What is the dimension of $l_p$-space, $1 leq p < infty$?

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What is the dimension of $l_p$-space, $1 leq p < infty$?





$l_p$ is a subspace of $Bbb K^Bbb N$ where $Bbb K = Bbb R$ or $Bbb C$. In $Bbb K^Bbb N$ the sequences $e_i, i in Bbb N$ are linearly independent where $e_i$ is the sequence whose $i$-th coordinate is $1$ and all other $0$. Since all these $e_i$'s are also in $l_p$ they are linearly independent in $l_p$ too. But dimension of $Bbb K^Bbb N$ is countably infinite. Since all the $e_i$'s are in $l_p$ so dimension of $l_p$ is also countably infinite. But $l_p$ is a Banach space. So it cannot have countably infinite elements in it's basis. So we get a contradiction. But why does that contradiction arise? What's going wrong in my argument? Please help me in this regard.



Thank you very much.







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    What is the dimension of $l_p$-space, $1 leq p < infty$?





    $l_p$ is a subspace of $Bbb K^Bbb N$ where $Bbb K = Bbb R$ or $Bbb C$. In $Bbb K^Bbb N$ the sequences $e_i, i in Bbb N$ are linearly independent where $e_i$ is the sequence whose $i$-th coordinate is $1$ and all other $0$. Since all these $e_i$'s are also in $l_p$ they are linearly independent in $l_p$ too. But dimension of $Bbb K^Bbb N$ is countably infinite. Since all the $e_i$'s are in $l_p$ so dimension of $l_p$ is also countably infinite. But $l_p$ is a Banach space. So it cannot have countably infinite elements in it's basis. So we get a contradiction. But why does that contradiction arise? What's going wrong in my argument? Please help me in this regard.



    Thank you very much.







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite













      What is the dimension of $l_p$-space, $1 leq p < infty$?





      $l_p$ is a subspace of $Bbb K^Bbb N$ where $Bbb K = Bbb R$ or $Bbb C$. In $Bbb K^Bbb N$ the sequences $e_i, i in Bbb N$ are linearly independent where $e_i$ is the sequence whose $i$-th coordinate is $1$ and all other $0$. Since all these $e_i$'s are also in $l_p$ they are linearly independent in $l_p$ too. But dimension of $Bbb K^Bbb N$ is countably infinite. Since all the $e_i$'s are in $l_p$ so dimension of $l_p$ is also countably infinite. But $l_p$ is a Banach space. So it cannot have countably infinite elements in it's basis. So we get a contradiction. But why does that contradiction arise? What's going wrong in my argument? Please help me in this regard.



      Thank you very much.







      share|cite|improve this question














      What is the dimension of $l_p$-space, $1 leq p < infty$?





      $l_p$ is a subspace of $Bbb K^Bbb N$ where $Bbb K = Bbb R$ or $Bbb C$. In $Bbb K^Bbb N$ the sequences $e_i, i in Bbb N$ are linearly independent where $e_i$ is the sequence whose $i$-th coordinate is $1$ and all other $0$. Since all these $e_i$'s are also in $l_p$ they are linearly independent in $l_p$ too. But dimension of $Bbb K^Bbb N$ is countably infinite. Since all the $e_i$'s are in $l_p$ so dimension of $l_p$ is also countably infinite. But $l_p$ is a Banach space. So it cannot have countably infinite elements in it's basis. So we get a contradiction. But why does that contradiction arise? What's going wrong in my argument? Please help me in this regard.



      Thank you very much.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 22 at 7:53









      Dbchatto67

      16912




      16912




















          2 Answers
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          The error lies in the sentence “But dimension of $Bbb K^mathbb N$ is countably infinite”. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.






          share|cite|improve this answer




















          • Is $l_p$ not generated by $e_i$'s?
            – Dbchatto67
            Aug 22 at 8:00










          • $l_p$ is a linear space over $Bbb K$. Isn't it so?
            – Dbchatto67
            Aug 22 at 8:01






          • 1




            No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
            – José Carlos Santos
            Aug 22 at 8:02










          • Yes, $ell_p$ is a vector space over $mathbb K$.
            – José Carlos Santos
            Aug 22 at 8:03











          • As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
            – Dbchatto67
            Aug 22 at 8:04

















          up vote
          1
          down vote













          The dimension of $Bbb K^Bbb N$ is not countably infinite ! The set $e_i: i in mathbb N$ is not a basis of $Bbb K^Bbb N$.






