Is $Bbb Q / Bbb Z$ discrete?
I would like to say that $Bbb Q / Bbb Z$ is not discrete (when $Bbb Q$ has euclidean topology), since $Bbb Z subset Bbb Q$ is not open. But OTOH we have
$$Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$$
which is a direct sum of discrete groups, so it should be a discrete group. Maybe the issue is that the above isomorphism is only as abstract groups, but not as topological groups.
Could anyone confirm/elaborate on this?
general-topology topological-groups
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
add a comment |
I would like to say that $Bbb Q / Bbb Z$ is not discrete (when $Bbb Q$ has euclidean topology), since $Bbb Z subset Bbb Q$ is not open. But OTOH we have
$$Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$$
which is a direct sum of discrete groups, so it should be a discrete group. Maybe the issue is that the above isomorphism is only as abstract groups, but not as topological groups.
Could anyone confirm/elaborate on this?
general-topology topological-groups
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
3
I think you're right, the above isomorphism is only an isomorphism of the underlying group structures.
– Brahadeesh
Nov 29 '18 at 9:53
2
Note that $Bbb Q$ with the Euclidean topology is isomorphic to $Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology.
– Asaf Karagila♦
Nov 29 '18 at 9:54
1
The completion of the metric space $mathbb{Q}/mathbb{Z}, d(a,b) = min_n |a-bn|$ is $mathbb{R}/mathbb{Z}$ while with the discrete metric $tilde{d}(a,b) = sup_p inf_n |a-b n|_p$ it is complete and $cong bigoplus_p Bbb Q_p / Bbb Z_p$.
– reuns
Nov 29 '18 at 10:20
@reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology.
– Paul Frost
Nov 29 '18 at 11:25
@PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ?
– reuns
Nov 29 '18 at 11:52
add a comment |
I would like to say that $Bbb Q / Bbb Z$ is not discrete (when $Bbb Q$ has euclidean topology), since $Bbb Z subset Bbb Q$ is not open. But OTOH we have
$$Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$$
which is a direct sum of discrete groups, so it should be a discrete group. Maybe the issue is that the above isomorphism is only as abstract groups, but not as topological groups.
Could anyone confirm/elaborate on this?
general-topology topological-groups
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
I would like to say that $Bbb Q / Bbb Z$ is not discrete (when $Bbb Q$ has euclidean topology), since $Bbb Z subset Bbb Q$ is not open. But OTOH we have
$$Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$$
which is a direct sum of discrete groups, so it should be a discrete group. Maybe the issue is that the above isomorphism is only as abstract groups, but not as topological groups.
Could anyone confirm/elaborate on this?
general-topology topological-groups
general-topology topological-groups
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
asked Nov 29 '18 at 9:44
AlphonseAlphonse
2,178623
2,178623
3
I think you're right, the above isomorphism is only an isomorphism of the underlying group structures.
– Brahadeesh
Nov 29 '18 at 9:53
2
Note that $Bbb Q$ with the Euclidean topology is isomorphic to $Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology.
– Asaf Karagila♦
Nov 29 '18 at 9:54
1
The completion of the metric space $mathbb{Q}/mathbb{Z}, d(a,b) = min_n |a-bn|$ is $mathbb{R}/mathbb{Z}$ while with the discrete metric $tilde{d}(a,b) = sup_p inf_n |a-b n|_p$ it is complete and $cong bigoplus_p Bbb Q_p / Bbb Z_p$.
– reuns
Nov 29 '18 at 10:20
@reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology.
– Paul Frost
Nov 29 '18 at 11:25
@PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ?
– reuns
Nov 29 '18 at 11:52
add a comment |
3
I think you're right, the above isomorphism is only an isomorphism of the underlying group structures.
– Brahadeesh
Nov 29 '18 at 9:53
2
Note that $Bbb Q$ with the Euclidean topology is isomorphic to $Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology.
– Asaf Karagila♦
Nov 29 '18 at 9:54
1
The completion of the metric space $mathbb{Q}/mathbb{Z}, d(a,b) = min_n |a-bn|$ is $mathbb{R}/mathbb{Z}$ while with the discrete metric $tilde{d}(a,b) = sup_p inf_n |a-b n|_p$ it is complete and $cong bigoplus_p Bbb Q_p / Bbb Z_p$.
– reuns
Nov 29 '18 at 10:20
@reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology.
– Paul Frost
Nov 29 '18 at 11:25
@PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ?
– reuns
Nov 29 '18 at 11:52
3
3
I think you're right, the above isomorphism is only an isomorphism of the underlying group structures.
– Brahadeesh
Nov 29 '18 at 9:53
I think you're right, the above isomorphism is only an isomorphism of the underlying group structures.
– Brahadeesh
Nov 29 '18 at 9:53
2
2
Note that $Bbb Q$ with the Euclidean topology is isomorphic to $Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology.
– Asaf Karagila♦
Nov 29 '18 at 9:54
Note that $Bbb Q$ with the Euclidean topology is isomorphic to $Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology.
