Is a finitely generated metrizable group discrete?












2














The question is in the title. A countable locally compact Hausdorff group is discrete, so saying that a finitely generated metrizable group is locally compact would be enough.



What if the group is a subgroup of a compact metrizable group? An open or closed subspace of a compact Hausdorff space is locally compact, but here there is no hypothesis on opennes or closure.






abstract-algebra group-theory topological-groups finitely-generated locally-compact-groups







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    What about $mathbb{Z}$ with the $p$-adic metric ?
    – reuns
    Nov 29 at 13:35
















2














The question is in the title. A countable locally compact Hausdorff group is discrete, so saying that a finitely generated metrizable group is locally compact would be enough.



What if the group is a subgroup of a compact metrizable group? An open or closed subspace of a compact Hausdorff space is locally compact, but here there is no hypothesis on opennes or closure.






abstract-algebra group-theory topological-groups finitely-generated locally-compact-groups







share|cite|improve this question


















  • 2




    What about $mathbb{Z}$ with the $p$-adic metric ?
    – reuns
    Nov 29 at 13:35














2












2








2


1





The question is in the title. A countable locally compact Hausdorff group is discrete, so saying that a finitely generated metrizable group is locally compact would be enough.



What if the group is a subgroup of a compact metrizable group? An open or closed subspace of a compact Hausdorff space is locally compact, but here there is no hypothesis on opennes or closure.






abstract-algebra group-theory topological-groups finitely-generated locally-compact-groups







share|cite|improve this question













The question is in the title. A countable locally compact Hausdorff group is discrete, so saying that a finitely generated metrizable group is locally compact would be enough.



What if the group is a subgroup of a compact metrizable group? An open or closed subspace of a compact Hausdorff space is locally compact, but here there is no hypothesis on opennes or closure.






abstract-algebra group-theory topological-groups finitely-generated locally-compact-groups




abstract-algebra group-theory topological-groups finitely-generated locally-compact-groups






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asked Nov 29 at 12:59









frafour

841212




841212








  • 2




    What about $mathbb{Z}$ with the $p$-adic metric ?
    – reuns
    Nov 29 at 13:35














  • 2




    What about $mathbb{Z}$ with the $p$-adic metric ?
    – reuns
    Nov 29 at 13:35








2




2




What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 at 13:35




What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 at 13:35










2 Answers
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2














The $p$-adic topology on $mathbb{Z}$, having as a basis of neighborhoods the subgroups of $mathbb{Z}$ of the form $p^nmathbb{Z}$ ($p$ a prime) is metrizable and non discrete. The completion is the group (or ring) of $p$-adic integers, which is compact and metrizable.






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    2














    No. Consider for instance the subgroup of $mathbb{R}/mathbb{Z}$ generated $a + mathbb{Z}$ where $a$ is irrational, with the natural metric inherited from $mathbb{R}/mathbb{Z}$.



    It is true though that every countable completely metrizable topological group is discrete, which follows from the more general fact that every countable (edit) homogeneous (/edit) complete metric space is discrete.






    share|cite|improve this answer



















    • 1




      Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
      – YCor
      Nov 29 at 21:54












    • Sorry yes, I meant to say 'homogeneous' in there.
      – Colin
      Nov 29 at 22:29











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    2 Answers
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    2 Answers
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    2














    The $p$-adic topology on $mathbb{Z}$, having as a basis of neighborhoods the subgroups of $mathbb{Z}$ of the form $p^nmathbb{Z}$ ($p$ a prime) is metrizable and non discrete. The completion is the group (or ring) of $p$-adic integers, which is compact and metrizable.






    share|cite|improve this answer


























      2














      The $p$-adic topology on $mathbb{Z}$, having as a basis of neighborhoods the subgroups of $mathbb{Z}$ of the form $p^nmathbb{Z}$ ($p$ a prime) is metrizable and non discrete. The completion is the group (or ring) of $p$-adic integers, which is compact and metrizable.






      share|cite|improve this answer
























        2












        2








        2






        The $p$-adic topology on $mathbb{Z}$, having as a basis of neighborhoods the subgroups of $mathbb{Z}$ of the form $p^nmathbb{Z}$ ($p$ a prime) is metrizable and non discrete. The completion is the group (or ring) of $p$-adic integers, which is compact and metrizable.






        share|cite|improve this answer












        The $p$-adic topology on $mathbb{Z}$, having as a basis of neighborhoods the subgroups of $mathbb{Z}$ of the form $p^nmathbb{Z}$ ($p$ a prime) is metrizable and non discrete. The completion is the group (or ring) of $p$-adic integers, which is compact and metrizable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 14:28









        egreg

        177k1484198




        177k1484198























            2














            No. Consider for instance the subgroup of $mathbb{R}/mathbb{Z}$ generated $a + mathbb{Z}$ where $a$ is irrational, with the natural metric inherited from $mathbb{R}/mathbb{Z}$.



            It is true though that every countable completely metrizable topological group is discrete, which follows from the more general fact that every countable (edit) homogeneous (/edit) complete metric space is discrete.






            share|cite|improve this answer



















            • 1




              Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
              – YCor
              Nov 29 at 21:54












            • Sorry yes, I meant to say 'homogeneous' in there.
              – Colin
              Nov 29 at 22:29
















            2














            No. Consider for instance the subgroup of $mathbb{R}/mathbb{Z}$ generated $a + mathbb{Z}$ where $a$ is irrational, with the natural metric inherited from $mathbb{R}/mathbb{Z}$.



            It is true though that every countable completely metrizable topological group is discrete, which follows from the more general fact that every countable (edit) homogeneous (/edit) complete metric space is discrete.






            share|cite|improve this answer



















            • 1




              Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
              – YCor
              Nov 29 at 21:54












            • Sorry yes, I meant to say 'homogeneous' in there.
              – Colin
              Nov 29 at 22:29














            2












            2








            2






            No. Consider for instance the subgroup of $mathbb{R}/mathbb{Z}$ generated $a + mathbb{Z}$ where $a$ is irrational, with the natural metric inherited from $mathbb{R}/mathbb{Z}$.



            It is true though that every countable completely metrizable topological group is discrete, which follows from the more general fact that every countable (edit) homogeneous (/edit) complete metric space is discrete.






            share|cite|improve this answer














            No. Consider for instance the subgroup of $mathbb{R}/mathbb{Z}$ generated $a + mathbb{Z}$ where $a$ is irrational, with the natural metric inherited from $mathbb{R}/mathbb{Z}$.



            It is true though that every countable completely metrizable topological group is discrete, which follows from the more general fact that every countable (edit) homogeneous (/edit) complete metric space is discrete.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 29 at 22:30

























            answered Nov 29 at 13:28









            Colin

            26317




            26317








            • 1




              Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
              – YCor
              Nov 29 at 21:54












            • Sorry yes, I meant to say 'homogeneous' in there.
              – Colin
              Nov 29 at 22:29














            • 1




              Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
              – YCor
              Nov 29 at 21:54












            • Sorry yes, I meant to say 'homogeneous' in there.
              – Colin
              Nov 29 at 22:29








            1




            1




            Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
            – YCor
            Nov 29 at 21:54






            Obviously not every countable complete metric space is discrete! However, every countable complete metric space has an isolated point (Baire). Hence if its self-homeomorphism group acts transitively (which holds for topological groups) it has to be discrete.
            – YCor
            Nov 29 at 21:54














            Sorry yes, I meant to say 'homogeneous' in there.
            – Colin
            Nov 29 at 22:29




            Sorry yes, I meant to say 'homogeneous' in there.
            – Colin
            Nov 29 at 22:29


















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