Are there any sets that are not complete metric spaces under all possible metrics?
$begingroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
$endgroup$
add a comment |
$begingroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
$endgroup$
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
add a comment |
$begingroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
$endgroup$
I don't have any particular set in mind but this seemed interesthing since completeness depends on the metric.
general-topology metric-spaces complete-spaces
general-topology metric-spaces complete-spaces
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
asked Dec 19 '18 at 14:42
2chromatic2chromatic
1829
1829
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
add a comment |
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
$endgroup$
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
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%2f3046458%2fare-there-any-sets-that-are-not-complete-metric-spaces-under-all-possible-metric%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
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
$endgroup$
add a comment |
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
$endgroup$
add a comment |
$begingroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
$endgroup$
Since every set can be given the discrete metric $d(x,y)=left{begin{array}{ll} 1 & text{if $xneq y$}\0 &text{if $x = y$}end{array}right.$, and that for this metric every Cauchy sequence is stationnary and so convergent, you can't find a such example.
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
answered Dec 19 '18 at 14:45
BalloonBalloon
4,710822
4,710822
add a comment |
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
$begingroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
Yes, the set of rationals $mathbb{Q}$ in its usual topology is not a Baire space, so whatever compatible metric you put on it, it will be non-complete.
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
answered Dec 19 '18 at 14:44
Henno BrandsmaHenno Brandsma
115k349125
115k349125
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
add a comment |
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
$begingroup$
This depends on what the question is asking, precisely. If we view $mathbb{Q}$ just as a set then obviously we can make it a complete metric space (and I think this is how the question leans, but I'm not certain).
$endgroup$
– Noah Schweber
Dec 19 '18 at 14:48
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.
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%2f3046458%2fare-there-any-sets-that-are-not-complete-metric-spaces-under-all-possible-metric%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
$begingroup$
Do you really mean set here, or topological space? That is, do you want to consider any metric on a given set, or only those compatible with a given topology?
$endgroup$
– Alex Kruckman
Dec 19 '18 at 18:27
$begingroup$
@AlexKruckman First one but I forgot about the discrete metric and Balloon's answer wrapped things up. Now, I think it would be better to talk of a norm, rather than a metric.
$endgroup$
– 2chromatic
Dec 19 '18 at 18:31