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












2












$begingroup$


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











$endgroup$

















    2












    $begingroup$


    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











    $endgroup$















      2












      2








      2





      $begingroup$


      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 28 '18 at 0:51







      ShaoyuPei

















      asked Dec 28 '18 at 0:43









      ShaoyuPeiShaoyuPei

      1778




      1778






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          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











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054478%2fexterior-covering-number-of-epsilon-2-is-greater-of-equal-than-covering-numb%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            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











            $endgroup$


















              1












              $begingroup$

              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











              $endgroup$
















                1












                1








                1





                $begingroup$

                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











                $endgroup$



                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














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 28 '18 at 1:33

























                answered Dec 28 '18 at 1:01









                Kenny WongKenny Wong

                19k21440




                19k21440






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054478%2fexterior-covering-number-of-epsilon-2-is-greater-of-equal-than-covering-numb%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                    Read more