Existence of subset with given Hausdorff dimensionQuestion on geometric measure theorySubsets of sets of...



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 $mathbb{R}^n$Hausdorff approximating measures and Borel setsWhen is Hausdorff measure locally finite?A quantity that distinguishes finer than Hausdorff dimension













3












$begingroup$


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 borel and compact.






reference-request geometric-measure-theory







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    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 borel and compact.






    reference-request geometric-measure-theory







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      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 borel and 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 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 20 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
            15 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 $Asubsetmathbb{R}^n$ 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
            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
            53 mins ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            15 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
            15 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
            15 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
            15 mins ago


















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
















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




          $begingroup$
          That's exactly what I was looking for, thanks very much.
          $endgroup$
          – Severin Schraven
          15 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 $Asubsetmathbb{R}^n$ 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
            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
            53 mins ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            15 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 $Asubsetmathbb{R}^n$ 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
            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
            53 mins ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            15 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 $Asubsetmathbb{R}^n$ 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 51 mins 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
            53 mins ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            15 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
            53 mins ago












          • $begingroup$
            @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
            $endgroup$
            – Severin Schraven
            15 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
          53 mins 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
          53 mins ago














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




          $begingroup$
          @PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
          $endgroup$
          – Severin Schraven
          15 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

          Installing LyX: “No textclass is found.”LyX installation error- text class not found- 'Reconfigure' or...

          Image
          How is it possible for user to changed after storage was encrypted? (on OS X, Android) How can bays and straits be determined in a procedurally generated map? Why Is Death Allowed In the Matrix? A function which translates a sentence to title-case How is it possible to have an ability score that is less than 3? TGV timetables / schedules? What is the command to reset a PC without deleting any files Why don't electron-positron collisions release infinite energy? How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? Continuity at a point in terms of closure ...
          Read more

          (1602) Indiana Índice Designación y nombre Características orbitales Véase...

          Asteroides del cinturón principalObjetos astronómicos descubiertos por el Indiana Asteroid ProgramObjetos astronómicos descubiertos desde el Observatorio Goethe LinkObjetos astronómicos descubiertos en 1950 asteroidecinturón de asteroidesIndiana Asteroid Programobservatorio Goethe LinkIndianaua (1602) Indiana Descubrimiento Descubridor Indiana Asteroid Program Fecha 14 de marzo de 1950 Lugar Brooklyn Designaciones 1943 DJ, 1950 GF, 1975 XR Nombre provisional 1950 GF Categoría Cinturón de asteroides Elementos orbitales Longitud del nodo ascendente 75,15° Inclinación 4,162° Argumento del periastro 73,23° Semieje mayor 2,245 ua Excentricidad 0,1041 Anomalía media 124,4° Elementos orbitales derivados Época 2457000,5 (09/12/2014) TDB [ 1 ] ​ Periastro o perihelio 2,011 ua Apoastro o afelio 2,478 ua Período orbital sideral 1228 días Características físicas Periodo de rotación 2,601 horas ...
          Read more

          Universidad Autónoma de Occidente Índice Historia Campus Facultades Programas Académicos Medios de...

          Image
          Universidades de CaliUniversidades privadas de ColombiaInstituciones de Educación Superior de Colombia acreditadas Ministerio de Educación de Colombia19996 de septiembreConsejo Nacional de AcreditaciónMinisterio de EducaciónColciencias198319861979199612 de julio1999Sociedad Interamericana de Prensa1993Universidad Carlos III de MadridUniversidad del Cauca197119781999Escuela de MonitoresOGEGrupo GescomGrupo GescomGrupo Gaia Universidad Autónoma de Occidente Acrónimo UAO Alias La Autónoma Lema Vive el conocimiento, construye tu vida [ 1 ] ​ Tipo Privada Fundación 20 de febrero de 1970 (49 años) Localización Dirección Cl 25 # 115-85 Km. 2 vía Cali - Jamundí   Cali,   Valle del Cauca, Colombia   Colombia Campus Valle del Lilí 122,253 kilómetros cuadrados (12 225,3 ha) Coordenadas 3°21′14″N 76°31′22″O  /  3.35389, -76.5228 Coordenadas: 3°21′14″N 76°31′22″O  /  3.35389, -76.5228 Otras sedes San Fernando (Cali) Candelaria El Cerrit...
          Read more