Characterisation of compactness in $mathbbR^n$ [duplicate]

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  • Inverse of Heine–Cantor theorem

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Let $E$ be a bounded measurable set of $mathbbR^n$.
If every countinuous function $f:EtomathbbR$ is also uniformly continuous, then $E$ is compact.



Since I do not know topological property of $f(E)$, it seems that I must use some measure property to show $E$ is closed. However, measure and topology seems not quite relevant.



Thank you!






real-analysis general-topology lebesgue-measure







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marked as duplicate by Ben, user91500, Jose Arnaldo Bebita Dris, Strants, rtybase Sep 5 at 17:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set?
    – Leonardo
    Sep 5 at 8:16











  • @Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact
    – Alessandro Codenotti
    Sep 5 at 9:46










  • You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all.
    – Leonardo
    Sep 5 at 14:57














up vote
3
down vote

favorite













This question already has an answer here:



  • Inverse of Heine–Cantor theorem

    2 answers



Let $E$ be a bounded measurable set of $mathbbR^n$.
If every countinuous function $f:EtomathbbR$ is also uniformly continuous, then $E$ is compact.



Since I do not know topological property of $f(E)$, it seems that I must use some measure property to show $E$ is closed. However, measure and topology seems not quite relevant.



Thank you!






real-analysis general-topology lebesgue-measure







share|cite|improve this question













marked as duplicate by Ben, user91500, Jose Arnaldo Bebita Dris, Strants, rtybase Sep 5 at 17:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set?
    – Leonardo
    Sep 5 at 8:16











  • @Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact
    – Alessandro Codenotti
    Sep 5 at 9:46










  • You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all.
    – Leonardo
    Sep 5 at 14:57












up vote
3
down vote

favorite









up vote
3
down vote

favorite












This question already has an answer here:



  • Inverse of Heine–Cantor theorem

    2 answers



Let $E$ be a bounded measurable set of $mathbbR^n$.
If every countinuous function $f:EtomathbbR$ is also uniformly continuous, then $E$ is compact.



Since I do not know topological property of $f(E)$, it seems that I must use some measure property to show $E$ is closed. However, measure and topology seems not quite relevant.



Thank you!






real-analysis general-topology lebesgue-measure







share|cite|improve this question














This question already has an answer here:



  • Inverse of Heine–Cantor theorem

    2 answers



Let $E$ be a bounded measurable set of $mathbbR^n$.
If every countinuous function $f:EtomathbbR$ is also uniformly continuous, then $E$ is compact.



Since I do not know topological property of $f(E)$, it seems that I must use some measure property to show $E$ is closed. However, measure and topology seems not quite relevant.



Thank you!





This question already has an answer here:



  • Inverse of Heine–Cantor theorem

    2 answers






real-analysis general-topology lebesgue-measure




real-analysis general-topology lebesgue-measure






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asked Sep 5 at 7:48









Leonardo

1387




1387




marked as duplicate by Ben, user91500, Jose Arnaldo Bebita Dris, Strants, rtybase Sep 5 at 17:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Ben, user91500, Jose Arnaldo Bebita Dris, Strants, rtybase Sep 5 at 17:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set?
    – Leonardo
    Sep 5 at 8:16











  • @Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact
    – Alessandro Codenotti
    Sep 5 at 9:46










  • You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all.
    – Leonardo
    Sep 5 at 14:57
















  • Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set?
    – Leonardo
    Sep 5 at 8:16











  • @Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact
    – Alessandro Codenotti
    Sep 5 at 9:46










  • You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all.
    – Leonardo
    Sep 5 at 14:57















Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set?
– Leonardo
Sep 5 at 8:16





Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set?
– Leonardo
Sep 5 at 8:16













@Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact
– Alessandro Codenotti
Sep 5 at 9:46




@Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact
– Alessandro Codenotti
Sep 5 at 9:46












You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all.
– Leonardo
Sep 5 at 14:57




