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












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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







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    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

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          3 Answers
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          $begingroup$

          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









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              3 Answers
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              3 Answers
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              active

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              active

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              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






























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