Distance to closed and convex sets that have intersection $C subseteq D rightarrow$ $ d_D(x^*) leq d_C(x^*) $












0














Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).



Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.



Let $D$ be a closed convex set containing $C$, i.e., $C subseteq D$.



Show that
$$
d_D(x^*) leq d_C(x^*)
$$



I do not know how to use $C subseteq D$ together with taking minimum.






optimization convex-optimization projective-geometry







share|cite|improve this question






















  • Assume the distance is strictly smaller, use minimum property and that every element in $C$ is an element in $D$. That leads to a contradiction.
    – B.Swan
    Nov 25 '18 at 7:57


















0














Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).



Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.



Let $D$ be a closed convex set containing $C$, i.e., $C subseteq D$.



Show that
$$
d_D(x^*) leq d_C(x^*)
$$



I do not know how to use $C subseteq D$ together with taking minimum.






optimization convex-optimization projective-geometry







share|cite|improve this question






















  • Assume the distance is strictly smaller, use minimum property and that every element in $C$ is an element in $D$. That leads to a contradiction.
    – B.Swan
    Nov 25 '18 at 7:57
















0












0








0







Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).



Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.



Let $D$ be a closed convex set containing $C$, i.e., $C subseteq D$.



Show that
$$
d_D(x^*) leq d_C(x^*)
$$



I do not know how to use $C subseteq D$ together with taking minimum.






optimization convex-optimization projective-geometry







share|cite|improve this question













Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).



Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.



Let $D$ be a closed convex set containing $C$, i.e., $C subseteq D$.



Show that
$$
d_D(x^*) leq d_C(x^*)
$$



I do not know how to use $C subseteq D$ together with taking minimum.






optimization convex-optimization projective-geometry




optimization convex-optimization projective-geometry






share|cite|improve this question













share|cite|improve this question











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asked Nov 25 '18 at 7:53









Saeed

651310




651310












  • Assume the distance is strictly smaller, use minimum property and that every element in $C$ is an element in $D$. That leads to a contradiction.
    – B.Swan
    Nov 25 '18 at 7:57




















  • Assume the distance is strictly smaller, use minimum property and that every element in $C$ is an element in $D$. That leads to a contradiction.
    – B.Swan
    Nov 25 '18 at 7:57


















Assume the distance is strictly smaller, use minimum property and that every element in $C$ is an element in $D$. That leads to a contradiction.
– B.Swan
Nov 25 '18 at 7:57






Assume the distance is strictly smaller, use minimum property and that every element in $C$ is an element in $D$. That leads to a contradiction.
– B.Swan
Nov 25 '18 at 7:57












2 Answers
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1














Since $C subseteq D$, the minimum over $C$ can only be greater or equal to the minimum over $D$, so



$$d_D(x^*)=min_{z in D}|z -x^*|_2 leq min_{z in C}|z -x^*|_2= d_D(x^*)$$






share|cite|improve this answer





























    0














    for $forall z in D$ and $yin C$, we have
    begin{align}
    d_D(x^*) &= min_{z in D} |z-x^*|_2 \
    &= min_{yin C} |(z_{best} -y)+ (y-x^*)|_2 \
    &leq min_{y in C}|z_{best}-y|_2 + min_{y in C}|y-x^*|_2 \
    & leq min_{yin C}|y-x^*|_2 \
    &=d_C(x^*)
    end{align}




    Answer above has mistakes. while, maybe we can try a new way to solve this.



    Notice that for any element $y in C$, because $C subseteq D$, so that $y in D$, and another element $z_{best} in D$, satisfy $d_D(x^*)=min_{z in D} |z-x^*|_2=|z_{best}-x^*|_2$, from this definition, it is quite clear to see
    $$ |y-x^*|_2 geq |z_{best}-x^*|_2, forall y in C$$
    which results in
    $$d_C(x^*) = min_{yin C}|y-x^*|_2 geq |z_{best}-x^*|_2 = d_D(x^*)$$






    share|cite|improve this answer























    • How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
      – Saeed
      Nov 25 '18 at 18:59










    • You are right! And I post a new solution under the original one. I hope it can help you.
      – Caldera
      Nov 26 '18 at 7:50











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Since $C subseteq D$, the minimum over $C$ can only be greater or equal to the minimum over $D$, so



    $$d_D(x^*)=min_{z in D}|z -x^*|_2 leq min_{z in C}|z -x^*|_2= d_D(x^*)$$






    share|cite|improve this answer


























      1














      Since $C subseteq D$, the minimum over $C$ can only be greater or equal to the minimum over $D$, so



      $$d_D(x^*)=min_{z in D}|z -x^*|_2 leq min_{z in C}|z -x^*|_2= d_D(x^*)$$






      share|cite|improve this answer
























        1












        1








        1






        Since $C subseteq D$, the minimum over $C$ can only be greater or equal to the minimum over $D$, so



        $$d_D(x^*)=min_{z in D}|z -x^*|_2 leq min_{z in C}|z -x^*|_2= d_D(x^*)$$






        share|cite|improve this answer












        Since $C subseteq D$, the minimum over $C$ can only be greater or equal to the minimum over $D$, so



        $$d_D(x^*)=min_{z in D}|z -x^*|_2 leq min_{z in C}|z -x^*|_2= d_D(x^*)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 8:05









        B.Swan

        1,0011719




        1,0011719























            0














            for $forall z in D$ and $yin C$, we have
            begin{align}
            d_D(x^*) &= min_{z in D} |z-x^*|_2 \
            &= min_{yin C} |(z_{best} -y)+ (y-x^*)|_2 \
            &leq min_{y in C}|z_{best}-y|_2 + min_{y in C}|y-x^*|_2 \
            & leq min_{yin C}|y-x^*|_2 \
            &=d_C(x^*)
            end{align}




            Answer above has mistakes. while, maybe we can try a new way to solve this.