– Asaf Karagila♦
Nov 29 '18 at 9:54
1
1
The completion of the metric space $mathbb{Q}/mathbb{Z}, d(a,b) = min_n |a-bn|$ is $mathbb{R}/mathbb{Z}$ while with the discrete metric $tilde{d}(a,b) = sup_p inf_n |a-b n|_p$ it is complete and $cong bigoplus_p Bbb Q_p / Bbb Z_p$.
– reuns
Nov 29 '18 at 10:20
The completion of the metric space $mathbb{Q}/mathbb{Z}, d(a,b) = min_n |a-bn|$ is $mathbb{R}/mathbb{Z}$ while with the discrete metric $tilde{d}(a,b) = sup_p inf_n |a-b n|_p$ it is complete and $cong bigoplus_p Bbb Q_p / Bbb Z_p$.
– reuns
Nov 29 '18 at 10:20
@reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology.
– Paul Frost
Nov 29 '18 at 11:25
@reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology.
– Paul Frost
Nov 29 '18 at 11:25
@PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ?
– reuns
Nov 29 '18 at 11:52
@PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ?
– reuns
Nov 29 '18 at 11:52
add a comment |
1 Answer
1
active
oldest
votes
A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.
However, you consider $G = Bbb Q / Bbb Z$ and emphasize that $Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.
The isomorphism $Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.
Edited: I just considered which topology is given to an infinite sum $bigoplus_{alpha in A} G_alpha$ of abelian topological groups. There are various approaches, see for example
Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.
https://core.ac.uk/download/pdf/82771298.pdf
Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.506.7942&rep=rep1&type=pdf
In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $bigoplus_p Bbb Q_p / Bbb Z_p$ makes it isomorphic as a topological group to $Bbb Q / Bbb Z$ is not known to me.
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
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%2f3018414%2fis-bbb-q-bbb-z-discrete%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.
However, you consider $G = Bbb Q / Bbb Z$ and emphasize that $Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.
The isomorphism $Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.
Edited: I just considered which topology is given to an infinite sum $bigoplus_{alpha in A} G_alpha$ of abelian topological groups. There are various approaches, see for example
Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.
https://core.ac.uk/download/pdf/82771298.pdf
Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.506.7942&rep=rep1&type=pdf
In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $bigoplus_p Bbb Q_p / Bbb Z_p$ makes it isomorphic as a topological group to $Bbb Q / Bbb Z$ is not known to me.
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
add a comment |
A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.
However, you consider $G = Bbb Q / Bbb Z$ and emphasize that $Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.
The isomorphism $Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.
Edited: I just considered which topology is given to an infinite sum $bigoplus_{alpha in A} G_alpha$ of abelian topological groups. There are various approaches, see for example
Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.
https://core.ac.uk/download/pdf/82771298.pdf
Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.506.7942&rep=rep1&type=pdf
In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $bigoplus_p Bbb Q_p / Bbb Z_p$ makes it isomorphic as a topological group to $Bbb Q / Bbb Z$ is not known to me.
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
add a comment |
A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.
However, you consider $G = Bbb Q / Bbb Z$ and emphasize that $Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.
The isomorphism $Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.
Edited: I just considered which topology is given to an infinite sum $bigoplus_{alpha in A} G_alpha$ of abelian topological groups. There are various approaches, see for example
Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.
https://core.ac.uk/download/pdf/82771298.pdf
Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.506.7942&rep=rep1&type=pdf
In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $bigoplus_p Bbb Q_p / Bbb Z_p$ makes it isomorphic as a topological group to $Bbb Q / Bbb Z$ is not known to me.
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.
However, you consider $G = Bbb Q / Bbb Z$ and emphasize that $Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.
The isomorphism $Bbb Q / Bbb Z cong bigoplus_p Bbb Q_p / Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.
Edited: I just considered which topology is given to an infinite sum $bigoplus_{alpha in A} G_alpha$ of abelian topological groups. There are various approaches, see for example
Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.
https://core.ac.uk/download/pdf/82771298.pdf
Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.506.7942&rep=rep1&type=pdf
In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $bigoplus_p Bbb Q_p / Bbb Z_p$ makes it isomorphic as a topological group to $Bbb Q / Bbb Z$ is not known to me.
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
edited Nov 29 '18 at 23:41
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
answered Nov 29 '18 at 11:47
Paul FrostPaul Frost
9,5002631
9,5002631
add a comment |
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%2f3018414%2fis-bbb-q-bbb-z-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
3
I think you're right, the above isomorphism is only an isomorphism of the underlying group structures.
– Brahadeesh
Nov 29 '18 at 9:53
2
Note that $Bbb Q$ with the Euclidean topology is isomorphic to $Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology.
– Asaf Karagila♦
Nov 29 '18 at 9:54
1
The completion of the metric space $mathbb{Q}/mathbb{Z}, d(a,b) = min_n |a-bn|$ is $mathbb{R}/mathbb{Z}$ while with the discrete metric $tilde{d}(a,b) = sup_p inf_n |a-b n|_p$ it is complete and $cong bigoplus_p Bbb Q_p / Bbb Z_p$.
– reuns
Nov 29 '18 at 10:20
@reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology.
– Paul Frost
Nov 29 '18 at 11:25
@PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ?
– reuns
Nov 29 '18 at 11:52