Exterior Covering Number of $epsilon /2$ is greater of equal than Covering Number of $epsilon$












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Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.



$$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$



Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$



Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$



then I was asked to prove that:



$$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$



how to see that ?



here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.






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    Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.



    $$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$



    Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$



    Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$



    then I was asked to prove that:



    $$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$



    how to see that ?



    here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.






    real-analysis probability general-topology functional-analysis







    share|cite|improve this question



























      2












      2








      2







      Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.



      $$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$



      Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$



      Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$



      then I was asked to prove that:



      $$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$



      how to see that ?



      here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.






      real-analysis probability general-topology functional-analysis







      share|cite|improve this question















      Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.



      $$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$



      Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$



      Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$



      then I was asked to prove that:



      $$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$



      how to see that ?



      here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.






      real-analysis probability general-topology functional-analysis




      real-analysis probability general-topology functional-analysis






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      edited Dec 28 '18 at 0:51

























      asked Dec 28 '18 at 0:43









      ShaoyuPei

      1698




      1698






















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          Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.



          For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).



          By the triangle inequality,
          $$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
          for each $i$, which means that
          $$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
          is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.






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            Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.



            For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).



            By the triangle inequality,
            $$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
            for each $i$, which means that
            $$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
            is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.






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              Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.



              For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).



              By the triangle inequality,
              $$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
              for each $i$, which means that
              $$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
              is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.






              share|cite|improve this answer


























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                Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.



                For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).



                By the triangle inequality,
                $$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
                for each $i$, which means that
                $$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
                is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.






                share|cite|improve this answer














                Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.



                For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).



                By the triangle inequality,
                $$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
                for each $i$, which means that
                $$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
                is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.







                share|cite|improve this answer














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                edited Dec 28 '18 at 1:33

























                answered Dec 28 '18 at 1:01









                Kenny Wong

                18k21438




                18k21438






























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