Is there a compact subset of $mathbb{R}$ which has positive Lebesgue-measure and no interior points?












2












$begingroup$


As the title says, I would like to know how to construct a compact subset of $mathbb{R}$ which does not have interior points but positive Lebesgue-measure?



As a hint, it is given $(0,1)setminus K$...






measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is an inner point? do you mean interior point?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:48










  • $begingroup$
    Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set.
    $endgroup$
    – math12
    Nov 6 '13 at 17:51






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… this may help you.
    $endgroup$
    – GA316
    Nov 6 '13 at 17:53










  • $begingroup$
    what is $k$ in the hint?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:54
















2












$begingroup$


As the title says, I would like to know how to construct a compact subset of $mathbb{R}$ which does not have interior points but positive Lebesgue-measure?



As a hint, it is given $(0,1)setminus K$...






measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is an inner point? do you mean interior point?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:48










  • $begingroup$
    Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set.
    $endgroup$
    – math12
    Nov 6 '13 at 17:51






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… this may help you.
    $endgroup$
    – GA316
    Nov 6 '13 at 17:53










  • $begingroup$
    what is $k$ in the hint?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:54














2












2








2





$begingroup$


As the title says, I would like to know how to construct a compact subset of $mathbb{R}$ which does not have interior points but positive Lebesgue-measure?



As a hint, it is given $(0,1)setminus K$...






measure-theory







share|cite|improve this question











$endgroup$




As the title says, I would like to know how to construct a compact subset of $mathbb{R}$ which does not have interior points but positive Lebesgue-measure?



As a hint, it is given $(0,1)setminus K$...






measure-theory




measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 6 '13 at 17:57









GA316

2,6691232




2,6691232










asked Nov 6 '13 at 17:43









math12math12

1,89411844




1,89411844












  • $begingroup$
    what is an inner point? do you mean interior point?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:48










  • $begingroup$
    Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set.
    $endgroup$
    – math12
    Nov 6 '13 at 17:51






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… this may help you.
    $endgroup$
    – GA316
    Nov 6 '13 at 17:53










  • $begingroup$
    what is $k$ in the hint?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:54


















  • $begingroup$
    what is an inner point? do you mean interior point?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:48










  • $begingroup$
    Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set.
    $endgroup$
    – math12
    Nov 6 '13 at 17:51






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… this may help you.
    $endgroup$
    – GA316
    Nov 6 '13 at 17:53










  • $begingroup$
    what is $k$ in the hint?
    $endgroup$
    – GA316
    Nov 6 '13 at 17:54
















$begingroup$
what is an inner point? do you mean interior point?
$endgroup$
– GA316
Nov 6 '13 at 17:48




$begingroup$
what is an inner point? do you mean interior point?
$endgroup$
– GA316
Nov 6 '13 at 17:48












$begingroup$
Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set.
$endgroup$
– math12
Nov 6 '13 at 17:51




$begingroup$
Yes, sorry. My english!.. I meant a point, for which there is a neighboorhoud within the set.
$endgroup$
– math12
Nov 6 '13 at 17:51




1




1




$begingroup$
en.wikipedia.org/wiki/… this may help you.
$endgroup$
– GA316
Nov 6 '13 at 17:53




$begingroup$
en.wikipedia.org/wiki/… this may help you.
$endgroup$
– GA316
Nov 6 '13 at 17:53












$begingroup$
what is $k$ in the hint?
$endgroup$
– GA316
Nov 6 '13 at 17:54




$begingroup$
what is $k$ in the hint?
$endgroup$
– GA316
Nov 6 '13 at 17:54










1 Answer
1






active

oldest

votes


















4












$begingroup$

Yes, indeed. A more generalized version of the construction of the Cantor set is something called a "fat Cantor set." As with the usual Cantor set, it is compact and has no interior points, but unlike the usual Cantor set, it has positive measure.



In fact, for any $0<a<1,$ there is a fat Cantor set of measure $1-a$. See Brian M. Scott's excellent answer here for more details, but the idea is to remove the middle $frac{a}3$ of each component interval at each stage, rather than the entire middle third.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
    $endgroup$
    – math12
    Nov 6 '13 at 18:05










  • $begingroup$
    Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:18






  • 1




    $begingroup$
    More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:21











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%2f554590%2fis-there-a-compact-subset-of-mathbbr-which-has-positive-lebesgue-measure-an%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









4












$begingroup$

Yes, indeed. A more generalized version of the construction of the Cantor set is something called a "fat Cantor set." As with the usual Cantor set, it is compact and has no interior points, but unlike the usual Cantor set, it has positive measure.



