Why is a polar cone a closed set?












5












$begingroup$


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







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$endgroup$








  • 2




    $begingroup$
    What does $u'x$ mean?
    $endgroup$
    – José Carlos Santos
    15 hours ago












  • $begingroup$
    probably inter product with $u'$ the tranpose
    $endgroup$
    – dmtri
    15 hours ago












  • $begingroup$
    @JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
    $endgroup$
    – Lecter
    15 hours ago






  • 1




    $begingroup$
    why the intersection of closed sets is not a close set?
    $endgroup$
    – dmtri
    15 hours ago






  • 1




    $begingroup$
    @dmtri It's done.
    $endgroup$
    – José Carlos Santos
    15 hours ago
















5












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What does $u'x$ mean?
    $endgroup$
    – José Carlos Santos
    15 hours ago












  • $begingroup$
    probably inter product with $u'$ the tranpose
    $endgroup$
    – dmtri
    15 hours ago












  • $begingroup$
    @JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
    $endgroup$
    – Lecter
    15 hours ago






  • 1




    $begingroup$
    why the intersection of closed sets is not a close set?
    $endgroup$
    – dmtri
    15 hours ago






  • 1




    $begingroup$
    @dmtri It's done.
    $endgroup$
    – José Carlos Santos
    15 hours ago














5












5








5


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 15 hours ago









Rodrigo de Azevedo

13.1k41960




13.1k41960










asked 15 hours ago









LecterLecter

11210




11210








  • 2




    $begingroup$
    What does $u'x$ mean?
    $endgroup$
    – José Carlos Santos
    15 hours ago












  • $begingroup$
    probably inter product with $u'$ the tranpose
    $endgroup$
    – dmtri
    15 hours ago












  • $begingroup$
    @JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
    $endgroup$
    – Lecter
    15 hours ago






  • 1




    $begingroup$
    why the intersection of closed sets is not a close set?
    $endgroup$
    – dmtri
    15 hours ago






  • 1




    $begingroup$
    @dmtri It's done.
    $endgroup$
    – José Carlos Santos
    15 hours ago














  • 2




    $begingroup$
    What does $u'x$ mean?
    $endgroup$
    – José Carlos Santos
    15 hours ago












  • $begingroup$
    probably inter product with $u'$ the tranpose
    $endgroup$
    – dmtri
    15 hours ago












  • $begingroup$
    @JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
    $endgroup$
    – Lecter
    15 hours ago






  • 1




    $begingroup$
    why the intersection of closed sets is not a close set?
    $endgroup$
    – dmtri
    15 hours ago






  • 1




    $begingroup$
    @dmtri It's done.
    $endgroup$
    – José Carlos Santos
    15 hours ago








2




2




$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
15 hours ago






$begingroup$
What does $u'x$ mean?
$endgroup$
– José Carlos Santos
15 hours ago














$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
15 hours ago






$begingroup$
probably inter product with $u'$ the tranpose
$endgroup$
– dmtri
15 hours ago














$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
15 hours ago




$begingroup$
@JoséCarlosSantos Usual product in $mathbb{R^n}$. Edited.
$endgroup$
– Lecter
15 hours ago




1




1




$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
15 hours ago




$begingroup$
why the intersection of closed sets is not a close set?
$endgroup$
– dmtri
15 hours ago




1




1




$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
15 hours ago




$begingroup$
@dmtri It's done.
$endgroup$
– José Carlos Santos
15 hours ago










2 Answers
2






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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






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      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 15 hours ago









          José Carlos SantosJosé Carlos Santos

          168k22132236




          168k22132236























              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 15 hours ago









                  Henning MakholmHenning Makholm

                  242k17308550




                  242k17308550






























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