Mixing
Showing the closed linear span is the closure of a linear span.
$begingroup$
I think I have this but it seems a little too easy.
Let $S$ be a collection of points ${y_j}$. Denote closure as $bar{S}$ and closed linear span as $cls(S)$. Suppose $x in bar{S}$. If $x$ is not a limit point then it's clearly in $cls(S)$ and there is nothing to prove. Suppose x is a limit point of $S$, then $x$ belongs to every closed linear subspace that contains $S$ so $x in cls(S)$.
Now suppose $x in cls(S)$, then $x in S$ or is a limit point of $S$, either way it's a point in $bar{S}$
Is this correct?
functional-analysis
asked Feb 7 '14 at 3:22
anonanon
61
$endgroup$
add a comment |
$begingroup$
I think I have this but it seems a little too easy.
Let $S$ be a collection of points ${y_j}$. Denote closure as $bar{S}$ and closed linear span as $cls(S)$. Suppose $x in bar{S}$. If $x$ is not a limit point then it's clearly in $cls(S)$ and there is nothing to prove. Suppose x is a limit point of $S$, then $x$ belongs to every closed linear subspace that contains $S$ so $x in cls(S)$.
Now suppose $x in cls(S)$, then $x in S$ or is a limit point of $S$, either way it's a point in $bar{S}$
Is this correct?
functional-analysis
asked Feb 7 '14 at 3:22
anonanon
61
$endgroup$
$begingroup$
The last section is wrong, suppose $S={(0,1)}subset mathbb{R}^2$. Then $(0,0)in cls(S)$, but $(0,0)notin S$ and $(0,0)$ is not a limit point of $S$. Your title suggests that you want to prove something else ...
$endgroup$
– Vobo
Feb 7 '14 at 16:50
add a comment |
$begingroup$
I think I have this but it seems a little too easy.
Let $S$ be a collection of points ${y_j}$. Denote closure as $bar{S}$ and closed linear span as $cls(S)$. Suppose $x in bar{S}$. If $x$ is not a limit point then it's clearly in $cls(S)$ and there is nothing to prove. Suppose x is a limit point of $S$, then $x$ belongs to every closed linear subspace that contains $S$ so $x in cls(S)$.
Now suppose $x in cls(S)$, then $x in S$ or is a limit point of $S$, either way it's a point in $bar{S}$
Is this correct?
functional-analysis
asked Feb 7 '14 at 3:22
anonanon
61
$endgroup$
I think I have this but it seems a little too easy.
Let $S$ be a collection of points ${y_j}$. Denote closure as $bar{S}$ and closed linear span as $cls(S)$. Suppose $x in bar{S}$. If $x$ is not a limit point then it's clearly in $cls(S)$ and there is nothing to prove. Suppose x is a limit point of $S$, then $x$ belongs to every closed linear subspace that contains $S$ so $x in cls(S)$.
Now suppose $x in cls(S)$, then $x in S$ or is a limit point of $S$, either way it's a point in $bar{S}$
Is this correct?
functional-analysis
functional-analysis
asked Feb 7 '14 at 3:22
anonanon
61
asked Feb 7 '14 at 3:22
anonanon
61
asked Feb 7 '14 at 3:22
anonanon
61
asked Feb 7 '14 at 3:22
anonanon
61
asked Feb 7 '14 at 3:22
anonanon
61
61
$begingroup$
The last section is wrong, suppose $S={(0,1)}subset mathbb{R}^2$. Then $(0,0)in cls(S)$, but $(0,0)notin S$ and $(0,0)$ is not a limit point of $S$. Your title suggests that you want to prove something else ...
$endgroup$
– Vobo
Feb 7 '14 at 16:50
add a comment |
$begingroup$
The last section is wrong, suppose $S={(0,1)}subset mathbb{R}^2$. Then $(0,0)in cls(S)$, but $(0,0)notin S$ and $(0,0)$ is not a limit point of $S$. Your title suggests that you want to prove something else ...
$endgroup$
– Vobo
Feb 7 '14 at 16:50
$begingroup$
The last section is wrong, suppose $S={(0,1)}subset mathbb{R}^2$. Then $(0,0)in cls(S)$, but $(0,0)notin S$ and $(0,0)$ is not a limit point of $S$. Your title suggests that you want to prove something else ...
$endgroup$
– Vobo
Feb 7 '14 at 16:50
$begingroup$
The last section is wrong, suppose $S={(0,1)}subset mathbb{R}^2$. Then $(0,0)in cls(S)$, but $(0,0)notin S$ and $(0,0)$ is not a limit point of $S$. Your title suggests that you want to prove something else ...
$endgroup$
– Vobo
Feb 7 '14 at 16:50
add a comment |
1 Answer
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$begingroup$
Here's my proof.
Let $A={A_j, jin J}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B={B_i, iin I}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $overline K$ the closure of $K$.
We know $bigcap limits_{iin I} B_i$ is a linear subspace. Thus, $overline {bigcap limits_{iin I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $overline{bigcap limits_{iin I} B_i} in A$. So, we know $bigcap limits_{jin J} A_j subseteq overline{bigcap limits_{iin I} B_i}$.
Next, we let $xin overline{bigcap limits_{iin I} B_i}$. Then, there exists a convergent sequence ${x_n}$ such that $x_nin bigcap limits_{iin I} B_i, x_nrightarrow x$. From the definitions of A and B, we know $bigcap limits_{iin I} B_i subseteq bigcap limits_{jin J} A_j$. So, $x_nin bigcap limits_{jin J} A_j$. Moreover, because $bigcap limits_{jin J} A_j$ is closed. Thus, $xinbigcap limits_{jin J} A_j$. Hence, $overline{bigcap limits_{iin I} B_i} subseteq bigcap limits_{jin J} A_j$.
