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













3












$begingroup$


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 borel and compact.






reference-request geometric-measure-theory







share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    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 borel and compact.






    reference-request geometric-measure-theory







    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      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 borel and 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 borel and 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 37 mins ago







      Severin Schraven

















      asked 3 hours ago









      Severin SchravenSeverin Schraven

      1997




      1997




















          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
            32 mins ago


















          2












          $begingroup$

          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
            1 hour 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
            33 mins ago










          Your Answer





          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "504"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325532%2fexistence-of-subset-with-given-hausdorff-dimension%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          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
            32 mins 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
            32 mins 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 1 hour ago









          SkeeveSkeeve

          1744




          1744











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
















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















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




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











          2












          $begingroup$

          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
            1 hour 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
            33 mins ago















          2












          $begingroup$

          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
            1 hour 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
            33 mins ago













          2












          2








          2





          $begingroup$

          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 1 hour ago









          Piotr HajlaszPiotr Hajlasz

          9,72843873




          9,72843873







          • 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
            1 hour 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
            33 mins 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
            1 hour 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
            33 mins 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
          1 hour 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
          1 hour 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
          33 mins ago




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

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to MathOverflow!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325532%2fexistence-of-subset-with-given-hausdorff-dimension%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

          Image
          Flight departed from the gate 5 min before scheduled departure time. Refund options How to show a density matrix is in a pure/mixed state? Can we cancel the equality mark here? How to ask rejected full-time candidates to apply to teach individual courses? Are there any AGPL-style licences that require source code modifications to be public? Why did Israel vote against lifting the American embargo on Cuba? What should one know about term logic before studying propositional and predicate logic? Why is Rajasthan pro BJP in the LS elections but not in the state elections? New Order #6: Easter Egg Question on Gÿongy' lemma proof How to resize main filesystem Is it OK if I do not take the receipt in Germany? My mentor says to set image to Fine instead of RAW — how is this different from JPG? Can I feed enough spin up electron to a black hole to affect it's angular momentum? How do I find my Spellcasting Ability for my D&D character? Is my guitar’s ac
          Read more

          2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

          Image
          Objetos do disco dispersoObjetos transnetunianos objeto transnetunianodisco dispersoSistema Solarcorpo celesteressonância orbitalplanetaplanetamagnitude absolutadiâmetro9 de abril2013Outer Solar System Origins Surveyórbitaexcentricidadesemieixo maiorUAperiélioSolafélio 2013 GY136 Data da descoberta 9 de abril de 2013 Descoberto por Outer Solar System Origins Survey Categoria Transnetuniano Objeto do disco disperso Elementos orbitais Semieixo maior 55,961 UA Perélio 32,595 UA Afélio 79,327 UA Excentricidade 0,418 Inclinação 10,9° Características físicas Dimensões 133 km Magnitude absoluta 7,6 ver 2013 GY 136 , também escrito como 2013 GY136 , é um objeto transnetuniano que está localizado no disco disperso, uma região do Sistema Solar. Este corpo celeste está em uma ressonância orbital de 2:5 com o planeta planeta. Ele possui uma magnitude absoluta de 7,6 [ 1 ] e tem um diâmetro estimado de 133 km. [ 2 ] Descoberta | 2013 GY 136 foi descoberto no dia 9 de abril de 2013 pelo Outer S
          Read more

          Button changing it's text & action. Good or terrible? The 2019 Stack Overflow Developer Survey Results Are Inchanging text on user mouseoverShould certain functions be “hard to find” for powerusers to discover?Custom liking function - do I need user login?Using different checkbox style for different checkbox behaviorBest Practices: Save and Exit in Software UIInteraction with remote validated formMore efficient UI to progress the user through a complicated process?Designing a popup notice for a gameShould bulk-editing functions be hidden until a table row is selected, or is there a better solution?Is it bad practice to disable (replace) the context menu?

          Image
          Delete all lines which don't have n characters before delimiter Why isn't the circumferential light around the M87 black hole's event horizon symmetric? Why did Acorn's A3000 have red function keys? Does the shape of a die affect the probability of a number being rolled? Worn-tile Scrabble Why do UK politicians seemingly ignore opinion polls on Brexit? If a Druid sees an animal’s corpse, can they wild shape into that animal? Is an up-to-date browser secure on an out-of-date OS? Loose spokes after only a few rides How are circuits which use complex ICs normally simulated? Multiply Two Integer Polynomials Resizing object distorts it (Illustrator CC 2018) Do these rules for Critical Successes and Critical Failures seem Fair? Why can Shazam fly? Does a dangling wire really electrocute me if I'm standing in water? Can we generate random numbers using irrational numbers like π and e? Why isn't airport relocation done gradually? What d
          Read more