Relation in distance between a set $A$ and boundary of a set $B$, in this particular case.












1














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







share|cite|improve this question
























  • Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
    – Aniruddha Deshmukh
    Jan 4 at 3:21










  • 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 $A times partial B subseteq A times C - int B.$
    – Vic
    Jan 4 at 3:31
















1














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







share|cite|improve this question
























  • Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
    – Aniruddha Deshmukh
    Jan 4 at 3:21










  • 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 $A times partial B subseteq A times C - int B.$
    – Vic
    Jan 4 at 3:31














1












1








1







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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 4:27







Vic

















asked Jan 4 at 2:55









VicVic

427




427












  • Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
    – Aniruddha Deshmukh
    Jan 4 at 3:21










  • 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 $A times partial B subseteq A times C - int B.$
    – Vic
    Jan 4 at 3:31


















  • Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
    – Aniruddha Deshmukh
    Jan 4 at 3:21










  • 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 $A times partial B subseteq A times C - int B.$
    – Vic
    Jan 4 at 3:31
















Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
– Aniruddha Deshmukh
Jan 4 at 3:21




Do we have a proof for $d left( A, C - text{int } B right) leq d left( A, partial B right)$?
– Aniruddha Deshmukh
Jan 4 at 3:21












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 $A times partial B subseteq A times C - int B.$
– Vic
Jan 4 at 3:31




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 $A times partial B subseteq A times C - int B.$
– Vic
Jan 4 at 3:31










1 Answer
1






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oldest

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1














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*}






share|cite|improve this answer























  • I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
    – Vic
    Jan 4 at 3:46












  • Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
    – Theo Bendit
    Jan 4 at 3:52










  • Right, I was thinking of it as a subset of the real line.
    – Vic
    Jan 4 at 3:56










  • @Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
    – Theo Bendit
    Jan 4 at 3:57










  • @Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
    – Theo Bendit
    Jan 4 at 3:58











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1














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*}






share|cite|improve this answer























  • I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
    – Vic
    Jan 4 at 3:46












  • Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
    – Theo Bendit
    Jan 4 at 3:52










  • Right, I was thinking of it as a subset of the real line.
    – Vic
    Jan 4 at 3:56










  • @Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
    – Theo Bendit
    Jan 4 at 3:57










  • @Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
    – Theo Bendit
    Jan 4 at 3:58
















1














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*}






share|cite|improve this answer























  • I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
    – Vic
    Jan 4 at 3:46












  • Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
    – Theo Bendit
    Jan 4 at 3:52










  • Right, I was thinking of it as a subset of the real line.
    – Vic
    Jan 4 at 3:56










  • @Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
    – Theo Bendit
    Jan 4 at 3:57










  • @Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
    – Theo Bendit
    Jan 4 at 3:58














1












1








1






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*}






share|cite|improve this answer














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*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 7:46

























answered Jan 4 at 3:35









Theo BenditTheo Bendit

16.7k12148




16.7k12148












  • I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
    – Vic
    Jan 4 at 3:46












  • Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
    – Theo Bendit
    Jan 4 at 3:52










  • Right, I was thinking of it as a subset of the real line.
    – Vic
    Jan 4 at 3:56










  • @Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
    – Theo Bendit
    Jan 4 at 3:57










  • @Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
    – Theo Bendit
    Jan 4 at 3:58


















  • I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
    – Vic
    Jan 4 at 3:46












  • Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
    – Theo Bendit
    Jan 4 at 3:52










  • Right, I was thinking of it as a subset of the real line.
    – Vic
    Jan 4 at 3:56










  • @Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
    – Theo Bendit
    Jan 4 at 3:57










  • @Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
    – Theo Bendit
    Jan 4 at 3:58
















I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
– Vic
Jan 4 at 3:46






I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks!
– Vic
Jan 4 at 3:46














Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
– Theo Bendit
Jan 4 at 3:52




Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$.
– Theo Bendit
Jan 4 at 3:52












Right, I was thinking of it as a subset of the real line.
– Vic
Jan 4 at 3:56




Right, I was thinking of it as a subset of the real line.
– Vic
Jan 4 at 3:56












@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
– Theo Bendit
Jan 4 at 3:57




@Vic The example doesn't work if you take $M = mathbb{R}$, as $C$ will only have the interior $(100, 103)$.
– Theo Bendit
Jan 4 at 3:57












@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
– Theo Bendit
Jan 4 at 3:58




@Vic In fact, the $[100, 103]$ bit was only included to prevent $partial B = emptyset$.
– Theo Bendit
Jan 4 at 3:58


















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