Mixing
Closure of the Unit Sphere
$begingroup$
I am trying to teach myself some basic tensor analysis and differential geometry. Right now, I am reading Abraham, Marsden, and Ratiu's Manifolds, Tensor Analysis, and Applications. In the first few sections, they define the boundary of a subset $A$ of a topological space $S$ as the intersection of the closure of $A$ with the closure of $S/A$. I was going through examples of this in my head and have run into a conceptual problem, possibly due to a misunderstanding. Perhaps someone can help?
Let $S$ be the usual topology of open sets on $mathbb{R}^3$ and let $A$ be the unit sphere. I want to find $bd(A)$. It is easy to see that $cl(A) = A$, but I do not understand how to find $cl(S/A)$. My first guess is that it is empty, since I cannot think of any closed sets which contain $S/A$ as a proper subset. But then that implies
$bd(A)=cl(A) cap cl(S/A)=Acapemptyset=emptyset$
which sort of goes against my intuition. Am I missing something here, or is this a true statement?
general-topology
asked Jan 26 at 2:02
J_PsiJ_Psi
837
$endgroup$
add a comment |
$begingroup$
I am trying to teach myself some basic tensor analysis and differential geometry. Right now, I am reading Abraham, Marsden, and Ratiu's Manifolds, Tensor Analysis, and Applications. In the first few sections, they define the boundary of a subset $A$ of a topological space $S$ as the intersection of the closure of $A$ with the closure of $S/A$. I was going through examples of this in my head and have run into a conceptual problem, possibly due to a misunderstanding. Perhaps someone can help?
Let $S$ be the usual topology of open sets on $mathbb{R}^3$ and let $A$ be the unit sphere. I want to find $bd(A)$. It is easy to see that $cl(A) = A$, but I do not understand how to find $cl(S/A)$. My first guess is that it is empty, since I cannot think of any closed sets which contain $S/A$ as a proper subset. But then that implies
$bd(A)=cl(A) cap cl(S/A)=Acapemptyset=emptyset$
which sort of goes against my intuition. Am I missing something here, or is this a true statement?
general-topology
asked Jan 26 at 2:02
J_PsiJ_Psi
837
$endgroup$
$begingroup$
You mean $S=Bbb R^3$? And "/" maybe should be a setminus, not a quotient?
$endgroup$
– Torsten Schoeneberg
Jan 26 at 2:06
$begingroup$
Yes, $S=mathbb{R}^3$ equipped with the topology of open sets. Also, that's how Abraham, et al. denotes $S$ but without $A$, so that's why I'm writing it here.
$endgroup$
– J_Psi
Jan 26 at 2:13
add a comment |
$begingroup$
I am trying to teach myself some basic tensor analysis and differential geometry. Right now, I am reading Abraham, Marsden, and Ratiu's Manifolds, Tensor Analysis, and Applications. In the first few sections, they define the boundary of a subset $A$ of a topological space $S$ as the intersection of the closure of $A$ with the closure of $S/A$. I was going through examples of this in my head and have run into a conceptual problem, possibly due to a misunderstanding. Perhaps someone can help?
Let $S$ be the usual topology of open sets on $mathbb{R}^3$ and let $A$ be the unit sphere. I want to find $bd(A)$. It is easy to see that $cl(A) = A$, but I do not understand how to find $cl(S/A)$. My first guess is that it is empty, since I cannot think of any closed sets which contain $S/A$ as a proper subset. But then that implies
$bd(A)=cl(A) cap cl(S/A)=Acapemptyset=emptyset$
which sort of goes against my intuition. Am I missing something here, or is this a true statement?
general-topology
asked Jan 26 at 2:02
J_PsiJ_Psi
837
$endgroup$
I am trying to teach myself some basic tensor analysis and differential geometry. Right now, I am reading Abraham, Marsden, and Ratiu's Manifolds, Tensor Analysis, and Applications. In the first few sections, they define the boundary of a subset $A$ of a topological space $S$ as the intersection of the closure of $A$ with the closure of $S/A$. I was going through examples of this in my head and have run into a conceptual problem, possibly due to a misunderstanding. Perhaps someone can help?
Let $S$ be the usual topology of open sets on $mathbb{R}^3$ and let $A$ be the unit sphere. I want to find $bd(A)$. It is easy to see that $cl(A) = A$, but I do not understand how to find $cl(S/A)$. My first guess is that it is empty, since I cannot think of any closed sets which contain $S/A$ as a proper subset. But then that implies
$bd(A)=cl(A) cap cl(S/A)=Acapemptyset=emptyset$
which sort of goes against my intuition. Am I missing something here, or is this a true statement?
general-topology
general-topology
asked Jan 26 at 2:02
J_PsiJ_Psi
837
asked Jan 26 at 2:02
J_PsiJ_Psi
837
edited Jan 26 at 2:10
J_Psi
asked Jan 26 at 2:02
J_PsiJ_Psi
837
asked Jan 26 at 2:02
J_PsiJ_Psi
837
asked Jan 26 at 2:02
J_PsiJ_Psi
837
837
$begingroup$
You mean $S=Bbb R^3$? And "/" maybe should be a setminus, not a quotient?
