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 at 12:59
frafour
841212
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 at 12:59
frafour
841212
2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 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 at 12:59
frafour
841212
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 at 12:59
frafour
841212
asked Nov 29 at 12:59
frafour
841212
asked Nov 29 at 12:59
frafour
841212
asked Nov 29 at 12:59
frafour
841212
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
add a comment |
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
add a comment |
2 Answers
2
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 at 14:28
egreg
177k1484198
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 at 13:28
Colin
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
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018588%2fis-a-finitely-generated-metrizable-group-discrete%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 at 14:28
egreg
177k1484198
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 at 14:28
egreg
177k1484198
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 at 14:28
egreg
177k1484198
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 at 14:28
egreg
177k1484198
answered Nov 29 at 14:28
egreg
177k1484198
answered Nov 29 at 14:28
egreg
177k1484198
answered Nov 29 at 14:28
egreg
177k1484198
177k1484198
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 at 13:28
Colin
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
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 at 13:28
Colin
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
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 at 13:28
Colin
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 at 13:28
Colin
26317
edited Nov 29 at 22:30
answered Nov 29 at 13:28
Colin
26317
answered Nov 29 at 13:28
Colin
26317
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
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 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
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018588%2fis-a-finitely-generated-metrizable-group-discrete%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2
What about $mathbb{Z}$ with the $p$-adic metric ?
– reuns
Nov 29 at 13:35