Existence of subset with given Hausdorff dimensionQuestion on geometric measure theoryHausdorff measure on the sphere is well defined?Subsets of sets of positive Hausdorff dimension with controlled upper Minkowski dimensionHow can dimension depend on the point?Multiplicity of a subcovering in spaces of given Hausdorff dimensionExistence of a measurable map between metric spacesHausdorff dimension of boundaries of open sets diffeomorphic to $mathbbR^n$Hausdorff approximating measures and Borel setsWhen is Hausdorff measure locally finite?Existence of a discrete subset

Existence of subset with given Hausdorff dimension


Question on geometric measure theoryHausdorff measure on the sphere is well defined?Subsets of sets of positive Hausdorff dimension with controlled upper Minkowski dimensionHow can dimension depend on the point?Multiplicity of a subcovering in spaces of given Hausdorff dimensionExistence of a measurable map between metric spacesHausdorff dimension of boundaries of open sets diffeomorphic to $mathbbR^n$Hausdorff approximating measures and Borel setsWhen is Hausdorff measure locally finite?Existence of a discrete subset













4












$begingroup$


Let $Asubseteq mathbbR$ be Lebesgue-measurable and let $0<alpha<1$ be its Hausdorff dimension.




For a given $0<beta <alpha$ can we find a subset $Bsubset A$ with Hausdorff dimension $beta$?




In case this is true, could you provide a reference for this statement?



Added: Actually I am happy if $A$ is compact.






reference-request geometric-measure-theory







share|cite|improve this question











$endgroup$
















    4












    $begingroup$


    Let $Asubseteq mathbbR$ be Lebesgue-measurable and let $0<alpha<1$ be its Hausdorff dimension.




    For a given $0<beta <alpha$ can we find a subset $Bsubset A$ with Hausdorff dimension $beta$?




    In case this is true, could you provide a reference for this statement?



    Added: Actually I am happy if $A$ is compact.






    reference-request geometric-measure-theory







    share|cite|improve this question











    $endgroup$














      4












      4








      4





      $begingroup$


      Let $Asubseteq mathbbR$ be Lebesgue-measurable and let $0<alpha<1$ be its Hausdorff dimension.




      For a given $0<beta <alpha$ can we find a subset $Bsubset A$ with Hausdorff dimension $beta$?




      In case this is true, could you provide a reference for this statement?



      Added: Actually I am happy if $A$ is compact.






      reference-request geometric-measure-theory







      share|cite|improve this question











      $endgroup$




      Let $Asubseteq mathbbR$ be Lebesgue-measurable and let $0<alpha<1$ be its Hausdorff dimension.




      For a given $0<beta <alpha$ can we find a subset $Bsubset A$ with Hausdorff dimension $beta$?




      In case this is true, could you provide a reference for this statement?



      Added: Actually I am happy if $A$ is compact.






      reference-request geometric-measure-theory




      reference-request geometric-measure-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 20 mins ago







      Severin Schraven

















      asked 4 hours ago









      Severin SchravenSeverin Schraven

      20417




      20417




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Here is a partial answer. First of all, $dim_H (A) = alpha$ iff $ H^k(A)=infty$ for all $k<beta$ and $H^k(A) = 0$ for all $k>beta$. Then $H^alpha(A) = infty$ for all $alpha in (0,beta)$. If $A$ is closed then by Theorem 5.4 from The Geometry of Fractal Sets by Falconer there is a compact $Ksubset A$ such that $0<H^alpha(K)<infty$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            1 hour ago


















          3












          $begingroup$

          The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.




          Theorem. If a compact set $AsubsetmathbbR^n$ has non-$sigma$-finite measure $mathcalH^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcalH^beta<infty$.




          [1] R.O. Davies,
          A theorem on the existence of non-σ-finite subsets.
          Mathematika 15 (1968), 60–62.






          share|cite|improve this answer











          $endgroup$








          • 2




            $begingroup$
            I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
            $endgroup$
            – Skeeve
            2 hours ago







          • 1




            $begingroup$
            @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
            $endgroup$
            – Piotr Hajlasz
            1 hour ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            1 hour ago










          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Here is a partial answer. First of all, $dim_H (A) = alpha$ iff $ H^k(A)=infty$ for all $k<beta$ and $H^k(A) = 0$ for all $k>beta$. Then $H^alpha(A) = infty$ for all $alpha in (0,beta)$. If $A$ is closed then by Theorem 5.4 from The Geometry of Fractal Sets by Falconer there is a compact $Ksubset A$ such that $0<H^alpha(K)<infty$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            1 hour ago















          2












          $begingroup$

          Here is a partial answer. First of all, $dim_H (A) = alpha$ iff $ H^k(A)=infty$ for all $k<beta$ and $H^k(A) = 0$ for all $k>beta$. Then $H^alpha(A) = infty$ for all $alpha in (0,beta)$. If $A$ is closed then by Theorem 5.4 from The Geometry of Fractal Sets by Falconer there is a compact $Ksubset A$ such that $0<H^alpha(K)<infty$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            1 hour ago













