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












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















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

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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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


















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









1












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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*}






share|cite|improve this answer











$endgroup$



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*}







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

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


















  • $begingroup$
    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


















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














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





















































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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more