Show that if two metrics induce the same topology, one metric space is compact iff the other one is.












0












$begingroup$


Question: Given two metric spaces $(X, d)$ and $(X,e)$, where $tau(X,d)$ = $tau(X,e)$, show that $(X,e)$ is compact if and only if $(X,d)$ is compact.



I'm aiming to prove this using the definition of compactness whereby a metric space is compact exactly when each sequence contains a convergent subsequence. I can't see how the existence of a convergent subsequence (of a sequence) in $(X,d)$ could guarantee one in $(X,e)$. Any idea what comes next?






general-topology metric-spaces compactness cauchy-sequences







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Question: Given two metric spaces $(X, d)$ and $(X,e)$, where $tau(X,d)$ = $tau(X,e)$, show that $(X,e)$ is compact if and only if $(X,d)$ is compact.



    I'm aiming to prove this using the definition of compactness whereby a metric space is compact exactly when each sequence contains a convergent subsequence. I can't see how the existence of a convergent subsequence (of a sequence) in $(X,d)$ could guarantee one in $(X,e)$. Any idea what comes next?






    general-topology metric-spaces compactness cauchy-sequences







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Question: Given two metric spaces $(X, d)$ and $(X,e)$, where $tau(X,d)$ = $tau(X,e)$, show that $(X,e)$ is compact if and only if $(X,d)$ is compact.



      I'm aiming to prove this using the definition of compactness whereby a metric space is compact exactly when each sequence contains a convergent subsequence. I can't see how the existence of a convergent subsequence (of a sequence) in $(X,d)$ could guarantee one in $(X,e)$. Any idea what comes next?






      general-topology metric-spaces compactness cauchy-sequences







      share|cite|improve this question











      $endgroup$




      Question: Given two metric spaces $(X, d)$ and $(X,e)$, where $tau(X,d)$ = $tau(X,e)$, show that $(X,e)$ is compact if and only if $(X,d)$ is compact.



      I'm aiming to prove this using the definition of compactness whereby a metric space is compact exactly when each sequence contains a convergent subsequence. I can't see how the existence of a convergent subsequence (of a sequence) in $(X,d)$ could guarantee one in $(X,e)$. Any idea what comes next?






      general-topology metric-spaces compactness cauchy-sequences




      general-topology metric-spaces compactness cauchy-sequences






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 15:54







      H H

















      asked Dec 16 '18 at 15:39









      H HH H

      185




      185






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          If two metrics $d$ and $e$ induce the same topology, then a sequence $(x_n)_{ninmathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$. Therefore, a sequence has a convergent subsequence in $(X,d)$ if and only if it has a convergent subsequence in $(X,e)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
            $endgroup$
            – H H
            Dec 16 '18 at 15:54










          • $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 15:54










          • $begingroup$
            Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
            $endgroup$
            – Matematleta
            Dec 16 '18 at 16:04










          • $begingroup$
            Of course, but the OP is interested in the caracterization of compacity through sequences.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 16:05



















          1












          $begingroup$

          $x_n to x$ in a topological space iff for all open subsets $O$ that contain $x$, $O$ contains all but finitely many $x_n$.



          So convergence only depends on the topology not on the metric, and so if $d$ and $e$ are equivalent metrics, they have the same open sets and thus the same sequences converge to the same points in either metric. So if $(X,d)$ is sequentially compact, so is $(X,e)$ and vice versa. That's really all there is to it.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            A sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_d(x,delta)$ in $(X,d)$ iff the sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_e(x,delta)$ in $(X,e)$ as the two metrics $d$ & $e$ generate the same topology. That is $(x_n)_{n=1}^{infty}$ is convergent in $(X,d)$ iff it is convergent in $(X,e)$. Hence if one of $(X,d)$ & $(X,e)$ is compact, then the other is also so.






