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 11 hours ago
Rodrigo de Azevedo
13k41960
asked 12 hours ago
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 11 hours ago
Rodrigo de Azevedo
13k41960
asked 12 hours ago
LecterLecter
11210
$endgroup$
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
12 hours ago
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
12 hours ago
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
12 hours ago
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
11 hours ago
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
11 hours ago
|
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 11 hours ago
Rodrigo de Azevedo
13k41960
asked 12 hours ago
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 11 hours ago
Rodrigo de Azevedo
13k41960
asked 12 hours ago
LecterLecter
11210
edited 11 hours ago
Rodrigo de Azevedo
13k41960
asked 12 hours ago
LecterLecter
11210
edited 11 hours ago
Rodrigo de Azevedo
13k41960
edited 11 hours ago
Rodrigo de Azevedo
13k41960
edited 11 hours ago
Rodrigo de Azevedo
13k41960
13k41960
asked 12 hours ago
LecterLecter
11210
asked 12 hours ago
LecterLecter
11210
asked 12 hours ago
LecterLecter
11210
11210
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
12 hours ago
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
12 hours ago
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
12 hours ago
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
11 hours ago
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
11 hours ago
|
show 1 more comment
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
12 hours ago
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
12 hours ago
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
12 hours ago
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
11 hours ago
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
11 hours ago
2
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
12 hours ago
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
12 hours ago
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
12 hours ago
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
12 hours ago
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
12 hours ago
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
12 hours ago
1
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
11 hours ago
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
11 hours ago
1
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
11 hours ago
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
11 hours ago
|
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 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
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 11 hours ago
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
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oldest
votes
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
answered 11 hours ago
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
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 11 hours ago
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 11 hours ago
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 11 hours ago
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 11 hours ago
Henning MakholmHenning Makholm
242k17308550
answered 11 hours ago
Henning MakholmHenning Makholm
242k17308550
answered 11 hours ago
Henning MakholmHenning Makholm
242k17308550
answered 11 hours ago
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
12 hours ago
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
12 hours ago
$begingroup$
@JoséCarlosSantos Usual product in $mathbbR^n$. Edited.
$endgroup$
– Lecter
12 hours ago
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
11 hours ago
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
11 hours ago