Non-complete inner product spaces with orthonormal basis

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It is known that every Hilbert space admits an orthonormal basis, but is it true that an inner product space which admits an orthonormal basis is necessarily complete as a metric space? Can you give a counter-example?




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It is known that every Hilbert space admits an orthonormal basis, but is it true that an inner product space which admits an orthonormal basis is necessarily complete as a metric space? Can you give a counter-example?




inner-product-space




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  • related question math.stackexchange.com/q/201119/442
    – GEdgar
    Jul 27 at 10:32












up vote
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up vote
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It is known that every Hilbert space admits an orthonormal basis, but is it true that an inner product space which admits an orthonormal basis is necessarily complete as a metric space? Can you give a counter-example?




inner-product-space




share|cite|improve this question













It is known that every Hilbert space admits an orthonormal basis, but is it true that an inner product space which admits an orthonormal basis is necessarily complete as a metric space? Can you give a counter-example?





inner-product-space





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edited Jul 27 at 10:37









Bernard

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  • related question math.stackexchange.com/q/201119/442
    – GEdgar
    Jul 27 at 10:32
















  • related question math.stackexchange.com/q/201119/442
    – GEdgar
    Jul 27 at 10:32















related question math.stackexchange.com/q/201119/442
– GEdgar
Jul 27 at 10:32




related question math.stackexchange.com/q/201119/442
– GEdgar
Jul 27 at 10:32










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In a Hilbert space, with orthonormal basis $e_n,n=1,2,3,dots$, let $X$ be the set of finite linear combinations $sum t_n e_n$. This will be an incomplete inner product space, and $e_n$ is still a complete orthonormal set.






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    The space $l_0$ of sequences with at most finitely many non-zero terms (as a subspace of $l^2$) with the usual basis elements $(1,0,0...),(0,1,0...)...$ is incomplete.






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      2 Answers
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      In a Hilbert space, with orthonormal basis $e_n,n=1,2,3,dots$, let $X$ be the set of finite linear combinations $sum t_n e_n$. This will be an incomplete inner product space, and $e_n$ is still a complete orthonormal set.






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        In a Hilbert space, with orthonormal basis $e_n,n=1,2,3,dots$, let $X$ be the set of finite linear combinations $sum t_n e_n$. This will be an incomplete inner product space, and $e_n$ is still a complete orthonormal set.






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          up vote
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          In a Hilbert space, with orthonormal basis $e_n,n=1,2,3,dots$, let $X$ be the set of finite linear combinations $sum t_n e_n$. This will be an incomplete inner product space, and $e_n$ is still a complete orthonormal set.






          share|cite|improve this answer













          In a Hilbert space, with orthonormal basis $e_n,n=1,2,3,dots$, let $X$ be the set of finite linear combinations $sum t_n e_n$. This will be an incomplete inner product space, and $e_n$ is still a complete orthonormal set.







          share|cite|improve this answer













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          answered Jul 27 at 10:31









          GEdgar

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              The space $l_0$ of sequences with at most finitely many non-zero terms (as a subspace of $l^2$) with the usual basis elements $(1,0,0...),(0,1,0...)...$ is incomplete.






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                The space $l_0$ of sequences with at most finitely many non-zero terms (as a subspace of $l^2$) with the usual basis elements $(1,0,0...),(0,1,0...)...$ is incomplete.






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                  The space $l_0$ of sequences with at most finitely many non-zero terms (as a subspace of $l^2$) with the usual basis elements $(1,0,0...),(0,1,0...)...$ is incomplete.






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                  The space $l_0$ of sequences with at most finitely many non-zero terms (as a subspace of $l^2$) with the usual basis elements $(1,0,0...),(0,1,0...)...$ is incomplete.







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                  answered Jul 27 at 10:32









                  Kavi Rama Murthy

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