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.


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.

  • $\begingroup$ Is $l_p$ not generated by $e_i$'s? $\endgroup$ – Dbchatto67 Aug 22 '18 at 8:00
  • $\begingroup$ $l_p$ is a linear space over $\Bbb K$. Isn't it so? $\endgroup$ – Dbchatto67 Aug 22 '18 at 8:01
  • 1
    $\begingroup$ No. The span of the $e_i$'s is the space of all sequences which are equal to $0$ if $i$ is large enough. $\endgroup$ – José Carlos Santos Aug 22 '18 at 8:02
  • $\begingroup$ Yes, $\ell_p$ is a vector space over $\mathbb K$. $\endgroup$ – José Carlos Santos Aug 22 '18 at 8:03
  • $\begingroup$ As any element of $\Bbb K^{\Bbb N}$ cannot be written by using finitely many $e_i$'s. Isn't it so? $\endgroup$ – Dbchatto67 Aug 22 '18 at 8:04

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


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