Supporting Hyperplanes to the intersection of half-spaces












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Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.



But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?



I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.










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    $begingroup$


    Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
    Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.



    But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?



    I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
      Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.



      But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?



      I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.










      share|cite|improve this question









      $endgroup$




      Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
      Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.



      But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?



      I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.







      convex-analysis






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      asked Nov 29 '18 at 16:06









      Mathew KnudsonMathew Knudson

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