            Notice that for any element $y in C$, because $C subseteq D$, so that $y in D$, and another element $z_{best} in D$, satisfy $d_D(x^*)=min_{z in D} |z-x^*|_2=|z_{best}-x^*|_2$, from this definition, it is quite clear to see
            $$ |y-x^*|_2 geq |z_{best}-x^*|_2, forall y in C$$
            which results in
            $$d_C(x^*) = min_{yin C}|y-x^*|_2 geq |z_{best}-x^*|_2 = d_D(x^*)$$






            share|cite|improve this answer























            • How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
              – Saeed
              Nov 25 '18 at 18:59










            • You are right! And I post a new solution under the original one. I hope it can help you.
              – Caldera
              Nov 26 '18 at 7:50
















            0














            for $forall z in D$ and $yin C$, we have
            begin{align}
            d_D(x^*) &= min_{z in D} |z-x^*|_2 \
            &= min_{yin C} |(z_{best} -y)+ (y-x^*)|_2 \
            &leq min_{y in C}|z_{best}-y|_2 + min_{y in C}|y-x^*|_2 \
            & leq min_{yin C}|y-x^*|_2 \
            &=d_C(x^*)
            end{align}




            Answer above has mistakes. while, maybe we can try a new way to solve this.



            Notice that for any element $y in C$, because $C subseteq D$, so that $y in D$, and another element $z_{best} in D$, satisfy $d_D(x^*)=min_{z in D} |z-x^*|_2=|z_{best}-x^*|_2$, from this definition, it is quite clear to see
            $$ |y-x^*|_2 geq |z_{best}-x^*|_2, forall y in C$$
            which results in
            $$d_C(x^*) = min_{yin C}|y-x^*|_2 geq |z_{best}-x^*|_2 = d_D(x^*)$$






            share|cite|improve this answer























            • How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
              – Saeed
              Nov 25 '18 at 18:59










            • You are right! And I post a new solution under the original one. I hope it can help you.
              – Caldera
              Nov 26 '18 at 7:50














            0












            0








            0






            for $forall z in D$ and $yin C$, we have
            begin{align}
            d_D(x^*) &= min_{z in D} |z-x^*|_2 \
            &= min_{yin C} |(z_{best} -y)+ (y-x^*)|_2 \
            &leq min_{y in C}|z_{best}-y|_2 + min_{y in C}|y-x^*|_2 \
            & leq min_{yin C}|y-x^*|_2 \
            &=d_C(x^*)
            end{align}




            Answer above has mistakes. while, maybe we can try a new way to solve this.



            Notice that for any element $y in C$, because $C subseteq D$, so that $y in D$, and another element $z_{best} in D$, satisfy $d_D(x^*)=min_{z in D} |z-x^*|_2=|z_{best}-x^*|_2$, from this definition, it is quite clear to see
            $$ |y-x^*|_2 geq |z_{best}-x^*|_2, forall y in C$$
            which results in
            $$d_C(x^*) = min_{yin C}|y-x^*|_2 geq |z_{best}-x^*|_2 = d_D(x^*)$$






            share|cite|improve this answer














            for $forall z in D$ and $yin C$, we have
            begin{align}
            d_D(x^*) &= min_{z in D} |z-x^*|_2 \
            &= min_{yin C} |(z_{best} -y)+ (y-x^*)|_2 \
            &leq min_{y in C}|z_{best}-y|_2 + min_{y in C}|y-x^*|_2 \
            & leq min_{yin C}|y-x^*|_2 \
            &=d_C(x^*)
            end{align}




            Answer above has mistakes. while, maybe we can try a new way to solve this.



            Notice that for any element $y in C$, because $C subseteq D$, so that $y in D$, and another element $z_{best} in D$, satisfy $d_D(x^*)=min_{z in D} |z-x^*|_2=|z_{best}-x^*|_2$, from this definition, it is quite clear to see
            $$ |y-x^*|_2 geq |z_{best}-x^*|_2, forall y in C$$
            which results in
            $$d_C(x^*) = min_{yin C}|y-x^*|_2 geq |z_{best}-x^*|_2 = d_D(x^*)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 '18 at 7:48

























            answered Nov 25 '18 at 8:17









            Caldera

            11




            11












            • How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
              – Saeed
              Nov 25 '18 at 18:59










            • You are right! And I post a new solution under the original one. I hope it can help you.
              – Caldera
              Nov 26 '18 at 7:50


















            • How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
              – Saeed
              Nov 25 '18 at 18:59










            • You are right! And I post a new solution under the original one. I hope it can help you.
              – Caldera
              Nov 26 '18 at 7:50
















            How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
            – Saeed
            Nov 25 '18 at 18:59




            How do you know when you ignore $min_{y in C}|z_{best}-y|_2$ inequality still holds?
            – Saeed
            Nov 25 '18 at 18:59












            You are right! And I post a new solution under the original one. I hope it can help you.
            – Caldera
            Nov 26 '18 at 7:50




            You are right! And I post a new solution under the original one. I hope it can help you.
            – Caldera
            Nov 26 '18 at 7:50


















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