Are there any vector spaces that cannot be given a norm that makes the vector space a complete metric space?












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And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.






general-topology metric-spaces normed-spaces baire-category







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    And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.






    general-topology metric-spaces normed-spaces baire-category







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.






      general-topology metric-spaces normed-spaces baire-category







      share|cite|improve this question









      $endgroup$




      And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.






      general-topology metric-spaces normed-spaces baire-category




      general-topology metric-spaces normed-spaces baire-category






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      asked Dec 19 '18 at 18:46









      2chromatic2chromatic

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          If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.






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          • $begingroup$
            A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
            $endgroup$
            – Max
            Dec 19 '18 at 20:50



















          4












          $begingroup$

          Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.






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            $begingroup$

            The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.



            It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).






            share|cite|improve this answer









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              3 Answers
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              3 Answers
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              active

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              4












              $begingroup$

              If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
                $endgroup$
                – Max
                Dec 19 '18 at 20:50
















              4












              $begingroup$

              If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
                $endgroup$
                – Max
                Dec 19 '18 at 20:50














              4












              4








              4





              $begingroup$

              If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.






              share|cite|improve this answer









              $endgroup$



              If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 19 '18 at 18:50









              ThomasThomas

              4,102510




              4,102510












              • $begingroup$
                A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
                $endgroup$
                – Max
                Dec 19 '18 at 20:50


















              • $begingroup$
                A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
                $endgroup$
                – Max
                Dec 19 '18 at 20:50
















              $begingroup$
              A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
              $endgroup$
              – Max
              Dec 19 '18 at 20:50




              $begingroup$
              A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
              $endgroup$
              – Max
              Dec 19 '18 at 20:50











              4












              $begingroup$

              Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.






                  share|cite|improve this answer









                  $endgroup$



                  Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 18:57









                  José Carlos SantosJosé Carlos Santos

                  164k22132235




                  164k22132235























                      1












                      $begingroup$

                      The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.



                      It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.



                        It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.



                          It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).






                          share|cite|improve this answer









                          $endgroup$



                          The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.



                          It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 19 '18 at 22:05









                          Henno BrandsmaHenno Brandsma

                          111k348118




                          111k348118






























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