            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%2f3042738%2fshow-that-if-two-metrics-induce-the-same-topology-one-metric-space-is-compact-i%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              If two metrics $d$ and $e$ induce the same topology, then a sequence $(x_n)_{ninmathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$. Therefore, a sequence has a convergent subsequence in $(X,d)$ if and only if it has a convergent subsequence in $(X,e)$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
                $endgroup$
                – H H
                Dec 16 '18 at 15:54










              • $begingroup$
                I've edited my answer. I hope that everything is clear now.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 15:54










              • $begingroup$
                Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
                $endgroup$
                – Matematleta
                Dec 16 '18 at 16:04










              • $begingroup$
                Of course, but the OP is interested in the caracterization of compacity through sequences.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 16:05
















              1












              $begingroup$

              If two metrics $d$ and $e$ induce the same topology, then a sequence $(x_n)_{ninmathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$. Therefore, a sequence has a convergent subsequence in $(X,d)$ if and only if it has a convergent subsequence in $(X,e)$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
                $endgroup$
                – H H
                Dec 16 '18 at 15:54










              • $begingroup$
                I've edited my answer. I hope that everything is clear now.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 15:54










              • $begingroup$
                Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
                $endgroup$
                – Matematleta
                Dec 16 '18 at 16:04










              • $begingroup$
                Of course, but the OP is interested in the caracterization of compacity through sequences.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 16:05














              1












              1








              1





              $begingroup$

              If two metrics $d$ and $e$ induce the same topology, then a sequence $(x_n)_{ninmathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$. Therefore, a sequence has a convergent subsequence in $(X,d)$ if and only if it has a convergent subsequence in $(X,e)$.






              share|cite|improve this answer











              $endgroup$



              If two metrics $d$ and $e$ induce the same topology, then a sequence $(x_n)_{ninmathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$. Therefore, a sequence has a convergent subsequence in $(X,d)$ if and only if it has a convergent subsequence in $(X,e)$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 16 '18 at 15:54

























              answered Dec 16 '18 at 15:44









              José Carlos SantosJosé Carlos Santos

              166k22132235




              166k22132235












              • $begingroup$
                I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
                $endgroup$
                – H H
                Dec 16 '18 at 15:54










              • $begingroup$
                I've edited my answer. I hope that everything is clear now.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 15:54










              • $begingroup$
                Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
                $endgroup$
                – Matematleta
                Dec 16 '18 at 16:04










              • $begingroup$
                Of course, but the OP is interested in the caracterization of compacity through sequences.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 16:05


















              • $begingroup$
                I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
                $endgroup$
                – H H
                Dec 16 '18 at 15:54










              • $begingroup$
                I've edited my answer. I hope that everything is clear now.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 15:54










              • $begingroup$
                Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
                $endgroup$
                – Matematleta
                Dec 16 '18 at 16:04










              • $begingroup$
                Of course, but the OP is interested in the caracterization of compacity through sequences.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 16:05
















              $begingroup$
              I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
              $endgroup$
              – H H
              Dec 16 '18 at 15:54




              $begingroup$
              I'm not sure I understand what you're saying I'm wrong about - isn't that the exact definition I just gave? Can you explain why it has to be the case that if two metrics induce the same topology then a sequence $x_n$ only converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$.
              $endgroup$
              – H H
              Dec 16 '18 at 15:54












              $begingroup$
              I've edited my answer. I hope that everything is clear now.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 15:54




              $begingroup$
              I've edited my answer. I hope that everything is clear now.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 15:54












              $begingroup$
              Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
              $endgroup$
              – Matematleta
              Dec 16 '18 at 16:04




              $begingroup$
              Or you can just observe that the open cover definition of compactness implies that it is a purely topological property
              $endgroup$
              – Matematleta
              Dec 16 '18 at 16:04












              $begingroup$
              Of course, but the OP is interested in the caracterization of compacity through sequences.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 16:05




              $begingroup$
              Of course, but the OP is interested in the caracterization of compacity through sequences.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 16:05











              1












              $begingroup$

              $x_n to x$ in a topological space iff for all open subsets $O$ that contain $x$, $O$ contains all but finitely many $x_n$.



