Mixing
Subset of a metric space is a metric space.
0
$begingroup$
I have a question? Why is it that every subset of a metric space is a metric space? I mean what if the subset is the empty set, then it can't be a metric space, right? because a metric space is by definition a non-empty set. So would I be correct to say that every non-empty subset of a metric space is a metric space? But i'm confused because since compact sets are metric spaces, and every finite set is a compact set, doesn't that mean that the empty set is compact? Therefore a metric space by definition?
real-analysis general-topology analysis proof-verification proof-explanation
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
$endgroup$
|
show 3 more comments
0
$begingroup$
I have a question? Why is it that every subset of a metric space is a metric space? I mean what if the subset is the empty set, then it can't be a metric space, right? because a metric space is by definition a non-empty set. So would I be correct to say that every non-empty subset of a metric space is a metric space? But i'm confused because since compact sets are metric spaces, and every finite set is a compact set, doesn't that mean that the empty set is compact? Therefore a metric space by definition?
real-analysis general-topology analysis proof-verification proof-explanation
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
$endgroup$
$begingroup$
I guess it's possible, but "my" definition of metric space allows $emptyset$. Not that it matters.
$endgroup$
– Randall
Jan 20 at 1:35
$begingroup$
I suppose it depends on your definition of metric space. The definition that I am familiar with allows the empty set. Either way I think this is not an important question to ask since the empty set is not an interesting space...
$endgroup$
– angryavian
Jan 20 at 1:36
$begingroup$
I will always laugh at whoever it was here that had that story from their grad school years about "the keeper of the empty set."
$endgroup$
– Randall
Jan 20 at 1:37
$begingroup$
@angryavian what if I wanted to prove that a subset of a metric space has a following property that for every element in the set P(x), how would I do that if the empty set is a subset? would I be incorrect to say that if I have something that said Let S $subset$ X. Prove $forall x in S$ P(x), then I must only consider the non- empty sets? or is it still true vacously? what about $forall x in S exists$? or $exists$.....
$endgroup$
– mathsssislife
Jan 20 at 1:41
$begingroup$
It's true vacuously, but no one cares. It's not going to suddenly prove some great big theorem because you can apply it to the empty set.
$endgroup$
– Randall
Jan 20 at 1:44
|
show 3 more comments
0
0
0
$begingroup$
I have a question? Why is it that every subset of a metric space is a metric space? I mean what if the subset is the empty set, then it can't be a metric space, right? because a metric space is by definition a non-empty set. So would I be correct to say that every non-empty subset of a metric space is a metric space? But i'm confused because since compact sets are metric spaces, and every finite set is a compact set, doesn't that mean that the empty set is compact? Therefore a metric space by definition?
real-analysis general-topology analysis proof-verification proof-explanation
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
$endgroup$
I have a question? Why is it that every subset of a metric space is a metric space? I mean what if the subset is the empty set, then it can't be a metric space, right? because a metric space is by definition a non-empty set. So would I be correct to say that every non-empty subset of a metric space is a metric space? But i'm confused because since compact sets are metric spaces, and every finite set is a compact set, doesn't that mean that the empty set is compact? Therefore a metric space by definition?
real-analysis general-topology analysis proof-verification proof-explanation
real-analysis general-topology analysis proof-verification proof-explanation
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
edited Jan 20 at 1:36
mathsssislife
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
asked Jan 20 at 1:33
mathsssislifemathsssislife
448
448
$begingroup$
I guess it's possible, but "my" definition of metric space allows $emptyset$. Not that it matters.
$endgroup$
– Randall
Jan 20 at 1:35
$begingroup$
I suppose it depends on your definition of metric space. The definition that I am familiar with allows the empty set. Either way I think this is not an important question to ask since the empty set is not an interesting space...
$endgroup$
– angryavian
Jan 20 at 1:36
$begingroup$
I will always laugh at whoever it was here that had that story from their grad school years about "the keeper of the empty set."
$endgroup$
– Randall
Jan 20 at 1:37
$begingroup$
@angryavian what if I wanted to prove that a subset of a metric space has a following property that for every element in the set P(x), how would I do that if the empty set is a subset? would I be incorrect to say that if I have something that said Let S $subset$ X. Prove $forall x in S$ P(x), then I must only consider the non- empty sets? or is it still true vacously? what about $forall x in S exists$? or $exists$.....
$endgroup$
– mathsssislife
Jan 20 at 1:41
$begingroup$
It's true vacuously, but no one cares. It's not going to suddenly prove some great big theorem because you can apply it to the empty set.
$endgroup$
– Randall
Jan 20 at 1:44
|
show 3 more comments
$begingroup$
I guess it's possible, but "my" definition of metric space allows $emptyset$. Not that it matters.
$endgroup$
– Randall
Jan 20 at 1:35
$begingroup$
I suppose it depends on your definition of metric space. The definition that I am familiar with allows the empty set. Either way I think this is not an important question to ask since the empty set is not an interesting space...
$endgroup$
– angryavian
Jan 20 at 1:36
$begingroup$
I will always laugh at whoever it was here that had that story from their grad school years about "the keeper of the empty set."
