Why is a polar cone a closed set?Excercise of Convex AnalysisWhat is the name of this object?Find the Polar of a set.Polar of revolution coneHow to prove the following cone theoremThe polar of an unit disc is itselfFinding the polar cone of the given conePolar cone of the Polar cone of $K$ a closed convex cone is again $K$Question about dual coneConvex cones in $Bbb R^n$
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Why is a polar cone a closed set?
Excercise of Convex AnalysisWhat is the name of this object?Find the Polar of a set.Polar of revolution coneHow to prove the following cone theoremThe polar of an unit disc is itselfFinding the polar cone of the given conePolar cone of the Polar cone of $K$ a closed convex cone is again $K$Question about dual coneConvex cones in $Bbb R^n$
$begingroup$
Let $X subset mathbbR^n$. We define the polar cone as
$$Xº:=,langle u,xrangleleq 0,forall uin X$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that $,langle u,xrangleleq 0$ is a closed halfspace; but if $X$ is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
general-topology convex-analysis
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
LecterLecter
11210
$endgroup$
|
show 1 more comment
$begingroup$
Let $X subset mathbbR^n$. We define the polar cone as
$$Xº:=,langle u,xrangleleq 0,forall uin X$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that $,langle u,xrangleleq 0$ is a closed halfspace; but if $X$ is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
general-topology convex-analysis
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
LecterLecter
11210
$endgroup$
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
yesterday
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
yesterday
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
yesterday
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
yesterday
|
show 1 more comment
$begingroup$
Let $X subset mathbbR^n$. We define the polar cone as
$$Xº:=,langle u,xrangleleq 0,forall uin X$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that $,langle u,xrangleleq 0$ is a closed halfspace; but if $X$ is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
general-topology convex-analysis
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
LecterLecter
11210
$endgroup$
Let $X subset mathbbR^n$. We define the polar cone as
$$Xº:=,langle u,xrangleleq 0,forall uin X$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that $,langle u,xrangleleq 0$ is a closed halfspace; but if $X$ is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
general-topology convex-analysis
general-topology convex-analysis
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
LecterLecter
11210
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
LecterLecter
11210
edited yesterday
Rodrigo de Azevedo
13.1k41960
edited yesterday
Rodrigo de Azevedo
13.1k41960
edited yesterday
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked yesterday
LecterLecter
11210
asked yesterday
LecterLecter
11210
asked yesterday
LecterLecter
11210
11210
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
yesterday
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
yesterday
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
yesterday
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
yesterday
|
show 1 more comment
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
yesterday
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
yesterday
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
yesterday
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
yesterday
2
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
yesterday
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
yesterday
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
yesterday
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
yesterday
1
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
yesterday
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
yesterday
1
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
yesterday
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
if 𝑋 is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
Yes you can. One of the axioms of topology is that any union of open sets is open. (This is easy to show directly for the standard topology on a metric space).
Taking complements, you get that any intersection of closed sets is closed.
answered yesterday
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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votes
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
answered yesterday
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
add a comment |
add a comment |
$begingroup$
if 𝑋 is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
Yes you can. One of the axioms of topology is that any union of open sets is open. (This is easy to show directly for the standard topology on a metric space).
Taking complements, you get that any intersection of closed sets is closed.
answered yesterday
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
$begingroup$
if 𝑋 is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
Yes you can. One of the axioms of topology is that any union of open sets is open. (This is easy to show directly for the standard topology on a metric space).
Taking complements, you get that any intersection of closed sets is closed.
answered yesterday
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
$begingroup$
if 𝑋 is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
Yes you can. One of the axioms of topology is that any union of open sets is open. (This is easy to show directly for the standard topology on a metric space).
Taking complements, you get that any intersection of closed sets is closed.
answered yesterday
Henning MakholmHenning Makholm
242k17308550
$endgroup$
if 𝑋 is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).
Yes you can. One of the axioms of topology is that any union of open sets is open. (This is easy to show directly for the standard topology on a metric space).
Taking complements, you get that any intersection of closed sets is closed.
answered yesterday
Henning MakholmHenning Makholm
242k17308550
answered yesterday
Henning MakholmHenning Makholm
242k17308550
answered yesterday
Henning MakholmHenning Makholm
242k17308550
answered yesterday
Henning MakholmHenning Makholm
242k17308550
242k17308550
add a comment |
add a comment |
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2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
yesterday
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
yesterday
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
yesterday
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
yesterday