Proof that in a smooth manifold every open set has same cardinality?
$begingroup$
The manifold of $Bbb R^n$ with its usual topology certainly has this property, every open set has the same size as every other open set, so there exists a bijection between any two open sets. But, does the fact that a smooth manifold is locally homeomorphic to some euclidean space imply that every open set has same cardinality?
I'm not able to prove it because the definition says that it is locally homeomorphic, but I think it is because every open set has a representation, in one chart or another so it must be the same as some open set in $Bbb R^n$. Is this right?
smooth-manifolds
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
$endgroup$
add a comment |
$begingroup$
The manifold of $Bbb R^n$ with its usual topology certainly has this property, every open set has the same size as every other open set, so there exists a bijection between any two open sets. But, does the fact that a smooth manifold is locally homeomorphic to some euclidean space imply that every open set has same cardinality?
I'm not able to prove it because the definition says that it is locally homeomorphic, but I think it is because every open set has a representation, in one chart or another so it must be the same as some open set in $Bbb R^n$. Is this right?
smooth-manifolds
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
$endgroup$
1
$begingroup$
Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something?
$endgroup$
– Dave L. Renfro
Dec 1 '18 at 19:31
add a comment |
$begingroup$
The manifold of $Bbb R^n$ with its usual topology certainly has this property, every open set has the same size as every other open set, so there exists a bijection between any two open sets. But, does the fact that a smooth manifold is locally homeomorphic to some euclidean space imply that every open set has same cardinality?
I'm not able to prove it because the definition says that it is locally homeomorphic, but I think it is because every open set has a representation, in one chart or another so it must be the same as some open set in $Bbb R^n$. Is this right?
smooth-manifolds
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
$endgroup$
The manifold of $Bbb R^n$ with its usual topology certainly has this property, every open set has the same size as every other open set, so there exists a bijection between any two open sets. But, does the fact that a smooth manifold is locally homeomorphic to some euclidean space imply that every open set has same cardinality?
I'm not able to prove it because the definition says that it is locally homeomorphic, but I think it is because every open set has a representation, in one chart or another so it must be the same as some open set in $Bbb R^n$. Is this right?
smooth-manifolds
smooth-manifolds
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
asked Dec 1 '18 at 18:39
GarmekainGarmekain
1,323720
1,323720
1
$begingroup$
Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something?
$endgroup$
– Dave L. Renfro
Dec 1 '18 at 19:31
add a comment |
1
$begingroup$
Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something?
$endgroup$
– Dave L. Renfro
Dec 1 '18 at 19:31
1
1
$begingroup$
Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something?
$endgroup$
– Dave L. Renfro
Dec 1 '18 at 19:31
$begingroup$
Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something?
$endgroup$
– Dave L. Renfro
Dec 1 '18 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming you require your manifolds to be second-countable, then it is true that every nonempty open subset of a manifold of positive dimension has cardinality $2^{aleph_0}$. Indeed, suppose $M$ is a manifold of dimension $n>0$. Since $M$ is a manifold, it has a basis of open sets that are homeomorphic to $mathbb{R}^n$ (and thus have cardinality $2^{aleph_0}$. By second-countability, this basis can be chosen to be countable. Any open subset of $M$ is a union of countably many basis sets. So, any open subset has cardinality at most $aleph_0cdot 2^{aleph_0}=2^{aleph_0}$. Conversely, any nonempty open subset contains at least one basis set and so has cardinality at least $2^{aleph_0}$.
(Of course, this is false for $0$-dimensional manifolds since you could take a discrete space of any countable cardinality which has open subsets of all smaller cardinalities. It also fails if you do not require second-countability since then you could take a disjoint union of more than $2^{aleph_0}$ copies of $mathbb{R}^n$.)
