Existence of subset with given Hausdorff dimension












6












$begingroup$


Let $Asubseteq mathbb{R}$ 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











$endgroup$

















    6












    $begingroup$


    Let $Asubseteq mathbb{R}$ 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











    $endgroup$















      6












      6








      6





      $begingroup$


      Let $Asubseteq mathbb{R}$ 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 $Asubseteq mathbb{R}$ 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 2 hours ago







      Severin Schraven

















      asked 6 hours ago









      Severin SchravenSeverin Schraven

      21418




      21418






















          2 Answers
          2






          active

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          3












          $begingroup$

          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
            3 hours ago










          • $begingroup$
            Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
            $endgroup$
            – Severin Schraven
            1 hour ago



















          4












          $begingroup$

          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 $Asubsetmathbb{R}^n$ has non-$sigma$-finite measure $mathcal{H}^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcal{H}^beta<infty$.




          [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
            4 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
            4 hours ago












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











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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          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
            3 hours ago










          • $begingroup$
            Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
            $endgroup$
            – Severin Schraven
            1 hour ago
















          3












          $begingroup$

          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
            3 hours ago










          • $begingroup$
            Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
            $endgroup$
            – Severin Schraven
            1 hour ago














          3












          3








          3





          $begingroup$

          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 4 hours ago









          SkeeveSkeeve

          30914




          30914












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










          • $begingroup$
            Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
            $endgroup$
            – Severin Schraven
            1 hour ago


















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










          • $begingroup$
            Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
            $endgroup$
            – Severin Schraven
            1 hour ago
















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




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












          $begingroup$
          Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
          $endgroup$
          – Severin Schraven
          1 hour ago




          $begingroup$
          Actually it is worth mentioning that in the same reference in Theorem 5.6, this is generalized to Souslin spaces.
          $endgroup$
          – Severin Schraven
          1 hour ago











          4












          $begingroup$

          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 $Asubsetmathbb{R}^n$ has non-$sigma$-finite measure $mathcal{H}^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcal{H}^beta<infty$.




          [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
            4 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
            4 hours ago












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
















          4












          $begingroup$

          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 $Asubsetmathbb{R}^n$ has non-$sigma$-finite measure $mathcal{H}^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcal{H}^beta<infty$.




          [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
            4 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
            4 hours ago












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














          4












          4








          4





          $begingroup$

          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 $Asubsetmathbb{R}^n$ has non-$sigma$-finite measure $mathcal{H}^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcal{H}^beta<infty$.




          [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 $Asubsetmathbb{R}^n$ has non-$sigma$-finite measure $mathcal{H}^beta$, then there us a
          subset $Bsubset A$ such that $0<mathcal{H}^beta<infty$.




          [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 4 hours ago

























          answered 4 hours ago









          Piotr HajlaszPiotr Hajlasz

          9,75843873




          9,75843873








          • 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
            4 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
            4 hours ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            3 hours 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
            4 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
            4 hours ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            3 hours 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
          4 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
          4 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
          4 hours 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
          4 hours ago














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




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


















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