You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all.
– Leonardo
Sep 5 at 14:57










1 Answer
1






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Suppose $x_n in E, x_n to x$ and $x notin E$. Let $f(y)=frac 1 y-x$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $|x_n-x_m| to 0$ it follows from uniform continuity that $frac 1 -frac 1 to 0$. Since Cauchy sequences in $mathbb R^n$ are bounded it follows that $frac 1 $ is bounded. But this sequence tends to $infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.






share|cite|improve this answer
















  • 1




    Thank you! I am still thinking whether measurabiluty is necessary.
    – Leonardo
    Sep 5 at 8:18






  • 1




    Measurability is absolutely unnecessary.
    – Kavi Rama Murthy
    Sep 5 at 8:19

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote













Suppose $x_n in E, x_n to x$ and $x notin E$. Let $f(y)=frac 1 y-x$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $|x_n-x_m| to 0$ it follows from uniform continuity that $frac 1 -frac 1 to 0$. Since Cauchy sequences in $mathbb R^n$ are bounded it follows that $frac 1 $ is bounded. But this sequence tends to $infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.






share|cite|improve this answer
















  • 1




    Thank you! I am still thinking whether measurabiluty is necessary.
    – Leonardo
    Sep 5 at 8:18






  • 1




    Measurability is absolutely unnecessary.
    – Kavi Rama Murthy
    Sep 5 at 8:19














up vote
4
down vote













Suppose $x_n in E, x_n to x$ and $x notin E$. Let $f(y)=frac 1 y-x$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $|x_n-x_m| to 0$ it follows from uniform continuity that $frac 1 -frac 1 to 0$. Since Cauchy sequences in $mathbb R^n$ are bounded it follows that $frac 1 $ is bounded. But this sequence tends to $infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.






share|cite|improve this answer
















  • 1




    Thank you! I am still thinking whether measurabiluty is necessary.
    – Leonardo
    Sep 5 at 8:18






  • 1




    Measurability is absolutely unnecessary.
    – Kavi Rama Murthy
    Sep 5 at 8:19












up vote
4
down vote










up vote
4
down vote









Suppose $x_n in E, x_n to x$ and $x notin E$. Let $f(y)=frac 1 y-x$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $|x_n-x_m| to 0$ it follows from uniform continuity that $frac 1 -frac 1 to 0$. Since Cauchy sequences in $mathbb R^n$ are bounded it follows that $frac 1 $ is bounded. But this sequence tends to $infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.






share|cite|improve this answer












Suppose $x_n in E, x_n to x$ and $x notin E$. Let $f(y)=frac 1 y-x$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $|x_n-x_m| to 0$ it follows from uniform continuity that $frac 1 -frac 1 to 0$. Since Cauchy sequences in $mathbb R^n$ are bounded it follows that $frac 1 $ is bounded. But this sequence tends to $infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 5 at 7:56









Kavi Rama Murthy

26.3k31437




26.3k31437







  • 1




    Thank you! I am still thinking whether measurabiluty is necessary.
    – Leonardo
    Sep 5 at 8:18






  • 1




    Measurability is absolutely unnecessary.
    – Kavi Rama Murthy
    Sep 5 at 8:19












  • 1




    Thank you! I am still thinking whether measurabiluty is necessary.
    – Leonardo
    Sep 5 at 8:18






  • 1




    Measurability is absolutely unnecessary.
    – Kavi Rama Murthy
    Sep 5 at 8:19







1




1




Thank you! I am still thinking whether measurabiluty is necessary.
– Leonardo
Sep 5 at 8:18




Thank you! I am still thinking whether measurabiluty is necessary.
– Leonardo
Sep 5 at 8:18




1




1




Measurability is absolutely unnecessary.
– Kavi Rama Murthy
Sep 5 at 8:19




Measurability is absolutely unnecessary.
– Kavi Rama Murthy
Sep 5 at 8:19


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