Therefore, we have $overline{bigcap limits_{iin I} B_i} = bigcap limits_{jin J} A_j$. Done.
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Here's my proof.
Let $A={A_j, jin J}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B={B_i, iin I}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $overline K$ the closure of $K$.
We know $bigcap limits_{iin I} B_i$ is a linear subspace. Thus, $overline {bigcap limits_{iin I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $overline{bigcap limits_{iin I} B_i} in A$. So, we know $bigcap limits_{jin J} A_j subseteq overline{bigcap limits_{iin I} B_i}$.
Next, we let $xin overline{bigcap limits_{iin I} B_i}$. Then, there exists a convergent sequence ${x_n}$ such that $x_nin bigcap limits_{iin I} B_i, x_nrightarrow x$. From the definitions of A and B, we know $bigcap limits_{iin I} B_i subseteq bigcap limits_{jin J} A_j$. So, $x_nin bigcap limits_{jin J} A_j$. Moreover, because $bigcap limits_{jin J} A_j$ is closed. Thus, $xinbigcap limits_{jin J} A_j$. Hence, $overline{bigcap limits_{iin I} B_i} subseteq bigcap limits_{jin J} A_j$.
Therefore, we have $overline{bigcap limits_{iin I} B_i} = bigcap limits_{jin J} A_j$. Done.
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
$endgroup$
add a comment |
$begingroup$
Here's my proof.
Let $A={A_j, jin J}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B={B_i, iin I}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $overline K$ the closure of $K$.
We know $bigcap limits_{iin I} B_i$ is a linear subspace. Thus, $overline {bigcap limits_{iin I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $overline{bigcap limits_{iin I} B_i} in A$. So, we know $bigcap limits_{jin J} A_j subseteq overline{bigcap limits_{iin I} B_i}$.
Next, we let $xin overline{bigcap limits_{iin I} B_i}$. Then, there exists a convergent sequence ${x_n}$ such that $x_nin bigcap limits_{iin I} B_i, x_nrightarrow x$. From the definitions of A and B, we know $bigcap limits_{iin I} B_i subseteq bigcap limits_{jin J} A_j$. So, $x_nin bigcap limits_{jin J} A_j$. Moreover, because $bigcap limits_{jin J} A_j$ is closed. Thus, $xinbigcap limits_{jin J} A_j$. Hence, $overline{bigcap limits_{iin I} B_i} subseteq bigcap limits_{jin J} A_j$.
Therefore, we have $overline{bigcap limits_{iin I} B_i} = bigcap limits_{jin J} A_j$. Done.
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
$endgroup$
add a comment |
$begingroup$
Here's my proof.
Let $A={A_j, jin J}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B={B_i, iin I}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $overline K$ the closure of $K$.
We know $bigcap limits_{iin I} B_i$ is a linear subspace. Thus, $overline {bigcap limits_{iin I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $overline{bigcap limits_{iin I} B_i} in A$. So, we know $bigcap limits_{jin J} A_j subseteq overline{bigcap limits_{iin I} B_i}$.
Next, we let $xin overline{bigcap limits_{iin I} B_i}$. Then, there exists a convergent sequence ${x_n}$ such that $x_nin bigcap limits_{iin I} B_i, x_nrightarrow x$. From the definitions of A and B, we know $bigcap limits_{iin I} B_i subseteq bigcap limits_{jin J} A_j$. So, $x_nin bigcap limits_{jin J} A_j$. Moreover, because $bigcap limits_{jin J} A_j$ is closed. Thus, $xinbigcap limits_{jin J} A_j$. Hence, $overline{bigcap limits_{iin I} B_i} subseteq bigcap limits_{jin J} A_j$.
Therefore, we have $overline{bigcap limits_{iin I} B_i} = bigcap limits_{jin J} A_j$. Done.
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
$endgroup$
Here's my proof.
Let $A={A_j, jin J}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B={B_i, iin I}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $overline K$ the closure of $K$.
We know $bigcap limits_{iin I} B_i$ is a linear subspace. Thus, $overline {bigcap limits_{iin I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $overline{bigcap limits_{iin I} B_i} in A$. So, we know $bigcap limits_{jin J} A_j subseteq overline{bigcap limits_{iin I} B_i}$.
Next, we let $xin overline{bigcap limits_{iin I} B_i}$. Then, there exists a convergent sequence ${x_n}$ such that $x_nin bigcap limits_{iin I} B_i, x_nrightarrow x$. From the definitions of A and B, we know $bigcap limits_{iin I} B_i subseteq bigcap limits_{jin J} A_j$. So, $x_nin bigcap limits_{jin J} A_j$. Moreover, because $bigcap limits_{jin J} A_j$ is closed. Thus, $xinbigcap limits_{jin J} A_j$. Hence, $overline{bigcap limits_{iin I} B_i} subseteq bigcap limits_{jin J} A_j$.
Therefore, we have $overline{bigcap limits_{iin I} B_i} = bigcap limits_{jin J} A_j$. Done.
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
answered Jul 4 '16 at 21:05
Xianjin YangXianjin Yang
475210
475210
add a comment |
add a comment |
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$begingroup$
The last section is wrong, suppose $S={(0,1)}subset mathbb{R}^2$. Then $(0,0)in cls(S)$, but $(0,0)notin S$ and $(0,0)$ is not a limit point of $S$. Your title suggests that you want to prove something else ...
$endgroup$
– Vobo
Feb 7 '14 at 16:50