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











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1 Answer
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1 Answer
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active

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active

oldest

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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


















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