$endgroup$
– Randall
Jan 20 at 1:37
$begingroup$
@angryavian what if I wanted to prove that a subset of a metric space has a following property that for every element in the set P(x), how would I do that if the empty set is a subset? would I be incorrect to say that if I have something that said Let S $subset$ X. Prove $forall x in S$ P(x), then I must only consider the non- empty sets? or is it still true vacously? what about $forall x in S exists$? or $exists$.....
$endgroup$
– mathsssislife
Jan 20 at 1:41
$begingroup$
It's true vacuously, but no one cares. It's not going to suddenly prove some great big theorem because you can apply it to the empty set.
$endgroup$
– Randall
Jan 20 at 1:44
$begingroup$
I guess it's possible, but "my" definition of metric space allows $emptyset$. Not that it matters.
$endgroup$
– Randall
Jan 20 at 1:35
$begingroup$
I guess it's possible, but "my" definition of metric space allows $emptyset$. Not that it matters.
$endgroup$
– Randall
Jan 20 at 1:35
$begingroup$
I suppose it depends on your definition of metric space. The definition that I am familiar with allows the empty set. Either way I think this is not an important question to ask since the empty set is not an interesting space...
$endgroup$
– angryavian
Jan 20 at 1:36
$begingroup$
I suppose it depends on your definition of metric space. The definition that I am familiar with allows the empty set. Either way I think this is not an important question to ask since the empty set is not an interesting space...
$endgroup$
– angryavian
Jan 20 at 1:36
$begingroup$
I will always laugh at whoever it was here that had that story from their grad school years about "the keeper of the empty set."
$endgroup$
– Randall
Jan 20 at 1:37
$begingroup$
I will always laugh at whoever it was here that had that story from their grad school years about "the keeper of the empty set."
$endgroup$
– Randall
Jan 20 at 1:37
$begingroup$
@angryavian what if I wanted to prove that a subset of a metric space has a following property that for every element in the set P(x), how would I do that if the empty set is a subset? would I be incorrect to say that if I have something that said Let S $subset$ X. Prove $forall x in S$ P(x), then I must only consider the non- empty sets? or is it still true vacously? what about $forall x in S exists$? or $exists$.....
$endgroup$
– mathsssislife
Jan 20 at 1:41
$begingroup$
@angryavian what if I wanted to prove that a subset of a metric space has a following property that for every element in the set P(x), how would I do that if the empty set is a subset? would I be incorrect to say that if I have something that said Let S $subset$ X. Prove $forall x in S$ P(x), then I must only consider the non- empty sets? or is it still true vacously? what about $forall x in S exists$? or $exists$.....
$endgroup$
– mathsssislife
Jan 20 at 1:41
$begingroup$
It's true vacuously, but no one cares. It's not going to suddenly prove some great big theorem because you can apply it to the empty set.
$endgroup$
– Randall
Jan 20 at 1:44
$begingroup$
It's true vacuously, but no one cares. It's not going to suddenly prove some great big theorem because you can apply it to the empty set.
$endgroup$
– Randall
Jan 20 at 1:44
|
show 3 more comments
1 Answer
1
active
oldest
votes
1
$begingroup$
I think a metric space can be empty, see Wikipedia.
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
$endgroup$
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
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
});
}
});
draft saved
draft discarded
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%2f3080064%2fsubset-of-a-metric-space-is-a-metric-space%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
1
$begingroup$
I think a metric space can be empty, see Wikipedia.
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
$endgroup$
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
add a comment |
1
$begingroup$
I think a metric space can be empty, see Wikipedia.
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
$endgroup$
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
add a comment |
1
1
1
$begingroup$
I think a metric space can be empty, see Wikipedia.
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
$endgroup$
I think a metric space can be empty, see Wikipedia.
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
answered Jan 20 at 1:37
YuiTo ChengYuiTo Cheng
1,9142633
1,9142633
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
add a comment |
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
$begingroup$
Yes. Or see General Topology by R. Engelking, or almost any random textbook on the subject.
$endgroup$
– DanielWainfleet
Jan 20 at 9:49
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3080064%2fsubset-of-a-metric-space-is-a-metric-space%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
$begingroup$
I guess it's possible, but "my" definition of metric space allows $emptyset$. Not that it matters.
$endgroup$
– Randall
Jan 20 at 1:35
$begingroup$
I suppose it depends on your definition of metric space. The definition that I am familiar with allows the empty set. Either way I think this is not an important question to ask since the empty set is not an interesting space...
$endgroup$
– angryavian
Jan 20 at 1:36
$begingroup$
I will always laugh at whoever it was here that had that story from their grad school years about "the keeper of the empty set."
$endgroup$
– Randall
Jan 20 at 1:37
$begingroup$
@angryavian what if I wanted to prove that a subset of a metric space has a following property that for every element in the set P(x), how would I do that if the empty set is a subset? would I be incorrect to say that if I have something that said Let S $subset$ X. Prove $forall x in S$ P(x), then I must only consider the non- empty sets? or is it still true vacously? what about $forall x in S exists$? or $exists$.....
$endgroup$
– mathsssislife
Jan 20 at 1:41
$begingroup$
It's true vacuously, but no one cares. It's not going to suddenly prove some great big theorem because you can apply it to the empty set.
$endgroup$
– Randall
Jan 20 at 1:44