Existence of subset with given Hausdorff dimensionQuestion on geometric measure theorySubsets 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 dimensionHausdorff dimension of sequence spaceConstruction of null sets with prescribed Hausdorff dimension and generalizationsHausdorff 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 theorySubsets 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 dimensionHausdorff dimension of sequence spaceConstruction of null sets with prescribed Hausdorff dimension and generalizationsHausdorff 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













9












$begingroup$


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











$endgroup$
















    9












    $begingroup$


    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











    $endgroup$














      9












      9








      9





      $begingroup$


      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 8 hours ago







      Severin Schraven

















      asked 12 hours ago









      Severin SchravenSeverin Schraven

      22918




      22918




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          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
            8 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
            6 hours ago


















          5












          $begingroup$

          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
            10 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
            9 hours ago











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









          4












          $begingroup$

          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
            8 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
            6 hours ago















          4












          $begingroup$

          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
            8 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
            6 hours ago













          4












          4








          4





          $begingroup$

          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 10 hours ago









          SkeeveSkeeve

          31915




          31915











          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            8 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
            6 hours ago
















          • $begingroup$
            That's exactly what I was looking for, thanks very much.
            $endgroup$
            – Severin Schraven
            8 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
            6 hours ago















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




          $begingroup$
          That's exactly what I was looking for, thanks very much.
          $endgroup$
          – Severin Schraven
          8 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
          6 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
          6 hours ago











          5












          $begingroup$

          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
            10 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
            9 hours ago











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















          5












          $begingroup$

          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
            10 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
            9 hours ago











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













          5












          5








          5





          $begingroup$

          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 9 hours ago

























          answered 10 hours ago









          Piotr HajlaszPiotr Hajlasz

          9,76843873




          9,76843873







          • 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
            10 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
            9 hours ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            8 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
            10 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
            9 hours ago











          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            8 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
          10 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
          10 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
          9 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
          9 hours ago













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




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

          How should I use the fbox command correctly to avoid producing a Bad Box message?How to put a long piece of text in a box?How to specify height and width of fboxIs there an arrayrulecolor-like command to change the rule color of fbox?What is the command to highlight bad boxes in pdf?Why does fbox sometimes place the box *over* the graphic image?how to put the text in the boxHow to create command for a box where text inside the box can automatically adjust?how can I make an fbox like command with certain color, shape and width of border?how to use fbox in align modeFbox increase the spacing between the box and it content (inner margin)how to change the box height of an equationWhat is the use of the hbox in a newcommand command?

          Image
          Is there a problem with hiding "forgot password" until it's needed? Time travel short story where a man arrives in the late 19th century in a time machine and then sends the machine back into the past Go Pregnant or Go Home Why Were Madagascar and New Zealand Discovered So Late? How can my private key be revealed if I use the same nonce while generating the signature? Modify casing of marked letters Trouble understanding overseas colleagues Best way to store options for panels Was Spock the First Vulcan in Starfleet? How do I keep an essay about "feeling flat" from feeling flat? How does a character multiclassing into warlock get a focus? Why does John Bercow say “unlock” after reading out the results of a vote? The Riley Riddle Mine Print name if parameter passed to function How can I replace every global instance of "x[2]" with "x_2" Hide Select Output from T-SQL Can I convert a rim brake wheel to a disc brak
          Read more

          152 Atala Notae | Nexus externi | Tabula navigationis"Discovery Circumstances: Numbered Minor Planets"2000152Small-Body Database

          Image
          L +1Latinitas +1 (bona)Asteroides anno 1875 repertiAsteroides Cinguli PrincipalisAsteroides a rebus ficticiis appellati Atalaasteroidessystematis solarisasteroidibus Cinguli PrincipalisAstronomismagnitudinem absolutam2 Novembris1875Paulo-Petro HenryObservatorio Lutetiaelitteris Francogallicis1801Castrobrientiiepochaeunitatum astronomicarumquam minimequam maximeInclinatio orbitalis°anomalia media [1]Small-Body DatabaseMercuriusVenusTellus (Terra)MarsIuppiterSaturnusUranusNeptunusSatellitesLunaAsteroidesindicesCorpora transneptuniana Vide etiam paginam discretivam: Atala 152 Atala Genus Asteroides Cinguli Principalis Res orbitales epochae 4 Ianuarii 2010 Excentricitas 0.0754396 Axis semimaior perihelion aphelion 3.1399791 UA 2.9031003 UA 3.3768579 UA Inclinatio orbitalis 12.11235° Longitudo nodi ascendentis 40.00824° Argumentum perihelii 58.52961° Anomalia media 348.89681° Periodus orbitalis 5.5641436375086 anni Res physicae Diameter km Massa kg Densitas media g/cm 3 Periodus rotation
          Read more

          Doxepinum Nexus interni Notae | Tabula navigationis3158DB01142WHOa682390"Structural Analysis of the Histamine H1 Receptor""Transdermal and Topical Drug Administration in the Treatment of Pain""Antidepressants as antipruritic agents: A review"

          Image
          Antidepressiva depressionuminsomniareceptorium H 1 agonista inversapruritusdermatitis atopicaelichenis simplicis chronici (neurodermatitis) Cave: notitiae huius paginae nec praescriptiones nec consilia medica sunt. Doxepinum Natura chemica Formula chemica C 19 H 21 NO Massa molaris 279.376 g/mol PubChem 3158 DrugBank DB01142 Natura pharmacologica Codex ATC N06AA12 ( WHO ) Tempus semivitae biologicum 17 (8-24) h Nor-Doxepinum: 31 h Metabolismus iecore (hepaticus): CYP2D6 CYP2C19 Metabolitus Nor-Doxepinum Excretio renibus (ca. 50%), faecibus: minus Ad usum therapeuticum Applicatio per os, i.v., i.m. MedlinePlus a682390 .mw-parser-output .existinglinksgray a,.mw-parser-output .existinglinksgray a:visitedcolor:gray.mw-parser-output .existinglinksgray a.newcolor:#ba0000.mw-parser-output .existinglinksgray a.new:visitedcolor:#a55858 (Anglice) Doxepinum est substantia psychoanaleptica et antidepressiva, ergo medicamentum tractationis depressionum, praesertim cum insomnia praesente, effectu
          Read more