Relation in distance between a set $A$ and boundary of a set $B$, in this particular case.
$begingroup$
In a metric space $(M,d)$, I have compact nonempty sets $A, B$ and $C$ with $Asubset operatorname{int} B$ and $B subset operatorname{int} C$, where $operatorname{int}$ denotes the interior of a set and $partial$ its boundary. I'm trying to find out if $d(A,C - operatorname{int} B) = d(A, partial B).$ If this were to be true it would solve a problem I'm working on, but can't prove it nor find a counterexample. It is always true that $d(A,C - operatorname{int} B) leq d(A, partial B),$ since $partial B subseteq C- operatorname{int} B, $ and the reverse inequality seems true because taking a point in $C - operatorname{int} B$ which is not in the boundary of $B$, it would be "further away" from $A$ than a point in the boundary of $B,$ but I'm thinking geometrically on the plane, and I can't work on the triangle inequality to give me this result.
Is what I'm trying prove to even true? Any hints are appreciated, thanks.
(OBS: the metric space is also a smooth manifold in my problem, if any extra structure helps).
(OBS2: in the original question, $M$ could be a topological manifold, since I thought this problem could be solved with topology alone. User @theo-bendit showed this is false in general, but more research made me see this may be true for a smooth manifold, as in this other question When is distance to the boundary always less than that to the exterior? . In not used to work with riemann metrics so I can't be sure).
general-topology metric-spaces
asked Jan 4 at 2:55
VicVic
4217
$endgroup$
add a comment |
$begingroup$
In a metric space $(M,d)$, I have compact nonempty sets $A, B$ and $C$ with $Asubset operatorname{int} B$ and $B subset operatorname{int} C$, where $operatorname{int}$ denotes the interior of a set and $partial$ its boundary. I'm trying to find out if $d(A,C - operatorname{int} B) = d(A, partial B).$ If this were to be true it would solve a problem I'm working on, but can't prove it nor find a counterexample. It is always true that $d(A,C - operatorname{int} B) leq d(A, partial B),$ since $partial B subseteq C- operatorname{int} B, $ and the reverse inequality seems true because taking a point in $C - operatorname{int} B$ which is not in the boundary of $B$, it would be "further away" from $A$ than a point in the boundary of $B,$ but I'm thinking geometrically on the plane, and I can't work on the triangle inequality to give me this result.
Is what I'm trying prove to even true? Any hints are appreciated, thanks.
(OBS: the metric space is also a smooth manifold in my problem, if any extra structure helps).
(OBS2: in the original question, $M$ could be a topological manifold, since I thought this problem could be solved with topology alone. User @theo-bendit showed this is false in general, but more research made me see this may be true for a smooth manifold, as in this other question When is distance to the boundary always less than that to the exterior? . In not used to work with riemann metrics so I can't be sure).
general-topology metric-spaces
asked Jan 4 at 2:55
VicVic
4217
$endgroup$
$begingroup$
Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
$endgroup$
– Aniruddha Deshmukh
Jan 4 at 3:21
$begingroup$
Sure. Since $B subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int B cup partial B,$ where the union is disjoint, therefore $partial B subseteq C - int B.$ Now the result follows since the distance is the infimum of $d$ on $A times C - int B$ and $A times partial B subseteq A times C - int B.$
$endgroup$
– Vic
Jan 4 at 3:31
add a comment |
$begingroup$
In a metric space $(M,d)$, I have compact nonempty sets $A, B$ and $C$ with $Asubset operatorname{int} B$ and $B subset operatorname{int} C$, where $operatorname{int}$ denotes the interior of a set and $partial$ its boundary. I'm trying to find out if $d(A,C - operatorname{int} B) = d(A, partial B).$ If this were to be true it would solve a problem I'm working on, but can't prove it nor find a counterexample. It is always true that $d(A,C - operatorname{int} B) leq d(A, partial B),$ since $partial B subseteq C- operatorname{int} B, $ and the reverse inequality seems true because taking a point in $C - operatorname{int} B$ which is not in the boundary of $B$, it would be "further away" from $A$ than a point in the boundary of $B,$ but I'm thinking geometrically on the plane, and I can't work on the triangle inequality to give me this result.
Is what I'm trying prove to even true? Any hints are appreciated, thanks.
