Is every metrizable topology induced by a Heine-Borel metric?












2












$begingroup$


A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?



If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?






general-topology metric-spaces compactness examples-counterexamples







share|cite|improve this question









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    2












    $begingroup$


    A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?



    If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?






    general-topology metric-spaces compactness examples-counterexamples







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?



      If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?






      general-topology metric-spaces compactness examples-counterexamples







      share|cite|improve this question









      $endgroup$




      A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?



      If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?






      general-topology metric-spaces compactness examples-counterexamples




      general-topology metric-spaces compactness examples-counterexamples






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 19 '18 at 6:03









      Keshav SrinivasanKeshav Srinivasan

      2,39121446




      2,39121446






















          1 Answer
          1






          active

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          3












          $begingroup$

          Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Or any uncountable set with the discrete topology, for the same reason.
            $endgroup$
            – A. Pongrácz
            Dec 19 '18 at 6:16










          • $begingroup$
            Why does the Heine-Borel property imply sigma-compactness?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:19










          • $begingroup$
            $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
            $endgroup$
            – Kavi Rama Murthy
            Dec 19 '18 at 6:21












          • $begingroup$
            OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:29










          • $begingroup$
            It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
            $endgroup$
            – bof
            Dec 19 '18 at 9:17












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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Or any uncountable set with the discrete topology, for the same reason.
            $endgroup$
            – A. Pongrácz
            Dec 19 '18 at 6:16










          • $begingroup$
            Why does the Heine-Borel property imply sigma-compactness?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:19










          • $begingroup$
            $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
            $endgroup$
            – Kavi Rama Murthy
            Dec 19 '18 at 6:21












          • $begingroup$
            OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:29










          • $begingroup$
            It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
            $endgroup$
            – bof
            Dec 19 '18 at 9:17
















          3












          $begingroup$

          Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Or any uncountable set with the discrete topology, for the same reason.
            $endgroup$
            – A. Pongrácz
            Dec 19 '18 at 6:16










          • $begingroup$
            Why does the Heine-Borel property imply sigma-compactness?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:19










          • $begingroup$
            $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
            $endgroup$
            – Kavi Rama Murthy
            Dec 19 '18 at 6:21












          • $begingroup$
            OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:29










          • $begingroup$
            It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
            $endgroup$
            – bof
            Dec 19 '18 at 9:17














          3












          3








          3





          $begingroup$

          Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.






          share|cite|improve this answer









          $endgroup$



          Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 6:13









          Kavi Rama MurthyKavi Rama Murthy

          71.5k53170




          71.5k53170












          • $begingroup$
            Or any uncountable set with the discrete topology, for the same reason.
            $endgroup$
            – A. Pongrácz
            Dec 19 '18 at 6:16










          • $begingroup$
            Why does the Heine-Borel property imply sigma-compactness?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:19










          • $begingroup$
            $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
            $endgroup$
            – Kavi Rama Murthy
            Dec 19 '18 at 6:21












          • $begingroup$
            OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:29










          • $begingroup$
            It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
            $endgroup$
            – bof
            Dec 19 '18 at 9:17


















          • $begingroup$
            Or any uncountable set with the discrete topology, for the same reason.
            $endgroup$
            – A. Pongrácz
            Dec 19 '18 at 6:16










          • $begingroup$
            Why does the Heine-Borel property imply sigma-compactness?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:19










          • $begingroup$
            $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
            $endgroup$
            – Kavi Rama Murthy
            Dec 19 '18 at 6:21












          • $begingroup$
            OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
            $endgroup$
            – Keshav Srinivasan
            Dec 19 '18 at 6:29










          • $begingroup$
            It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
            $endgroup$
            – bof
            Dec 19 '18 at 9:17
















          $begingroup$
          Or any uncountable set with the discrete topology, for the same reason.
          $endgroup$
          – A. Pongrácz
          Dec 19 '18 at 6:16




          $begingroup$
          Or any uncountable set with the discrete topology, for the same reason.
          $endgroup$
          – A. Pongrácz
          Dec 19 '18 at 6:16












          $begingroup$
          Why does the Heine-Borel property imply sigma-compactness?
          $endgroup$
          – Keshav Srinivasan
          Dec 19 '18 at 6:19




          $begingroup$
          Why does the Heine-Borel property imply sigma-compactness?
          $endgroup$
          – Keshav Srinivasan
          Dec 19 '18 at 6:19












          $begingroup$
          $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
          $endgroup$
          – Kavi Rama Murthy
          Dec 19 '18 at 6:21






          $begingroup$
          $X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
          $endgroup$
          – Kavi Rama Murthy
          Dec 19 '18 at 6:21














          $begingroup$
          OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
          $endgroup$
          – Keshav Srinivasan
          Dec 19 '18 at 6:29




          $begingroup$
          OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
          $endgroup$
          – Keshav Srinivasan
          Dec 19 '18 at 6:29












          $begingroup$
          It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
          $endgroup$
          – bof
          Dec 19 '18 at 9:17




          $begingroup$
          It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
          $endgroup$
          – bof
          Dec 19 '18 at 9:17


















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