanks in advance!
vector-spaces norm supremum-and-infimum quotient-spaces
asked Nov 15 at 12:12
Nikola
709618
add a comment |
up vote
0
down vote
favorite
If we have a quotient space $Ebackslash L_0$ where $E$ is a linear normed space and $L_0$ it's subspace the norm of an element $L$ in $Ebackslash L_0$ is defined as $$lVert LrVert = inf_{x in L}{lVert xrVert}$$ What I'm trying to understand better is why we can find an element $xin L$ s.t. $lVert x rVert < lVert L rVert + epsilon$. I understand that such element is not going to be an infimum, but how are we sure that this element is going to be in $L$? Does is have something to do with the fact that $L$ is a closed set?
Thanks in advance!
vector-spaces norm supremum-and-infimum quotient-spaces
asked Nov 15 at 12:12
Nikola
709618
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we have a quotient space $Ebackslash L_0$ where $E$ is a linear normed space and $L_0$ it's subspace the norm of an element $L$ in $Ebackslash L_0$ is defined as $$lVert LrVert = inf_{x in L}{lVert xrVert}$$ What I'm trying to understand better is why we can find an element $xin L$ s.t. $lVert x rVert < lVert L rVert + epsilon$. I understand that such element is not going to be an infimum, but how are we sure that this element is going to be in $L$? Does is have something to do with the fact that $L$ is a closed set?
Thanks in advance!
vector-spaces norm supremum-and-infimum quotient-spaces
asked Nov 15 at 12:12
Nikola
709618
If we have a quotient space $Ebackslash L_0$ where $E$ is a linear normed space and $L_0$ it's subspace the norm of an element $L$ in $Ebackslash L_0$ is defined as $$lVert LrVert = inf_{x in L}{lVert xrVert}$$ What I'm trying to understand better is why we can find an element $xin L$ s.t. $lVert x rVert < lVert L rVert + epsilon$. I understand that such element is not going to be an infimum, but how are we sure that this element is going to be in $L$? Does is have something to do with the fact that $L$ is a closed set?
Thanks in advance!
vector-spaces norm supremum-and-infimum quotient-spaces
vector-spaces norm supremum-and-infimum quotient-spaces
asked Nov 15 at 12:12
Nikola
709618
asked Nov 15 at 12:12
Nikola
709618
asked Nov 15 at 12:12
Nikola
709618
asked Nov 15 at 12:12
Nikola
709618
asked Nov 15 at 12:12
Nikola
709618
709618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
This is immediate from definition of infimum. If this is not true then $|x| geq |L|+epsilon$ for all $x in L$ which means $|L|+epsilon$ is a lower bound; hence the greatest lower bound $|L|$ must be equal to or exceed this, i.e. $|L| geq |L|+epsilon$ which is a contardiction.
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
1
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is immediate from definition of infimum. If this is not true then $|x| geq |L|+epsilon$ for all $x in L$ which means $|L|+epsilon$ is a lower bound; hence the greatest lower bound $|L|$ must be equal to or exceed this, i.e. $|L| geq |L|+epsilon$ which is a contardiction.
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
1
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
add a comment |
up vote
2
down vote
accepted
This is immediate from definition of infimum. If this is not true then $|x| geq |L|+epsilon$ for all $x in L$ which means $|L|+epsilon$ is a lower bound; hence the greatest lower bound $|L|$ must be equal to or exceed this, i.e. $|L| geq |L|+epsilon$ which is a contardiction.
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
1
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is immediate from definition of infimum. If this is not true then $|x| geq |L|+epsilon$ for all $x in L$ which means $|L|+epsilon$ is a lower bound; hence the greatest lower bound $|L|$ must be equal to or exceed this, i.e. $|L| geq |L|+epsilon$ which is a contardiction.
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
This is immediate from definition of infimum. If this is not true then $|x| geq |L|+epsilon$ for all $x in L$ which means $|L|+epsilon$ is a lower bound; hence the greatest lower bound $|L|$ must be equal to or exceed this, i.e. $|L| geq |L|+epsilon$ which is a contardiction.
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
answered Nov 15 at 12:17
Kavi Rama Murthy
43k31751
43k31751
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
1
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
add a comment |
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
1
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
I get that it's coming from the definition of an infimum. What confuses me is that we can always find an element $x in L$ satisfying every $epsilon > 0$. It's like there are no gaps between the elements in $L$.
– Nikola
Nov 15 at 12:20
1
1
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
Ask yourself what happens if there is no such $x$. You will get a contradiction. That is what I have done in my answer.
– Kavi Rama Murthy
Nov 15 at 12:21
add a comment |
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