Is a finitely generated metrizable group discrete?
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
asked Nov 29 '18 at 12:59
frafourfrafour
878312
add a comment |
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
asked Nov 29 '18 at 12:59
frafourfrafour
878312
2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 '18 at 13:35
add a comment |
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
asked Nov 29 '18 at 12:59
frafourfrafour
878312
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
asked Nov 29 '18 at 12:59
frafourfrafour
878312
asked Nov 29 '18 at 12:59
frafourfrafour
878312
asked Nov 29 '18 at 12:59
frafourfrafour
878312
asked Nov 29 '18 at 12:59
frafourfrafour
878312
asked Nov 29 '18 at 12:59
frafourfrafour
878312
878312
2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 '18 at 13:35
add a comment |
2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 '18 at 13:35
2
2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 '18 at 13:35
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 '18 at 13:35
add a comment |
2 Answers
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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.
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
add a comment |
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.
answered Nov 29 '18 at 13:28
ColinColin
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 '18 at 21:54
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 at 22:29
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
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.
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
add a comment |
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.
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
add a comment |
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.
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
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.
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
answered Nov 29 '18 at 14:28
egregegreg
179k1485202
179k1485202
add a comment |
add a comment |
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.
answered Nov 29 '18 at 13:28
ColinColin
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 '18 at 21:54
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 at 22:29
add a comment |
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.
answered Nov 29 '18 at 13:28
ColinColin
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 '18 at 21:54
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 at 22:29
add a comment |
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.
answered Nov 29 '18 at 13:28
ColinColin
26317
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.
answered Nov 29 '18 at 13:28
ColinColin
26317
edited Nov 29 '18 at 22:30
answered Nov 29 '18 at 13:28
ColinColin
26317
answered Nov 29 '18 at 13:28
ColinColin
26317
answered Nov 29 '18 at 13:28
ColinColin
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 '18 at 21:54
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 at 22:29
add a comment |
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 '18 at 21:54
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 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 '18 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 '18 at 21:54
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 at 22:29
Sorry yes, I meant to say 'homogeneous' in there.
– Colin
Nov 29 '18 at 22:29
add a comment |
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2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 '18 at 13:35