(OBS: the metric space is also a smooth manifold in my problem, if any extra structure helps).
(OBS2: in the original question, $M$ could be a topological manifold, since I thought this problem could be solved with topology alone. User @theo-bendit showed this is false in general, but more research made me see this may be true for a smooth manifold, as in this other question When is distance to the boundary always less than that to the exterior? . In not used to work with riemann metrics so I can't be sure).
general-topology metric-spaces
asked Jan 4 at 2:55
VicVic
4217
$endgroup$
In a metric space $(M,d)$, I have compact nonempty sets $A, B$ and $C$ with $Asubset operatorname{int} B$ and $B subset operatorname{int} C$, where $operatorname{int}$ denotes the interior of a set and $partial$ its boundary. I'm trying to find out if $d(A,C - operatorname{int} B) = d(A, partial B).$ If this were to be true it would solve a problem I'm working on, but can't prove it nor find a counterexample. It is always true that $d(A,C - operatorname{int} B) leq d(A, partial B),$ since $partial B subseteq C- operatorname{int} B, $ and the reverse inequality seems true because taking a point in $C - operatorname{int} B$ which is not in the boundary of $B$, it would be "further away" from $A$ than a point in the boundary of $B,$ but I'm thinking geometrically on the plane, and I can't work on the triangle inequality to give me this result.
Is what I'm trying prove to even true? Any hints are appreciated, thanks.
(OBS: the metric space is also a smooth manifold in my problem, if any extra structure helps).
(OBS2: in the original question, $M$ could be a topological manifold, since I thought this problem could be solved with topology alone. User @theo-bendit showed this is false in general, but more research made me see this may be true for a smooth manifold, as in this other question When is distance to the boundary always less than that to the exterior? . In not used to work with riemann metrics so I can't be sure).
general-topology metric-spaces
general-topology metric-spaces
asked Jan 4 at 2:55
VicVic
4217
asked Jan 4 at 2:55
VicVic
4217
edited Jan 4 at 4:27
Vic
asked Jan 4 at 2:55
VicVic
4217
asked Jan 4 at 2:55
VicVic
4217
asked Jan 4 at 2:55
VicVic
4217
4217
$begingroup$
Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
$endgroup$
– Aniruddha Deshmukh
Jan 4 at 3:21
$begingroup$
Sure. Since $B subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int B cup partial B,$ where the union is disjoint, therefore $partial B subseteq C - int B.$ Now the result follows since the distance is the infimum of $d$ on $A times C - int B$ and $A times partial B subseteq A times C - int B.$
$endgroup$
– Vic
Jan 4 at 3:31
add a comment |
$begingroup$
Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
$endgroup$
– Aniruddha Deshmukh
Jan 4 at 3:21
$begingroup$
Sure. Since $B subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int B cup partial B,$ where the union is disjoint, therefore $partial B subseteq C - int B.$ Now the result follows since the distance is the infimum of $d$ on $A times C - int B$ and $A times partial B subseteq A times C - int B.$
$endgroup$
– Vic
Jan 4 at 3:31
$begingroup$
Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
$endgroup$
– Aniruddha Deshmukh
Jan 4 at 3:21
$begingroup$
Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
$endgroup$
– Aniruddha Deshmukh
Jan 4 at 3:21
$begingroup$
Sure. Since $B subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int B cup partial B,$ where the union is disjoint, therefore $partial B subseteq C - int B.$ Now the result follows since the distance is the infimum of $d$ on $A times C - int B$ and $A times partial B subseteq A times C - int B.$
$endgroup$
– Vic
Jan 4 at 3:31
$begingroup$
Sure. Since $B subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int B cup partial B,$ where the union is disjoint, therefore $partial B subseteq C - int B.$ Now the result follows since the distance is the infimum of $d$ on $A times C - int B$ and $A times partial B subseteq A times C - int B.$
$endgroup$
– Vic
Jan 4 at 3:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I can't comment on the topological manifold part here, but in general, no, this is not the case.
