intersection of sets corresponding to free ultrafiler
1
$begingroup$
If $omega in beta Bbb Nsetminus Bbb N$,we define $S_{omega}={(x_n) in prod M_n(Bbb C):lim_{n to omega}tr_n(x_n)=0}$
Is the intersection $cap_{omega in beta Bbb N setminus Bbb N}S_{omega}$ empty?
general-topology functional-analysis elementary-set-theory c-star-algebras filters
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
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add a comment |
1
$begingroup$
If $omega in beta Bbb Nsetminus Bbb N$,we define $S_{omega}={(x_n) in prod M_n(Bbb C):lim_{n to omega}tr_n(x_n)=0}$
Is the intersection $cap_{omega in beta Bbb N setminus Bbb N}S_{omega}$ empty?
general-topology functional-analysis elementary-set-theory c-star-algebras filters
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
$endgroup$
add a comment |
1
1
1
$begingroup$
If $omega in beta Bbb Nsetminus Bbb N$,we define $S_{omega}={(x_n) in prod M_n(Bbb C):lim_{n to omega}tr_n(x_n)=0}$
Is the intersection $cap_{omega in beta Bbb N setminus Bbb N}S_{omega}$ empty?
general-topology functional-analysis elementary-set-theory c-star-algebras filters
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
$endgroup$
If $omega in beta Bbb Nsetminus Bbb N$,we define $S_{omega}={(x_n) in prod M_n(Bbb C):lim_{n to omega}tr_n(x_n)=0}$
Is the intersection $cap_{omega in beta Bbb N setminus Bbb N}S_{omega}$ empty?
general-topology functional-analysis elementary-set-theory c-star-algebras filters
general-topology functional-analysis elementary-set-theory c-star-algebras filters
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
edited Dec 7 '18 at 0:01
mathrookie
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
asked Dec 6 '18 at 23:55
mathrookiemathrookie
887512
887512
add a comment |
add a comment |
1 Answer
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No. Take $x=(x_n)$ where $x_n=tfrac1n,I_n$. Then $operatorname{tr}_n(x_n)=tfrac1nto0$, so $xin S_omega$ for all $omega$.
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
add a comment |
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1 Answer
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1 Answer
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No. Take $x=(x_n)$ where $x_n=tfrac1n,I_n$. Then $operatorname{tr}_n(x_n)=tfrac1nto0$, so $xin S_omega$ for all $omega$.
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
add a comment |
1
$begingroup$
No. Take $x=(x_n)$ where $x_n=tfrac1n,I_n$. Then $operatorname{tr}_n(x_n)=tfrac1nto0$, so $xin S_omega$ for all $omega$.
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
add a comment |
1
1
1
$begingroup$
No. Take $x=(x_n)$ where $x_n=tfrac1n,I_n$. Then $operatorname{tr}_n(x_n)=tfrac1nto0$, so $xin S_omega$ for all $omega$.
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
No. Take $x=(x_n)$ where $x_n=tfrac1n,I_n$. Then $operatorname{tr}_n(x_n)=tfrac1nto0$, so $xin S_omega$ for all $omega$.
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 1:16
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
add a comment |
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
An ultrafilter over $mathbb N$ always contains $mathbb N$. Not sure how it relates to the question, though.
$endgroup$
– Martin Argerami
Dec 7 '18 at 1:58
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
$begingroup$
My thought was wrong:I mistook the intersections of $S_{omega}$ as empty set.What is the intersetions of $S_{omega}$?Is it the $S_{mathcal{F}}$,where $mathcal{F}$ is the cofinite ultrafilter on $Bbb N$?
$endgroup$
– mathrookie
Dec 7 '18 at 2:05
add a comment |
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