Explanation of proof that $c_{0}$ doesn't complemented in $l^{infty}$
$begingroup$
I've stuck with a problem. We want to prove that $c_{0}$ doesn't complemented in $l^{infty}$.
There is a proof given in Complement of $c_{0}$ in $l^{infty}$.
The main problem I have connected with the second part of proof :
"Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J subset I_{n,k}$ be finite and consider $y = sum_{j in J} operatorname{sign}{[(Px_j)(n)]} cdot x_j$. Note that
$$
(Py)(n) = sum_{j in J} operatorname{sign}{[(Px_j)(n)]}cdot (Px_j)(n) geq frac{# J}{k}
$$
by our choice of $y$. Since $A_i cap A_j$ is finite for $i neq j$, we can write $y = f + z$ where $f$ has finite support and $|z| leq 1$. Thus $P(y) = P(f) + P(z) = P(z)$ by hypothesis on $P$ and therefore $|P(y)| leq |P| |z| leq |P|$. This yields
$$# J leq |P| k$$
whence the absurdity that $I_{n,k}$ must be finite".
As it said we consider finite subset $J subset I_{n,k}$. And then prove that $#J le |P| k$, so it shows that $J$ is finite. It confuses me. And even if we show that $J$ is finite, how it shows that $I_{n,k}$ is finite?
real-analysis functional-analysis proof-explanation banach-spaces
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
$endgroup$
add a comment |
$begingroup$
I've stuck with a problem. We want to prove that $c_{0}$ doesn't complemented in $l^{infty}$.
There is a proof given in Complement of $c_{0}$ in $l^{infty}$.
The main problem I have connected with the second part of proof :
"Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J subset I_{n,k}$ be finite and consider $y = sum_{j in J} operatorname{sign}{[(Px_j)(n)]} cdot x_j$. Note that
$$
(Py)(n) = sum_{j in J} operatorname{sign}{[(Px_j)(n)]}cdot (Px_j)(n) geq frac{# J}{k}
$$
by our choice of $y$. Since $A_i cap A_j$ is finite for $i neq j$, we can write $y = f + z$ where $f$ has finite support and $|z| leq 1$. Thus $P(y) = P(f) + P(z) = P(z)$ by hypothesis on $P$ and therefore $|P(y)| leq |P| |z| leq |P|$. This yields
$$# J leq |P| k$$
whence the absurdity that $I_{n,k}$ must be finite".
As it said we consider finite subset $J subset I_{n,k}$. And then prove that $#J le |P| k$, so it shows that $J$ is finite. It confuses me. And even if we show that $J$ is finite, how it shows that $I_{n,k}$ is finite?
real-analysis functional-analysis proof-explanation banach-spaces
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
$endgroup$
add a comment |
$begingroup$
I've stuck with a problem. We want to prove that $c_{0}$ doesn't complemented in $l^{infty}$.
There is a proof given in Complement of $c_{0}$ in $l^{infty}$.
The main problem I have connected with the second part of proof :
"Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J subset I_{n,k}$ be finite and consider $y = sum_{j in J} operatorname{sign}{[(Px_j)(n)]} cdot x_j$. Note that
$$
(Py)(n) = sum_{j in J} operatorname{sign}{[(Px_j)(n)]}cdot (Px_j)(n) geq frac{# J}{k}
$$
by our choice of $y$. Since $A_i cap A_j$ is finite for $i neq j$, we can write $y = f + z$ where $f$ has finite support and $|z| leq 1$. Thus $P(y) = P(f) + P(z) = P(z)$ by hypothesis on $P$ and therefore $|P(y)| leq |P| |z| leq |P|$. This yields
$$# J leq |P| k$$
whence the absurdity that $I_{n,k}$ must be finite".
As it said we consider finite subset $J subset I_{n,k}$. And then prove that $#J le |P| k$, so it shows that $J$ is finite. It confuses me. And even if we show that $J$ is finite, how it shows that $I_{n,k}$ is finite?
real-analysis functional-analysis proof-explanation banach-spaces
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
$endgroup$
I've stuck with a problem. We want to prove that $c_{0}$ doesn't complemented in $l^{infty}$.
There is a proof given in Complement of $c_{0}$ in $l^{infty}$.
The main problem I have connected with the second part of proof :
"Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J subset I_{n,k}$ be finite and consider $y = sum_{j in J} operatorname{sign}{[(Px_j)(n)]} cdot x_j$. Note that
$$
(Py)(n) = sum_{j in J} operatorname{sign}{[(Px_j)(n)]}cdot (Px_j)(n) geq frac{# J}{k}
$$
by our choice of $y$. Since $A_i cap A_j$ is finite for $i neq j$, we can write $y = f + z$ where $f$ has finite support and $|z| leq 1$. Thus $P(y) = P(f) + P(z) = P(z)$ by hypothesis on $P$ and therefore $|P(y)| leq |P| |z| leq |P|$. This yields
$$# J leq |P| k$$
whence the absurdity that $I_{n,k}$ must be finite".
As it said we consider finite subset $J subset I_{n,k}$. And then prove that $#J le |P| k$, so it shows that $J$ is finite. It confuses me. And even if we show that $J$ is finite, how it shows that $I_{n,k}$ is finite?
real-analysis functional-analysis proof-explanation banach-spaces
real-analysis functional-analysis proof-explanation banach-spaces
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
asked Dec 19 '18 at 22:37
openspaceopenspace
3,3062822
3,3062822
add a comment |
add a comment |
1 Answer
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The situation is the following: there is an uncountable set $S$ (here $I_{n,k}$). It is proven that there exists a constant $C$ (here $leftlVert PrightrVert k$) such that each finite subset of $S$ has at most $C$ elements. This proves that $S$ is finite; otherwise, pick a finite subset with for example $lfloor Crfloor +2$ elements.
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
The situation is the following: there is an uncountable set $S$ (here $I_{n,k}$). It is proven that there exists a constant $C$ (here $leftlVert PrightrVert k$) such that each finite subset of $S$ has at most $C$ elements. This proves that $S$ is finite; otherwise, pick a finite subset with for example $lfloor Crfloor +2$ elements.
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
add a comment |
$begingroup$
The situation is the following: there is an uncountable set $S$ (here $I_{n,k}$). It is proven that there exists a constant $C$ (here $leftlVert PrightrVert k$) such that each finite subset of $S$ has at most $C$ elements. This proves that $S$ is finite; otherwise, pick a finite subset with for example $lfloor Crfloor +2$ elements.
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
add a comment |
$begingroup$
The situation is the following: there is an uncountable set $S$ (here $I_{n,k}$). It is proven that there exists a constant $C$ (here $leftlVert PrightrVert k$) such that each finite subset of $S$ has at most $C$ elements. This proves that $S$ is finite; otherwise, pick a finite subset with for example $lfloor Crfloor +2$ elements.
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
The situation is the following: there is an uncountable set $S$ (here $I_{n,k}$). It is proven that there exists a constant $C$ (here $leftlVert PrightrVert k$) such that each finite subset of $S$ has at most $C$ elements. This proves that $S$ is finite; otherwise, pick a finite subset with for example $lfloor Crfloor +2$ elements.
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
answered Dec 19 '18 at 23:10
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
add a comment |
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
$begingroup$
Many thanks to you!!!
$endgroup$
– openspace
Dec 19 '18 at 23:18
add a comment |
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