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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  • 6




    Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
    – Kavi Rama Murthy
    Aug 14 at 7:29










  • Yeah I have understood sir. Thank you very much for your help.
    – Debabrata Chattopadhyay.
    Aug 14 at 7:56














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




share|cite|improve this question
















  • 6




    Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
    – Kavi Rama Murthy
    Aug 14 at 7:29










  • Yeah I have understood sir. Thank you very much for your help.
    – Debabrata Chattopadhyay.
    Aug 14 at 7:56












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




share|cite|improve this question














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





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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.
    – Kavi Rama Murthy
    Aug 14 at 7:29










  • Yeah I have understood sir. Thank you very much for your help.
    – Debabrata Chattopadhyay.
    Aug 14 at 7:56












  • 6




    Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
    – Kavi Rama Murthy
    Aug 14 at 7:29










  • Yeah I have understood sir. Thank you very much for your help.
    – 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.
– Kavi Rama Murthy
Aug 14 at 7:29




Subspace in this context is a linear subspace. Your set $A$ is not a vector space.
– Kavi Rama Murthy
Aug 14 at 7:29












Yeah I have understood sir. Thank you very much for your help.
– Debabrata Chattopadhyay.
Aug 14 at 7:56




Yeah I have understood sir. Thank you very much for your help.
– Debabrata Chattopadhyay.
Aug 14 at 7:56















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