$U={xin mathbb{R^n} :g_i(x)leq 0 : text{for} quad i=1,…,m }$ is closed












1












$begingroup$


Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.



with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.



we want to prove that $U$ is a closed set.










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$endgroup$












  • $begingroup$
    Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
    $endgroup$
    – TrostAft
    Dec 19 '18 at 7:23
















1












$begingroup$


Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.



with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.



we want to prove that $U$ is a closed set.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
    $endgroup$
    – TrostAft
    Dec 19 '18 at 7:23














1












1








1





$begingroup$


Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.



with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.



we want to prove that $U$ is a closed set.










share|cite|improve this question











$endgroup$




Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.



with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.



we want to prove that $U$ is a closed set.







general-topology convex-analysis






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edited Dec 19 '18 at 7:46









Hossien Sahebjame

817




817










asked Dec 19 '18 at 7:21









Abderrahmane OuachouachAbderrahmane Ouachouach

61




61












  • $begingroup$
    Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
    $endgroup$
    – TrostAft
    Dec 19 '18 at 7:23


















  • $begingroup$
    Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
    $endgroup$
    – TrostAft
    Dec 19 '18 at 7:23
















$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23




$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23










2 Answers
2






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4












$begingroup$

Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.






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$endgroup$





















    2












    $begingroup$

    I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.



    Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.



    Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.






          share|cite|improve this answer









          $endgroup$



          Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 7:29









          Kavi Rama MurthyKavi Rama Murthy

          56k42158




          56k42158























              2












              $begingroup$

              I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.



              Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.



              Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.



                Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.



                Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.



                  Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.



                  Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.






                  share|cite|improve this answer









                  $endgroup$



                  I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.



                  Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.



                  Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 7:30









                  FredFred

                  45.2k1847




                  45.2k1847






























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