          share|cite|improve this answer




















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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes








            up vote
            1
            down vote













            The error lies in the sentence “But dimension of $Bbb K^mathbb N$ is countably infinite”. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.






            share|cite|improve this answer




















            • Is $l_p$ not generated by $e_i$'s?
              – Dbchatto67
              Aug 22 at 8:00










            • $l_p$ is a linear space over $Bbb K$. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:01






            • 1




              No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
              – José Carlos Santos
              Aug 22 at 8:02










            • Yes, $ell_p$ is a vector space over $mathbb K$.
              – José Carlos Santos
              Aug 22 at 8:03











            • As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:04














            up vote
            1
            down vote













            The error lies in the sentence “But dimension of $Bbb K^mathbb N$ is countably infinite”. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.






            share|cite|improve this answer




















            • Is $l_p$ not generated by $e_i$'s?
              – Dbchatto67
              Aug 22 at 8:00










            • $l_p$ is a linear space over $Bbb K$. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:01






            • 1




              No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
              – José Carlos Santos
              Aug 22 at 8:02










            • Yes, $ell_p$ is a vector space over $mathbb K$.
              – José Carlos Santos
              Aug 22 at 8:03











            • As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:04












            up vote
            1
            down vote










            up vote
            1
            down vote









            The error lies in the sentence “But dimension of $Bbb K^mathbb N$ is countably infinite”. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.






            share|cite|improve this answer












            The error lies in the sentence “But dimension of $Bbb K^mathbb N$ is countably infinite”. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 22 at 7:57









            José Carlos Santos

            118k16101181




            118k16101181











            • Is $l_p$ not generated by $e_i$'s?
              – Dbchatto67
              Aug 22 at 8:00










            • $l_p$ is a linear space over $Bbb K$. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:01






            • 1




              No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
              – José Carlos Santos
              Aug 22 at 8:02










            • Yes, $ell_p$ is a vector space over $mathbb K$.
              – José Carlos Santos
              Aug 22 at 8:03











            • As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:04
















            • Is $l_p$ not generated by $e_i$'s?
              – Dbchatto67
              Aug 22 at 8:00










            • $l_p$ is a linear space over $Bbb K$. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:01






            • 1




              No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
              – José Carlos Santos
              Aug 22 at 8:02










            • Yes, $ell_p$ is a vector space over $mathbb K$.
              – José Carlos Santos
              Aug 22 at 8:03











            • As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
              – Dbchatto67
              Aug 22 at 8:04















            Is $l_p$ not generated by $e_i$'s?
            – Dbchatto67
            Aug 22 at 8:00




            Is $l_p$ not generated by $e_i$'s?
            – Dbchatto67
            Aug 22 at 8:00












            $l_p$ is a linear space over $Bbb K$. Isn't it so?
            – Dbchatto67
            Aug 22 at 8:01




            $l_p$ is a linear space over $Bbb K$. Isn't it so?
            – Dbchatto67
            Aug 22 at 8:01




            1




            1




            No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
            – José Carlos Santos
            Aug 22 at 8:02




            No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough.
            – José Carlos Santos
            Aug 22 at 8:02












            Yes, $ell_p$ is a vector space over $mathbb K$.
            – José Carlos Santos
            Aug 22 at 8:03





            Yes, $ell_p$ is a vector space over $mathbb K$.
            – José Carlos Santos
            Aug 22 at 8:03













            As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
            – Dbchatto67
            Aug 22 at 8:04




            As any element of $Bbb K^Bbb N$ cannot be written by using finitely many $e_i$'s. Isn't it so?
            – Dbchatto67
            Aug 22 at 8:04










            up vote
            1
            down vote













            The dimension of $Bbb K^Bbb N$ is not countably infinite ! The set $e_i: i in mathbb N$ is not a basis of $Bbb K^Bbb N$.






            share|cite|improve this answer
























              up vote
              1
              down vote













              The dimension of $Bbb K^Bbb N$ is not countably infinite ! The set $e_i: i in mathbb N$ is not a basis of $Bbb K^Bbb N$.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                The dimension of $Bbb K^Bbb N$ is not countably infinite ! The set $e_i: i in mathbb N$ is not a basis of $Bbb K^Bbb N$.






                share|cite|improve this answer












                The dimension of $Bbb K^Bbb N$ is not countably infinite ! The set $e_i: i in mathbb N$ is not a basis of $Bbb K^Bbb N$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 22 at 7:58









                Fred

                38.2k1238




                38.2k1238






















                     

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