Every proper subspace of a normed linear space is either dense or nowhere dense.
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Every proper subspace of a normed linear space is either dense or nowhere dense.
Our instructor proved this theorem in our class. I have understood each and every step of the proof. But still I have a confusion. Confusion arose due the following example $:$
Consider the real line $Bbb R$ with the Euclidean norm. Consider the proper subspace $A=[-1,1] cap Bbb Q$ then this is neither dense nor nowhere dense. But if the example is true that would definitely violate the above theorem. What's going wrong here? Inspite of my effort I couldn't find out why is it happening. Please help me in this regard.
Thank you very much.
functional-analysis normed-spaces
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up vote
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Every proper subspace of a normed linear space is either dense or nowhere dense.
Our instructor proved this theorem in our class. I have understood each and every step of the proof. But still I have a confusion. Confusion arose due the following example $:$
Consider the real line $Bbb R$ with the Euclidean norm. Consider the proper subspace $A=[-1,1] cap Bbb Q$ then this is neither dense nor nowhere dense. But if the example is true that would definitely violate the above theorem. What's going wrong here? Inspite of my effort I couldn't find out why is it happening. Please help me in this regard.
Thank you very much.
functional-analysis normed-spaces
6
Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
â Kavi Rama Murthy
Aug 14 at 7:29
Yeah I have understood sir. Thank you very much for your help.
â Debabrata Chattopadhyay.
Aug 14 at 7:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Every proper subspace of a normed linear space is either dense or nowhere dense.
Our instructor proved this theorem in our class. I have understood each and every step of the proof. But still I have a confusion. Confusion arose due the following example $:$
Consider the real line $Bbb R$ with the Euclidean norm. Consider the proper subspace $A=[-1,1] cap Bbb Q$ then this is neither dense nor nowhere dense. But if the example is true that would definitely violate the above theorem. What's going wrong here? Inspite of my effort I couldn't find out why is it happening. Please help me in this regard.
Thank you very much.
functional-analysis normed-spaces
Every proper subspace of a normed linear space is either dense or nowhere dense.
Our instructor proved this theorem in our class. I have understood each and every step of the proof. But still I have a confusion. Confusion arose due the following example $:$
Consider the real line $Bbb R$ with the Euclidean norm. Consider the proper subspace $A=[-1,1] cap Bbb Q$ then this is neither dense nor nowhere dense. But if the example is true that would definitely violate the above theorem. What's going wrong here? Inspite of my effort I couldn't find out why is it happening. Please help me in this regard.
Thank you very much.
functional-analysis normed-spaces
asked Aug 14 at 7:26
Debabrata Chattopadhyay.
15612
15612
6
Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
â Kavi Rama Murthy
Aug 14 at 7:29
Yeah I have understood sir. Thank you very much for your help.
â Debabrata Chattopadhyay.
Aug 14 at 7:56
add a comment |Â
6
Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
â Kavi Rama Murthy
Aug 14 at 7:29
Yeah I have understood sir. Thank you very much for your help.
â Debabrata Chattopadhyay.
Aug 14 at 7:56
6
6
Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
â Kavi Rama Murthy
Aug 14 at 7:29
Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
â Kavi Rama Murthy
Aug 14 at 7:29
Yeah I have understood sir. Thank you very much for your help.
â Debabrata Chattopadhyay.
Aug 14 at 7:56
Yeah I have understood sir. Thank you very much for your help.
â Debabrata Chattopadhyay.
Aug 14 at 7:56
add a comment |Â
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6
Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
â Kavi Rama Murthy
Aug 14 at 7:29
Yeah I have understood sir. Thank you very much for your help.
â Debabrata Chattopadhyay.
Aug 14 at 7:56