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
$endgroup$
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%2f3021666%2fproof-that-in-a-smooth-manifold-every-open-set-has-same-cardinality%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
$begingroup$
Assuming you require your manifolds to be second-countable, then it is true that every nonempty open subset of a manifold of positive dimension has cardinality $2^{aleph_0}$. Indeed, suppose $M$ is a manifold of dimension $n>0$. Since $M$ is a manifold, it has a basis of open sets that are homeomorphic to $mathbb{R}^n$ (and thus have cardinality $2^{aleph_0}$. By second-countability, this basis can be chosen to be countable. Any open subset of $M$ is a union of countably many basis sets. So, any open subset has cardinality at most $aleph_0cdot 2^{aleph_0}=2^{aleph_0}$. Conversely, any nonempty open subset contains at least one basis set and so has cardinality at least $2^{aleph_0}$.
(Of course, this is false for $0$-dimensional manifolds since you could take a discrete space of any countable cardinality which has open subsets of all smaller cardinalities. It also fails if you do not require second-countability since then you could take a disjoint union of more than $2^{aleph_0}$ copies of $mathbb{R}^n$.)
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
$begingroup$
Assuming you require your manifolds to be second-countable, then it is true that every nonempty open subset of a manifold of positive dimension has cardinality $2^{aleph_0}$. Indeed, suppose $M$ is a manifold of dimension $n>0$. Since $M$ is a manifold, it has a basis of open sets that are homeomorphic to $mathbb{R}^n$ (and thus have cardinality $2^{aleph_0}$. By second-countability, this basis can be chosen to be countable. Any open subset of $M$ is a union of countably many basis sets. So, any open subset has cardinality at most $aleph_0cdot 2^{aleph_0}=2^{aleph_0}$. Conversely, any nonempty open subset contains at least one basis set and so has cardinality at least $2^{aleph_0}$.
(Of course, this is false for $0$-dimensional manifolds since you could take a discrete space of any countable cardinality which has open subsets of all smaller cardinalities. It also fails if you do not require second-countability since then you could take a disjoint union of more than $2^{aleph_0}$ copies of $mathbb{R}^n$.)
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
$begingroup$
Assuming you require your manifolds to be second-countable, then it is true that every nonempty open subset of a manifold of positive dimension has cardinality $2^{aleph_0}$. Indeed, suppose $M$ is a manifold of dimension $n>0$. Since $M$ is a manifold, it has a basis of open sets that are homeomorphic to $mathbb{R}^n$ (and thus have cardinality $2^{aleph_0}$. By second-countability, this basis can be chosen to be countable. Any open subset of $M$ is a union of countably many basis sets. So, any open subset has cardinality at most $aleph_0cdot 2^{aleph_0}=2^{aleph_0}$. Conversely, any nonempty open subset contains at least one basis set and so has cardinality at least $2^{aleph_0}$.
(Of course, this is false for $0$-dimensional manifolds since you could take a discrete space of any countable cardinality which has open subsets of all smaller cardinalities. It also fails if you do not require second-countability since then you could take a disjoint union of more than $2^{aleph_0}$ copies of $mathbb{R}^n$.)
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
$endgroup$
Assuming you require your manifolds to be second-countable, then it is true that every nonempty open subset of a manifold of positive dimension has cardinality $2^{aleph_0}$. Indeed, suppose $M$ is a manifold of dimension $n>0$. Since $M$ is a manifold, it has a basis of open sets that are homeomorphic to $mathbb{R}^n$ (and thus have cardinality $2^{aleph_0}$. By second-countability, this basis can be chosen to be countable. Any open subset of $M$ is a union of countably many basis sets. So, any open subset has cardinality at most $aleph_0cdot 2^{aleph_0}=2^{aleph_0}$. Conversely, any nonempty open subset contains at least one basis set and so has cardinality at least $2^{aleph_0}$.
(Of course, this is false for $0$-dimensional manifolds since you could take a discrete space of any countable cardinality which has open subsets of all smaller cardinalities. It also fails if you do not require second-countability since then you could take a disjoint union of more than $2^{aleph_0}$ copies of $mathbb{R}^n$.)
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
answered Dec 1 '18 at 20:50
Eric WofseyEric Wofsey
181k12208336
181k12208336
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.
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%2f3021666%2fproof-that-in-a-smooth-manifold-every-open-set-has-same-cardinality%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
1
$begingroup$
Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something?
$endgroup$
– Dave L. Renfro
Dec 1 '18 at 19:31