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
$begingroup$
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
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
$endgroup$
add a comment |
$begingroup$
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
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
$endgroup$
add a comment |
$begingroup$
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
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
$endgroup$
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
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
edited 5 hours ago
Severin Schraven
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
asked 9 hours ago
Severin SchravenSeverin Schraven
22418
22418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $0<H^alpha(K)<infty$.
answered 7 hours ago
SkeeveSkeeve
30915
$endgroup$
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– 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
4 hours ago
add a comment |
$begingroup$
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.
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
$endgroup$
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
7 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
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $0<H^alpha(K)<infty$.
answered 7 hours ago
SkeeveSkeeve
30915
$endgroup$
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– 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
4 hours ago
add a comment |
$begingroup$
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 $0<H^alpha(K)<infty$.
answered 7 hours ago
SkeeveSkeeve
30915
$endgroup$
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– 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
4 hours ago
add a comment |
$begingroup$
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 $0<H^alpha(K)<infty$.
answered 7 hours ago
SkeeveSkeeve
30915
$endgroup$
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 $0<H^alpha(K)<infty$.
answered 7 hours ago
SkeeveSkeeve
30915
answered 7 hours ago
SkeeveSkeeve
30915
answered 7 hours ago
SkeeveSkeeve
30915
answered 7 hours ago
SkeeveSkeeve
30915
30915
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– 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
4 hours ago
add a comment |
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– 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
4 hours ago
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– Severin Schraven
6 hours ago
$begingroup$
That's exactly what I was looking for, thanks very much.
$endgroup$
– 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
4 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
4 hours ago
add a comment |
$begingroup$
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.
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
$endgroup$
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
7 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
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 hours ago
add a comment |
$begingroup$
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.
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
$endgroup$
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
7 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
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 hours ago
add a comment |
$begingroup$
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.
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
$endgroup$
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.
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
edited 6 hours ago
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
answered 7 hours ago
Piotr HajlaszPiotr Hajlasz
9,75843873
answered 7 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
7 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
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 hours ago
add a comment |
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
7 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
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 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
7 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
7 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
6 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
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 hours ago
$begingroup$
@PiotrHajlasz Indeed, the compact case is fine for me. Thanks for the answer.
$endgroup$
– Severin Schraven
6 hours ago
add a comment |
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