          2












          2








          2





          $begingroup$

          Here is a partial answer. First of all, $dim_H (A) = alpha$ iff $ H^k(A)=infty$ for all $k<beta$ and $H^k(A) = 0$ for all $k>beta$. Then $H^alpha(A) = infty$ for all $alpha in (0,beta)$. If $A$ is closed then by Theorem 5.4 from The Geometry of Fractal Sets by Falconer there is a compact $Ksubset A$ such that $0<H^alpha(K)<infty$.






          share|cite|improve this answer









          $endgroup$



          Here is a partial answer. First of all, $dim_H (A) = alpha$ iff $ H^k(A)=infty$ for all $k<beta$ and $H^k(A) = 0$ for all $k>beta$. Then $H^alpha(A) = infty$ for all $alpha in (0,beta)$. If $A$ is closed then by Theorem 5.4 from The Geometry of Fractal Sets by Falconer there is a compact $Ksubset A$ such that $0<H^alpha(K)<infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          SkeeveSkeeve

          1994




          1994











          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            1 hour ago
















          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            1 hour ago















          $begingroup$
          That's exactly what I was looking for, thanks very much.
          $endgroup$
          – Severin Schraven
          1 hour ago




          $begingroup$
          That's exactly what I was looking for, thanks very much.
          $endgroup$
          – Severin Schraven
          1 hour ago











          3












          $begingroup$

          The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.




          Theorem. If a compact set $AsubsetmathbbR^n$ has non-$sigma$-finite measure $mathcalH^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcalH^beta<infty$.




          [1] R.O. Davies,
          A theorem on the existence of non-σ-finite subsets.
          Mathematika 15 (1968), 60–62.






          share|cite|improve this answer











          $endgroup$








          • 2




            $begingroup$
            I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
            $endgroup$
            – Skeeve
            2 hours ago







          • 1




            $begingroup$
            @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
            $endgroup$
            – Piotr Hajlasz
            1 hour ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            1 hour ago















          3












          $begingroup$

          The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.




          Theorem. If a compact set $AsubsetmathbbR^n$ has non-$sigma$-finite measure $mathcalH^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcalH^beta<infty$.




          [1] R.O. Davies,
          A theorem on the existence of non-σ-finite subsets.
          Mathematika 15 (1968), 60–62.






          share|cite|improve this answer











          $endgroup$








          • 2




            $begingroup$
            I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
            $endgroup$
            – Skeeve
            2 hours ago







          • 1




            $begingroup$
            @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
            $endgroup$
            – Piotr Hajlasz
            1 hour ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            1 hour ago













          3












          3








          3





          $begingroup$

          The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.




          Theorem. If a compact set $AsubsetmathbbR^n$ has non-$sigma$-finite measure $mathcalH^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcalH^beta<infty$.




          [1] R.O. Davies,
          A theorem on the existence of non-σ-finite subsets.
          Mathematika 15 (1968), 60–62.






          share|cite|improve this answer











          $endgroup$



          The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.




          Theorem. If a compact set $AsubsetmathbbR^n$ has non-$sigma$-finite measure $mathcalH^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcalH^beta<infty$.




          [1] R.O. Davies,
          A theorem on the existence of non-σ-finite subsets.
          Mathematika 15 (1968), 60–62.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          Piotr HajlaszPiotr Hajlasz

          9,73843873




          9,73843873







          • 2




            $begingroup$
            I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
            $endgroup$
            – Skeeve
            2 hours ago







          • 1




            $begingroup$
            @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
            $endgroup$
            – Piotr Hajlasz
            1 hour ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            1 hour ago












          • 2




            $begingroup$
            I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
            $endgroup$
            – Skeeve
            2 hours ago







          • 1




            $begingroup$
            @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
            $endgroup$
            – Piotr Hajlasz
            1 hour ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            1 hour ago







          2




          2




          $begingroup$
          I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
          $endgroup$
          – Skeeve
          2 hours ago





          $begingroup$
          I actually have a question (which is the reason why I wrote that my answer is partial): in the OP it is assumed that $A$ is Lebesgue measurable. Strictly saying this does not imply that $A$ is Borel, so it is not immediate that the result we both eventually refer to can be used (if $A$ were Borel or at least Souslin then yes).
          $endgroup$
          – Skeeve
          2 hours ago





          1




          1




          $begingroup$
          @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
          $endgroup$
          – Piotr Hajlasz
          1 hour ago





          $begingroup$
          @Skeeve Good point. I do not know what happens in general, but perhaps the compact case is sufficient for the needs of OP. Let's hear from him. I will changes my answer to emphasize that it only answers the compact case. By the way you know a lot of geometric measure theory so you should be more active :) There are not too many of us.
          $endgroup$
          – Piotr Hajlasz
          1 hour ago













          $begingroup$
          @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
          $endgroup$
          – Severin Schraven
          1 hour ago




          $begingroup$
          @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
          $endgroup$
          – Severin Schraven
          1 hour ago

















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