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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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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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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