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.
functional-analysis normed-spaces lp-spaces
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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.
functional-analysis normed-spaces lp-spaces
add a comment |Â
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.
functional-analysis normed-spaces lp-spaces
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.
functional-analysis normed-spaces lp-spaces
asked Aug 22 at 7:53
Dbchatto67
16912
16912
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2 Answers
2
active
oldest
votes
up vote
1
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The error lies in the sentence âÂÂBut dimension of $Bbb K^mathbb N$ is countably infiniteâÂÂ. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â 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.
â José Carlos Santos
Aug 22 at 8:02
Yes, $ell_p$ is a vector space over $mathbb K$.
â 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?
â Dbchatto67
Aug 22 at 8:04
 |Â
show 3 more comments
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 $Bbb K^Bbb N$.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The error lies in the sentence âÂÂBut dimension of $Bbb K^mathbb N$ is countably infiniteâÂÂ. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â 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.
â José Carlos Santos
Aug 22 at 8:02
Yes, $ell_p$ is a vector space over $mathbb K$.
â 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?
â Dbchatto67
Aug 22 at 8:04
 |Â
show 3 more comments
up vote
1
down vote
The error lies in the sentence âÂÂBut dimension of $Bbb K^mathbb N$ is countably infiniteâÂÂ. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â 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.
â José Carlos Santos
Aug 22 at 8:02
Yes, $ell_p$ is a vector space over $mathbb K$.
â 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?
â Dbchatto67
Aug 22 at 8:04
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
The error lies in the sentence âÂÂBut dimension of $Bbb K^mathbb N$ is countably infiniteâÂÂ. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.
The error lies in the sentence âÂÂBut dimension of $Bbb K^mathbb N$ is countably infiniteâÂÂ. The dimension of this space is equal to the cardnal of $mathbb R$. The rest is fine.
answered Aug 22 at 7:57
José Carlos Santos
118k16101181
118k16101181
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â 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.
â José Carlos Santos
Aug 22 at 8:02
Yes, $ell_p$ is a vector space over $mathbb K$.
â 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?
â Dbchatto67
Aug 22 at 8:04
 |Â
show 3 more comments
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â 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.
â José Carlos Santos
Aug 22 at 8:02
Yes, $ell_p$ is a vector space over $mathbb K$.
â 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?
â Dbchatto67
Aug 22 at 8:04
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
Is $l_p$ not generated by $e_i$'s?
â Dbchatto67
Aug 22 at 8:00
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â Dbchatto67
Aug 22 at 8:01
$l_p$ is a linear space over $Bbb K$. Isn't it so?
â 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.
â 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.
â José Carlos Santos
Aug 22 at 8:02
Yes, $ell_p$ is a vector space over $mathbb K$.
â José Carlos Santos
Aug 22 at 8:03
Yes, $ell_p$ is a vector space over $mathbb K$.
â 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?
â 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?
â Dbchatto67
Aug 22 at 8:04
 |Â
show 3 more comments
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 $Bbb K^Bbb N$.
add a comment |Â
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 $Bbb K^Bbb N$.
add a comment |Â
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 $Bbb K^Bbb N$.
The dimension of $Bbb K^Bbb N$ is not countably infinite ! The set $e_i: i in mathbb N$ is not a basis of $Bbb K^Bbb N$.
answered Aug 22 at 7:58
Fred
38.2k1238
38.2k1238
add a comment |Â
add a comment |Â
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