“Fat” Cantor Set












9












$begingroup$


So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...



Are there other constraints that need to be made in order to accomplish this?






real-analysis cantor-set







share|cite|improve this question











$endgroup$








  • 11




    $begingroup$
    The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
    $endgroup$
    – Rahul
    Jan 27 '13 at 4:06












  • $begingroup$
    If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
    $endgroup$
    – Gerry Myerson
    Jan 27 '13 at 4:06










  • $begingroup$
    You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
    $endgroup$
    – user53153
    Jan 27 '13 at 4:09










  • $begingroup$
    You can, however, have outer measure arbitrarily close to one.
    $endgroup$
    – Brian M. Scott
    Jan 27 '13 at 4:10






  • 1




    $begingroup$
    Here are some details of what's contained on the Wikipedia page.
    $endgroup$
    – Martin
    Jan 27 '13 at 4:13


















9












$begingroup$


So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...



Are there other constraints that need to be made in order to accomplish this?






real-analysis cantor-set







share|cite|improve this question











$endgroup$








  • 11




    $begingroup$
    The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
    $endgroup$
    – Rahul
    Jan 27 '13 at 4:06












  • $begingroup$
    If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
    $endgroup$
    – Gerry Myerson
    Jan 27 '13 at 4:06










  • $begingroup$
    You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
    $endgroup$
    – user53153
    Jan 27 '13 at 4:09










  • $begingroup$
    You can, however, have outer measure arbitrarily close to one.
    $endgroup$
    – Brian M. Scott
    Jan 27 '13 at 4:10






  • 1




    $begingroup$
    Here are some details of what's contained on the Wikipedia page.
    $endgroup$
    – Martin
    Jan 27 '13 at 4:13
















9












9








9


3



$begingroup$


So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...



Are there other constraints that need to be made in order to accomplish this?






real-analysis cantor-set







share|cite|improve this question











$endgroup$




So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...



Are there other constraints that need to be made in order to accomplish this?






real-analysis cantor-set




real-analysis cantor-set






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 5 '15 at 8:56









Martin Sleziak

44.7k9117272




44.7k9117272










asked Jan 27 '13 at 4:03









FrankFrank

4612




4612








  • 11




    $begingroup$
    The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
    $endgroup$
    – Rahul
    Jan 27 '13 at 4:06












  • $begingroup$
    If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
    $endgroup$
    – Gerry Myerson
    Jan 27 '13 at 4:06










  • $begingroup$
    You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
    $endgroup$
    – user53153
    Jan 27 '13 at 4:09










  • $begingroup$
    You can, however, have outer measure arbitrarily close to one.
    $endgroup$
    – Brian M. Scott
    Jan 27 '13 at 4:10






  • 1




    $begingroup$
    Here are some details of what's contained on the Wikipedia page.
    $endgroup$
    – Martin
    Jan 27 '13 at 4:13
















  • 11




    $begingroup$
    The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
    $endgroup$
    – Rahul
    Jan 27 '13 at 4:06












  • $begingroup$
    If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
    $endgroup$
    – Gerry Myerson
    Jan 27 '13 at 4:06










  • $begingroup$
    You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
    $endgroup$
    – user53153
    Jan 27 '13 at 4:09










  • $begingroup$
    You can, however, have outer measure arbitrarily close to one.
    $endgroup$
    – Brian M. Scott
    Jan 27 '13 at 4:10






  • 1




    $begingroup$
    Here are some details of what's contained on the Wikipedia page.
    $endgroup$
    – Martin
    Jan 27 '13 at 4:13










11




11




$begingroup$
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
$endgroup$
– Rahul
Jan 27 '13 at 4:06






$begingroup$
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
$endgroup$
– Rahul
Jan 27 '13 at 4:06














$begingroup$
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
$endgroup$
– Gerry Myerson
Jan 27 '13 at 4:06




$begingroup$
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
$endgroup$
– Gerry Myerson
Jan 27 '13 at 4:06












$begingroup$
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
$endgroup$
– user53153
Jan 27 '13 at 4:09




$begingroup$
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
$endgroup$
– user53153
Jan 27 '13 at 4:09












$begingroup$
You can, however, have outer measure arbitrarily close to one.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 4:10




$begingroup$
You can, however, have outer measure arbitrarily close to one.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 4:10




1




1




$begingroup$
Here are some details of what's contained on the Wikipedia page.
$endgroup$
– Martin
Jan 27 '13 at 4:13






$begingroup$
Here are some details of what's contained on the Wikipedia page.
$endgroup$
– Martin
Jan 27 '13 at 4:13












1 Answer
1






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12












$begingroup$

Say you delete the middle third.



Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.



Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.



And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.






share|cite|improve this answer











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    1 Answer
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    active

    oldest

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    12












    $begingroup$

    Say you delete the middle third.



    Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.



    Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.



    And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.






    share|cite|improve this answer











    $endgroup$


















      12












      $begingroup$

      Say you delete the middle third.



      Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.



      Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.



      And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.






      share|cite|improve this answer











      $endgroup$
















        12












        12








        12





        $begingroup$

        Say you delete the middle third.



        Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.



        Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.



        And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.






        share|cite|improve this answer











        $endgroup$



        Say you delete the middle third.



        Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.



        Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.



        And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 9 '14 at 3:56

























        answered Jan 27 '13 at 4:23









        Michael HardyMichael Hardy

        1




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