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 $Bbb R^n$













5












$begingroup$


Let $X subset mathbbR^n$. We define the polar cone as



$$Xº:=xinmathbbR^n,$$



How can I show that this set is closed?



If I fix some $uin X$ then I have that $xinmathbbR^n,$ 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).










share|cite|improve this question











$endgroup$







  • 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















5












$begingroup$


Let $X subset mathbbR^n$. We define the polar cone as



$$Xº:=xinmathbbR^n,$$



How can I show that this set is closed?



If I fix some $uin X$ then I have that $xinmathbbR^n,$ 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).










share|cite|improve this question











$endgroup$







  • 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













5












5








5


1



$begingroup$


Let $X subset mathbbR^n$. We define the polar cone as



$$Xº:=xinmathbbR^n,$$



How can I show that this set is closed?



If I fix some $uin X$ then I have that $xinmathbbR^n,$ 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).










share|cite|improve this question











$endgroup$




Let $X subset mathbbR^n$. We define the polar cone as



$$Xº:=xinmathbbR^n,$$



How can I show that this set is closed?



If I fix some $uin X$ then I have that $xinmathbbR^n,$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Rodrigo de Azevedo

13.1k41960




13.1k41960










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












  • 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










2 Answers
2






active

oldest

votes


















7












$begingroup$

Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.






share|cite|improve this answer









$endgroup$




















    7












    $begingroup$


    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.






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.






      share|cite|improve this answer









      $endgroup$

















        7












        $begingroup$

        Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.






        share|cite|improve this answer









        $endgroup$















          7












          7








          7





          $begingroup$

          Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.






          share|cite|improve this answer









          $endgroup$



          Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          José Carlos SantosJosé Carlos Santos

          168k23132236




          168k23132236





















              7












              $begingroup$


              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.






              share|cite|improve this answer









              $endgroup$

















                7












                $begingroup$


                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.






                share|cite|improve this answer









                $endgroup$















                  7












                  7








                  7





                  $begingroup$


                  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.






                  share|cite|improve this answer









                  $endgroup$




                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Henning MakholmHenning Makholm

                  242k17308550




                  242k17308550



























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