In fact, for any $0<a<1,$ there is a fat Cantor set of measure $1-a$. See Brian M. Scott's excellent answer here for more details, but the idea is to remove the middle $frac{a}3$ of each component interval at each stage, rather than the entire middle third.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
    $endgroup$
    – math12
    Nov 6 '13 at 18:05










  • $begingroup$
    Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:18






  • 1




    $begingroup$
    More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:21
















4












$begingroup$

Yes, indeed. A more generalized version of the construction of the Cantor set is something called a "fat Cantor set." As with the usual Cantor set, it is compact and has no interior points, but unlike the usual Cantor set, it has positive measure.



In fact, for any $0<a<1,$ there is a fat Cantor set of measure $1-a$. See Brian M. Scott's excellent answer here for more details, but the idea is to remove the middle $frac{a}3$ of each component interval at each stage, rather than the entire middle third.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
    $endgroup$
    – math12
    Nov 6 '13 at 18:05










  • $begingroup$
    Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:18






  • 1




    $begingroup$
    More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:21














4












4








4





$begingroup$

Yes, indeed. A more generalized version of the construction of the Cantor set is something called a "fat Cantor set." As with the usual Cantor set, it is compact and has no interior points, but unlike the usual Cantor set, it has positive measure.



In fact, for any $0<a<1,$ there is a fat Cantor set of measure $1-a$. See Brian M. Scott's excellent answer here for more details, but the idea is to remove the middle $frac{a}3$ of each component interval at each stage, rather than the entire middle third.






share|cite|improve this answer











$endgroup$



Yes, indeed. A more generalized version of the construction of the Cantor set is something called a "fat Cantor set." As with the usual Cantor set, it is compact and has no interior points, but unlike the usual Cantor set, it has positive measure.



In fact, for any $0<a<1,$ there is a fat Cantor set of measure $1-a$. See Brian M. Scott's excellent answer here for more details, but the idea is to remove the middle $frac{a}3$ of each component interval at each stage, rather than the entire middle third.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 13:38

























answered Nov 6 '13 at 17:51









Cameron BuieCameron Buie

85.2k771155




85.2k771155












  • $begingroup$
    Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
    $endgroup$
    – math12
    Nov 6 '13 at 18:05










  • $begingroup$
    Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:18






  • 1




    $begingroup$
    More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:21


















  • $begingroup$
    Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
    $endgroup$
    – math12
    Nov 6 '13 at 18:05










  • $begingroup$
    Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:18






  • 1




    $begingroup$
    More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
    $endgroup$
    – Cameron Buie
    Nov 6 '13 at 18:21
















$begingroup$
Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
$endgroup$
– math12
Nov 6 '13 at 18:05




$begingroup$
Could you please have a look on this: classes.yale.edu/fractals/labs/paperfoldinglab/… Why does this set have no interior points?
$endgroup$
– math12
Nov 6 '13 at 18:05












$begingroup$
Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
$endgroup$
– Cameron Buie
Nov 6 '13 at 18:18




$begingroup$
Well, notice what happens at each stage. In the first stage, we remove an interval of length $frac14,$ leaving us two intervals of length $frac38.$ In the second stage, we remove an interval of length $frac16,$ leaving us $4$ intervals of length $frac5{32}$. In the third stage, we remove $4$ intervals of length $frac1{64},$ leaving us $8$ intervals of length $frac9{128}.$ In the fourth stage, we remove $8$ intervals of length $frac1{256},$ leaving $16$ intervals of length $frac{17}{512}$. And so on....
$endgroup$
– Cameron Buie
Nov 6 '13 at 18:18




1




1




$begingroup$
More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
$endgroup$
– Cameron Buie
Nov 6 '13 at 18:21




$begingroup$
More generally, it can be shown (by induction) that after the $n$th stage, there are $2^n$ intervals of length $dfrac{2^n+1}{2^{2n+1}}.$ It can be shown that $$lim_{ntoinfty}frac{2^n+1}{2^{2n+1}}=0,$$ so given any positive length $ell,$ there is some stage at which no interval of length $ell$ can lie within the set at that stage, and so no such interval can lie within the fat Cantor set. Since this holds for all positive lengths $ell,$ then the fat Cantor set contains no open interval.
$endgroup$
– Cameron Buie
Nov 6 '13 at 18:21


















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%2f554590%2fis-there-a-compact-subset-of-mathbbr-which-has-positive-lebesgue-measure-an%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

How do I know what Microsoft account the skydrive app is syncing to?

Image
up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
Read more

Grease: Live!

Image
A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
Read more

When does type information flow backwards in C++?

Image
up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
Read more