$endgroup$
– Torsten Schoeneberg
Jan 26 at 2:06
$begingroup$
Yes, $S=mathbb{R}^3$ equipped with the topology of open sets. Also, that's how Abraham, et al. denotes $S$ but without $A$, so that's why I'm writing it here.
$endgroup$
– J_Psi
Jan 26 at 2:13
add a comment |
$begingroup$
You mean $S=Bbb R^3$? And "/" maybe should be a setminus, not a quotient?
$endgroup$
– Torsten Schoeneberg
Jan 26 at 2:06
$begingroup$
Yes, $S=mathbb{R}^3$ equipped with the topology of open sets. Also, that's how Abraham, et al. denotes $S$ but without $A$, so that's why I'm writing it here.
$endgroup$
– J_Psi
Jan 26 at 2:13
$begingroup$
You mean $S=Bbb R^3$? And "/" maybe should be a setminus, not a quotient?
$endgroup$
– Torsten Schoeneberg
Jan 26 at 2:06
$begingroup$
You mean $S=Bbb R^3$? And "/" maybe should be a setminus, not a quotient?
$endgroup$
– Torsten Schoeneberg
Jan 26 at 2:06
$begingroup$
Yes, $S=mathbb{R}^3$ equipped with the topology of open sets. Also, that's how Abraham, et al. denotes $S$ but without $A$, so that's why I'm writing it here.
$endgroup$
– J_Psi
Jan 26 at 2:13
$begingroup$
Yes, $S=mathbb{R}^3$ equipped with the topology of open sets. Also, that's how Abraham, et al. denotes $S$ but without $A$, so that's why I'm writing it here.
$endgroup$
– J_Psi
Jan 26 at 2:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're correct that $A$ is closed. But the closure of $(mathbb{R}^3setminus A)$ is $mathbb{R}^3$. (In any topological space $X$, $X$ is a closed set. The only set $Y$ with $text{cl}(Y) = emptyset$ is $Y = emptyset$.)
So $text{bd}(A) = Acap mathbb{R}^3 = A$.
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
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%2f3087829%2fclosure-of-the-unit-sphere%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
$begingroup$
You're correct that $A$ is closed. But the closure of $(mathbb{R}^3setminus A)$ is $mathbb{R}^3$. (In any topological space $X$, $X$ is a closed set. The only set $Y$ with $text{cl}(Y) = emptyset$ is $Y = emptyset$.)
So $text{bd}(A) = Acap mathbb{R}^3 = A$.
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
add a comment |
$begingroup$
You're correct that $A$ is closed. But the closure of $(mathbb{R}^3setminus A)$ is $mathbb{R}^3$. (In any topological space $X$, $X$ is a closed set. The only set $Y$ with $text{cl}(Y) = emptyset$ is $Y = emptyset$.)
So $text{bd}(A) = Acap mathbb{R}^3 = A$.
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
add a comment |
$begingroup$
You're correct that $A$ is closed. But the closure of $(mathbb{R}^3setminus A)$ is $mathbb{R}^3$. (In any topological space $X$, $X$ is a closed set. The only set $Y$ with $text{cl}(Y) = emptyset$ is $Y = emptyset$.)
So $text{bd}(A) = Acap mathbb{R}^3 = A$.
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
You're correct that $A$ is closed. But the closure of $(mathbb{R}^3setminus A)$ is $mathbb{R}^3$. (In any topological space $X$, $X$ is a closed set. The only set $Y$ with $text{cl}(Y) = emptyset$ is $Y = emptyset$.)
So $text{bd}(A) = Acap mathbb{R}^3 = A$.
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
edited Jan 26 at 2:17
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
answered Jan 26 at 2:10
Alex KruckmanAlex Kruckman
28.1k32658
28.1k32658
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
add a comment |
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
$A$ is the surface of the unit ball, not the unit ball itself.
$endgroup$
– J_Psi
Jan 26 at 2:11
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
@J_Psi Oh, I misunderstood your question. I've edited.
$endgroup$
– Alex Kruckman
Jan 26 at 2:17
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
$begingroup$
Ah, that's right! I forgot that the space itself was included in the topology. Thank you for the help.
$endgroup$
– J_Psi
Jan 26 at 2:19
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%2f3087829%2fclosure-of-the-unit-sphere%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
$begingroup$
You mean $S=Bbb R^3$? And "/" maybe should be a setminus, not a quotient?
$endgroup$
– Torsten Schoeneberg
Jan 26 at 2:06
$begingroup$
Yes, $S=mathbb{R}^3$ equipped with the topology of open sets. Also, that's how Abraham, et al. denotes $S$ but without $A$, so that's why I'm writing it here.
$endgroup$
– J_Psi
Jan 26 at 2:13