Why is a polar cone a closed set?
$begingroup$
Let $X subset mathbb{R}^n$. We define the polar cone as
$$Xº:={xinmathbb{R}^n,|,langle u,xrangleleq 0,forall uin X}$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that ${xinmathbb{R}^n,|,langle u,xrangleleq 0}$ 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 Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
asked Mar 17 at 9:36
LecterLecter
11210
$endgroup$
|
show 1 more comment
$begingroup$
Let $X subset mathbb{R}^n$. We define the polar cone as
$$Xº:={xinmathbb{R}^n,|,langle u,xrangleleq 0,forall uin X}$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that ${xinmathbb{R}^n,|,langle u,xrangleleq 0}$ 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 Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
asked Mar 17 at 9:36
LecterLecter
11210
$endgroup$
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
Mar 17 at 9:39
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
Mar 17 at 9:41
$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
Mar 17 at 9:43
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
Mar 17 at 9:45
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
Mar 17 at 9:49
|
show 1 more comment
$begingroup$
Let $X subset mathbb{R}^n$. We define the polar cone as
$$Xº:={xinmathbb{R}^n,|,langle u,xrangleleq 0,forall uin X}$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that ${xinmathbb{R}^n,|,langle u,xrangleleq 0}$ 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 Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
asked Mar 17 at 9:36
LecterLecter
11210
$endgroup$
Let $X subset mathbb{R}^n$. We define the polar cone as
$$Xº:={xinmathbb{R}^n,|,langle u,xrangleleq 0,forall uin X}$$
How can I show that this set is closed?
If I fix some $uin X$ then I have that ${xinmathbb{R}^n,|,langle u,xrangleleq 0}$ 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 Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
asked Mar 17 at 9:36
LecterLecter
11210
edited Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
asked Mar 17 at 9:36
LecterLecter
11210
edited Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
edited Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
edited Mar 17 at 10:30
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Mar 17 at 9:36
LecterLecter
11210
asked Mar 17 at 9:36
LecterLecter
11210
asked Mar 17 at 9:36
LecterLecter
11210
11210
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
Mar 17 at 9:39
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
Mar 17 at 9:41
$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
Mar 17 at 9:43
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
Mar 17 at 9:45
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
Mar 17 at 9:49
|
show 1 more comment
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
Mar 17 at 9:39
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
Mar 17 at 9:41
$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
Mar 17 at 9:43
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
Mar 17 at 9:45
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
Mar 17 at 9:49
2
2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
Mar 17 at 9:39
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
Mar 17 at 9:39
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
Mar 17 at 9:41
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
Mar 17 at 9:41
$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
Mar 17 at 9:43
$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
Mar 17 at 9:43
1
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
Mar 17 at 9:45
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
Mar 17 at 9:45
1
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
Mar 17 at 9:49
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
Mar 17 at 9:49
|
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 Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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 Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 17 at 9:53
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
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 Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
$endgroup$
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 Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
$endgroup$
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 Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
$endgroup$
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 Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
answered Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
answered Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
answered Mar 17 at 9:55
Henning MakholmHenning Makholm
242k17308551
242k17308551
add a comment |
add a comment |
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2
$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
Mar 17 at 9:39
$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
Mar 17 at 9:41
$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
Mar 17 at 9:43
1
$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
Mar 17 at 9:45
1
$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
Mar 17 at 9:49