Let $K$ be the Cantor Middle Third set, and
begin{align*}
M &= C = K cup [100, 103] \
B &= left(K cap left[0, frac13right]right) cup [101,102] \
A &= K cap left[0, frac19right].
end{align*}
Then,
begin{align*}
C setminus operatorname{int} B &= left(K cap left[frac23, 1right]right) cup [100, 101] cup [102, 103] \
partial B &= { 101, 102} \
d(A, C setminus operatorname{int} B) &= frac59 \
d(A, partial B) &= frac{908}{9}.
end{align*}
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
$endgroup$
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
|
show 1 more 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%2f3061271%2frelation-in-distance-between-a-set-a-and-boundary-of-a-set-b-in-this-partic%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
$begingroup$
I can't comment on the topological manifold part here, but in general, no, this is not the case.
Let $K$ be the Cantor Middle Third set, and
begin{align*}
M &= C = K cup [100, 103] \
B &= left(K cap left[0, frac13right]right) cup [101,102] \
A &= K cap left[0, frac19right].
end{align*}
Then,
begin{align*}
C setminus operatorname{int} B &= left(K cap left[frac23, 1right]right) cup [100, 101] cup [102, 103] \
partial B &= { 101, 102} \
d(A, C setminus operatorname{int} B) &= frac59 \
d(A, partial B) &= frac{908}{9}.
end{align*}
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
$endgroup$
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
|
show 1 more comment
$begingroup$
I can't comment on the topological manifold part here, but in general, no, this is not the case.
Let $K$ be the Cantor Middle Third set, and
begin{align*}
M &= C = K cup [100, 103] \
B &= left(K cap left[0, frac13right]right) cup [101,102] \
A &= K cap left[0, frac19right].
end{align*}
Then,
begin{align*}
C setminus operatorname{int} B &= left(K cap left[frac23, 1right]right) cup [100, 101] cup [102, 103] \
partial B &= { 101, 102} \
d(A, C setminus operatorname{int} B) &= frac59 \
d(A, partial B) &= frac{908}{9}.
end{align*}
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
$endgroup$
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
|
show 1 more comment
$begingroup$
I can't comment on the topological manifold part here, but in general, no, this is not the case.
Let $K$ be the Cantor Middle Third set, and
begin{align*}
M &= C = K cup [100, 103] \
B &= left(K cap left[0, frac13right]right) cup [101,102] \
A &= K cap left[0, frac19right].
end{align*}
Then,
begin{align*}
C setminus operatorname{int} B &= left(K cap left[frac23, 1right]right) cup [100, 101] cup [102, 103] \
partial B &= { 101, 102} \
d(A, C setminus operatorname{int} B) &= frac59 \
d(A, partial B) &= frac{908}{9}.
end{align*}
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
$endgroup$
I can't comment on the topological manifold part here, but in general, no, this is not the case.
Let $K$ be the Cantor Middle Third set, and
begin{align*}
M &= C = K cup [100, 103] \
B &= left(K cap left[0, frac13right]right) cup [101,102] \
A &= K cap left[0, frac19right].
end{align*}
Then,
begin{align*}
C setminus operatorname{int} B &= left(K cap left[frac23, 1right]right) cup [100, 101] cup [102, 103] \
partial B &= { 101, 102} \
d(A, C setminus operatorname{int} B) &= frac59 \
d(A, partial B) &= frac{908}{9}.
end{align*}
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
edited Jan 4 at 7:46
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
answered Jan 4 at 3:35
Theo BenditTheo Bendit
19.5k12353
19.5k12353
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
|
show 1 more comment
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
$endgroup$
– Vic
Jan 4 at 3:46
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
$endgroup$
– Theo Bendit
Jan 4 at 3:52
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
Right, I was thinking of it as a subset of the real line.
$endgroup$
– Vic
Jan 4 at 3:56
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
$endgroup$
– Theo Bendit
Jan 4 at 3:57
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
$begingroup$
@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
$endgroup$
– Theo Bendit
Jan 4 at 3:58
|
show 1 more 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%2f3061271%2frelation-in-distance-between-a-set-a-and-boundary-of-a-set-b-in-this-partic%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 we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
$endgroup$
– Aniruddha Deshmukh
Jan 4 at 3:21
$begingroup$
Sure. Since $B subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int B cup partial B,$ where the union is disjoint, therefore $partial B subseteq C - int B.$ Now the result follows since the distance is the infimum of $d$ on $A times C - int B$ and $A times partial B subseteq A times C - int B.$
$endgroup$
– Vic
Jan 4 at 3:31