Finite subcover with special property












2















Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



$$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






general-topology metric-spaces compactness







share|cite|improve this question





























    2















    Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




    Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



    $$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



    How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






    general-topology metric-spaces compactness







    share|cite|improve this question



























      2












      2








      2








      Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




      Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



      $$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



      How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






      general-topology metric-spaces compactness







      share|cite|improve this question
















      Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.




      Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have



      $$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$



      How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?






      general-topology metric-spaces compactness




      general-topology metric-spaces compactness






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 6 hours ago









      Kenny Wong

      17k21135




      17k21135










      asked 6 hours ago









      Lucas Corrêa

      1,3451321




      1,3451321






















          1 Answer
          1






          active

          oldest

          votes


















          4














          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer





















          • Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago











          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%2f3053369%2ffinite-subcover-with-special-property%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









          4














          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer





















          • Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago
















          4














          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer





















          • Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago














          4












          4








          4






          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...






          share|cite|improve this answer












          Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.



          Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?



          Then use the compactness of $M$ to find a finite refinement of $mathcal V$...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          Kenny Wong

          17k21135




          17k21135












          • Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago


















          • Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
            – Lucas Corrêa
            6 hours ago










          • Is this a good idea?
            – Lucas Corrêa
            6 hours ago






          • 1




            @LucasCorrêa Yes, that's the right idea.
            – Kenny Wong
            6 hours ago
















          Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
          – Lucas Corrêa
          6 hours ago




          Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
          – Lucas Corrêa
          6 hours ago












          Is this a good idea?
          – Lucas Corrêa
          6 hours ago




          Is this a good idea?
          – Lucas Corrêa
          6 hours ago




          1




          1




          @LucasCorrêa Yes, that's the right idea.
          – Kenny Wong
          6 hours ago




          @LucasCorrêa Yes, that's the right idea.
          – Kenny Wong
          6 hours ago


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3053369%2ffinite-subcover-with-special-property%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

          What visual should I use to simply compare current year value vs last year in Power BI desktop

          Image
          0 I have column SubToBind that shows current year ratio and column PY SubToBind which shows previous year ratio. How can I simply create a card that would be displaying them both in a nice way? Something like that: I cannot achieve that with regular card, I also tried Card with States by OKViz but also unable to see both values. In KPI visual - I dont like the word "Goal", it will confuse end users. Are there any cards or other visuals that would do the job? Thanks powerbi dax powerbi-embedded share | improve this question edited Jun 11 '18 at 20:11 Oleg
          Read more

          Alexandru Averescu

          Image
          Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center
          Read more

          Trompette piccolo

          Image
          Ne doit pas être confondu avec trompette de poche. Trompette piccolo en si bémol, avec des branches d'embouchure différentes pour le si bémol (la plus courte) ou la (la plus longue) Trompette piccolo La trompette piccolo est la plus petite de la famille des trompettes. Elle joue une octave plus haut que la trompette standard en si bémol. La plupart des trompettes piccolo sont conçues pour jouer dans les deux tonalités de si bémol ou la, en utilisant une branche d'embouchure différente pour chaque tonalité. Le tube de la trompette piccolo en si bémol a une longueur moitié de celle de la trompette normale en si bémol. Des trompettes piccolo en sol, fa, ut sont également fabriquées, mais elles sont rares. La trompette soprano en ré, aussi connu comme la « trompette Bach », a été inventée vers 1890 par le facteur belge Victor-Charles Mahillon pour jouer les parties hautes de la trompette dans la musique de Bach et Haendel. La trompette piccolo moderne permet aux
          Read more