Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.












1












$begingroup$


Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00
















1












$begingroup$


Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00














1












1








1





$begingroup$


Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis







share|cite|improve this question











$endgroup$




Let $B$ be a compact subset of a metric space $X$. Let $x in X setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x in U$ and $B subseteq V$.



Is this an existence proof? If yes, do I get to define $U,V$?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 14:22







mathdaddy

















asked Dec 20 '18 at 2:25









mathdaddymathdaddy

133




133












  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00


















  • $begingroup$
    you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
    $endgroup$
    – Martund
    Dec 20 '18 at 2:39








  • 1




    $begingroup$
    Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
    $endgroup$
    – Clayton
    Dec 20 '18 at 3:00
















$begingroup$
you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
$endgroup$
– Martund
Dec 20 '18 at 2:39






$begingroup$
you cannot choose such an $varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ bin B:forall varepsilon>0 B_varepsilon(b)cap Bneq phi$
$endgroup$
– Martund
Dec 20 '18 at 2:39






1




1




$begingroup$
Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
$endgroup$
– Clayton
Dec 20 '18 at 3:00




$begingroup$
Am I missing something? If $bin B$ and $B_varepsilon(b)$ means the ball centered at $b$ of radius $varepsilon$, then $B_varepsilon(b)cap Bneqvarnothing$ no matter how big or small $varepsilon>0$. This is because $bin B_varepsilon(b)$ and $bin B$, so it is in the intersection.
$endgroup$
– Clayton
Dec 20 '18 at 3:00










2 Answers
2






active

oldest

votes


















0












$begingroup$

$B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



    Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



    Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






    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%2f3047081%2fprove-that-there-exists-disjoint-open-subsets-u-v-of-x-such-that-x-in-u-a%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



      Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



        Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



          Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.






          share|cite|improve this answer









          $endgroup$



          $B$ is compact so it is closed. Its complement is open so for any $xin B^c$ there exists $epsilongt 0$ such that the open ball of radius $epsilon$ around $x$ is a subset of $B^c$.



          Let $U$ be the open ball of radius $frac{epsilon}{2}$ around $x$ and let $V={y:d(x,y)gtfrac{epsilon}{2}}$. Then $xin U$ and $Bsubset{y:d(x,y)geepsilon}subset{y:d(x,y)gtfrac{epsilon}{2}}=V$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 3:16









          John DoumaJohn Douma

          5,62711420




          5,62711420























              0












              $begingroup$

              You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



              Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



              Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



                Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



                Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



                  Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



                  Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$






                  share|cite|improve this answer











                  $endgroup$



                  You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)setminus{b}cap Bneqvarnothing;forall;r>0.$



                  Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.



                  Instead choose $delta=inf{d(x,y):::yin B}$. Consider $U=Bleft(x,dfrac{delta}{2}right)$ and $displaystyle V=bigcup_{bin B} Bleft(b,frac{delta}{2}right)$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 20 '18 at 3:00

























                  answered Dec 20 '18 at 2:31









                  Yadati KiranYadati Kiran

                  2,1121622




                  2,1121622






























                      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%2f3047081%2fprove-that-there-exists-disjoint-open-subsets-u-v-of-x-such-that-x-in-u-a%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

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more

                      web3.py web3.isConnected() returns false always

                      Image
                      1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                      Read more