              So convergence only depends on the topology not on the metric, and so if $d$ and $e$ are equivalent metrics, they have the same open sets and thus the same sequences converge to the same points in either metric. So if $(X,d)$ is sequentially compact, so is $(X,e)$ and vice versa. That's really all there is to it.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $x_n to x$ in a topological space iff for all open subsets $O$ that contain $x$, $O$ contains all but finitely many $x_n$.



                So convergence only depends on the topology not on the metric, and so if $d$ and $e$ are equivalent metrics, they have the same open sets and thus the same sequences converge to the same points in either metric. So if $(X,d)$ is sequentially compact, so is $(X,e)$ and vice versa. That's really all there is to it.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $x_n to x$ in a topological space iff for all open subsets $O$ that contain $x$, $O$ contains all but finitely many $x_n$.



                  So convergence only depends on the topology not on the metric, and so if $d$ and $e$ are equivalent metrics, they have the same open sets and thus the same sequences converge to the same points in either metric. So if $(X,d)$ is sequentially compact, so is $(X,e)$ and vice versa. That's really all there is to it.






                  share|cite|improve this answer









                  $endgroup$



                  $x_n to x$ in a topological space iff for all open subsets $O$ that contain $x$, $O$ contains all but finitely many $x_n$.



                  So convergence only depends on the topology not on the metric, and so if $d$ and $e$ are equivalent metrics, they have the same open sets and thus the same sequences converge to the same points in either metric. So if $(X,d)$ is sequentially compact, so is $(X,e)$ and vice versa. That's really all there is to it.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 11:00









                  Henno BrandsmaHenno Brandsma

                  112k348120




                  112k348120























                      0












                      $begingroup$

                      A sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_d(x,delta)$ in $(X,d)$ iff the sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_e(x,delta)$ in $(X,e)$ as the two metrics $d$ & $e$ generate the same topology. That is $(x_n)_{n=1}^{infty}$ is convergent in $(X,d)$ iff it is convergent in $(X,e)$. Hence if one of $(X,d)$ & $(X,e)$ is compact, then the other is also so.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        A sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_d(x,delta)$ in $(X,d)$ iff the sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_e(x,delta)$ in $(X,e)$ as the two metrics $d$ & $e$ generate the same topology. That is $(x_n)_{n=1}^{infty}$ is convergent in $(X,d)$ iff it is convergent in $(X,e)$. Hence if one of $(X,d)$ & $(X,e)$ is compact, then the other is also so.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          A sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_d(x,delta)$ in $(X,d)$ iff the sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_e(x,delta)$ in $(X,e)$ as the two metrics $d$ & $e$ generate the same topology. That is $(x_n)_{n=1}^{infty}$ is convergent in $(X,d)$ iff it is convergent in $(X,e)$. Hence if one of $(X,d)$ & $(X,e)$ is compact, then the other is also so.






                          share|cite|improve this answer









                          $endgroup$



                          A sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_d(x,delta)$ in $(X,d)$ iff the sequence $(x_n)_{n=1}^{infty}$ is eventual in $B_e(x,delta)$ in $(X,e)$ as the two metrics $d$ & $e$ generate the same topology. That is $(x_n)_{n=1}^{infty}$ is convergent in $(X,d)$ iff it is convergent in $(X,e)$. Hence if one of $(X,d)$ & $(X,e)$ is compact, then the other is also so.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 17 '18 at 11:41









                          Arjun BanerjeeArjun Banerjee

                          550110




                          550110






























                              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%2f3042738%2fshow-that-if-two-metrics-induce-the-same-topology-one-metric-space-is-compact-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

                              Bundesstraße 106

                              Image
                              Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
                              Read more

                              Verónica Boquete

                              Image
                              Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
                              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