Does exist a compact connected $K' subset U$ such that $K subset K'$, if $U$ is an open connected and $K$ in...












1












$begingroup$


Let $U$ be an open and connected set in $mathbb{R}^n$.
Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?



I know that there exists a compact set $K' subset U$ such that $K subset K'$.
But how I can guarantee that $K'$ is connected?






real-analysis general-topology connectedness







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $U$ be an open and connected set in $mathbb{R}^n$.
    Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?



    I know that there exists a compact set $K' subset U$ such that $K subset K'$.
    But how I can guarantee that $K'$ is connected?






    real-analysis general-topology connectedness







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $U$ be an open and connected set in $mathbb{R}^n$.
      Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?



      I know that there exists a compact set $K' subset U$ such that $K subset K'$.
      But how I can guarantee that $K'$ is connected?






      real-analysis general-topology connectedness







      share|cite|improve this question











      $endgroup$




      Let $U$ be an open and connected set in $mathbb{R}^n$.
      Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?



      I know that there exists a compact set $K' subset U$ such that $K subset K'$.
      But how I can guarantee that $K'$ is connected?






      real-analysis general-topology connectedness




      real-analysis general-topology connectedness






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 11:01







      Santos

















      asked Dec 5 '18 at 10:55









      SantosSantos

      626514




      626514






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.



          Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.



          $U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.



          Then
          $K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
          is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.



            Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.



            Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.



            Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.



            By construction $K'subset U$.






            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%2f3026934%2fdoes-exist-a-compact-connected-k-subset-u-such-that-k-subset-k-if-u-i%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









              5












              $begingroup$

              Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.



              Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.



              $U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.



              Then
              $K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
              is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.



                Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.



                $U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.



                Then
                $K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
                is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.



                  Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.



                  $U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.



                  Then
                  $K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
                  is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.






                  share|cite|improve this answer









                  $endgroup$



                  Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.



                  Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.



                  $U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.



                  Then
                  $K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
                  is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 11:32









                  David MorrisDavid Morris

                  661




                  661























                      1












                      $begingroup$

                      For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.



                      Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.



                      Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.



                      Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.



                      By construction $K'subset U$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.



                        Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.



                        Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.



                        Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.



                        By construction $K'subset U$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.



                          Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.



                          Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.



                          Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.



                          By construction $K'subset U$.






                          share|cite|improve this answer









                          $endgroup$



                          For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.



                          Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.



                          Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.



                          Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.



                          By construction $K'subset U$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 5 '18 at 11:38









                          user126154user126154

                          5,378716




                          5,378716






























                              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%2f3026934%2fdoes-exist-a-compact-connected-k-subset-u-such-that-k-subset-k-if-u-i%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

                              Plaza Victoria

                              Image
                              Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                              Read more

                              Puebla de Zaragoza

                              Image
                              Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                              Read more

                              Musa

                              Image
                              Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                              Read more