Questions tagged [hausdorff-dimension]

Questions about dimensions of possibly highly irregular or "rough" sets, Hausdorff–Besicovitch dimension and related concepts such as box-counting or Minkowski–Bouligand dimension.

Filter by
Sorted by
Tagged with
0 votes
1 answer
95 views

Fractal sets and dimensions

Can we construct two sets $E$ and $F$ meeting the following criteria $\dim_H(E) = \dim_H(F) = \dim_H(E ∩ F)$ $\dim_P(E), \dim_P(F)$, and $\dim_P(E ∩ F)$ are distinct? Here $\dim_H$ denotes the ...
B-S's user avatar
  • 39
2 votes
0 answers
20 views

Kaplan-Yorke dimension when all Lyapunov exponents are positive or the system is 1-dimensional

The Lyapunov/Kaplan-Yorke dimension $D_{KY}$ can be calculated using the Lyapunov Exponents of an n-dimensional dynamical system, as shown in the relevant Scholarpedia entry. If $\lambda_1>\...
Claudiu Crăciun's user avatar
8 votes
0 answers
202 views

The Hausdorff dimension of the set of reals of inner models

Suppose that both $M$ and $N$ are models of $ZFC$ with $M\subseteq N$ so that $M$ is definable in $N$. Question Can $(\mathbb{R})^M$ have Hausdorff dimension strictly between $0$ and $1$ in $N$? How ...
喻 良's user avatar
  • 4,191
2 votes
1 answer
162 views

Hausdorff dimension of the curve of a continuous nowhere differentiable function

It is of course well-known that there are plenty of functions from $\mathbb R$ into itself which are continuous and nowhere differentiable. Although the Baire Category Theorem is enough to prove the ...
Bazin's user avatar
  • 15k
15 votes
4 answers
1k views

Steinhaus theorem and Hausdorff dimension

Assume for simplicity that sets $A_i\subset\mathbb{R}$ are compact. If $A_1$ and $A_2$ have positive length, then $A_1+A_2$ contains an interval. That is a variant of the classical Steinhaus theorem ...
Piotr Hajlasz's user avatar
2 votes
1 answer
62 views

Existence of an $\alpha$-regular measure with positive measure on a binary digits do not have a limiting frequency

let $$X=\left\{ \sum_{n=1}^{\infty}a_{n}2^{-n}:a_{n}\in\left\{ 0,1\right\} ,\liminf\frac{1}{n}\sum_{i=1}^{n}a_{i}<\limsup\frac{1}{n}\sum_{i=1}^{n}a_{i}\right\} $$ I'm studying fractal geometry and ...
Simple Conjugate's user avatar
0 votes
1 answer
288 views

Finding examples of functions which are infinite or undefined with current extensions of the expected value?

Preliminaries Consider the expectations desribed in this paper, which is an extension of the Lebesgue density theorem; this paper which is an extension of the Hausdorff measure, using Hyperbolic ...
Arbuja's user avatar
  • 1
0 votes
0 answers
68 views

Using programming to measure the uniformity of measurable subsets of the unit square?

This is a follow up to this post using this answer: Let $S:=[0,1]^2$ be the unit square. "Partition" $S$ naturally into four congruent squares $S_{1,j}$ (with side length $1/2$ each), where ...
Arbuja's user avatar
  • 1
1 vote
0 answers
111 views

Are the extensions of the expected value, below, finite for all functions in only a shy subset of all measurable functions?

This is a follow up to this post, where I wish to verify whether one of the statements (in the post) is true but first let's recap the definitions: Let $(X,d)$ be a metric space. If set $A\subseteq X$,...
Arbuja's user avatar
  • 1
7 votes
1 answer
150 views

Plane curve with continuously increasing Hausdorff dimension

In a recent paper, we required the following fact. Proposition 1. There exists a simple closed curve $\gamma\subset\mathbb{C}$ with the following property. If $\phi$ is a biholomorphic map, defined on ...
Lasse Rempe's user avatar
  • 6,455
1 vote
0 answers
722 views

Finding a unique and finite expected value for almost all measurable functions?

Let $(X,d)$ be a metric space. If set $A\subseteq X$, let $H^{\alpha}$ be the $\alpha$-dimensional Hausdorff measure on $A$, where $\alpha\in[0,+\infty)$ and $\text{dim}_{\text{H}}(A)$ is the ...
Arbuja's user avatar
  • 1
2 votes
2 answers
782 views

Defining a measure of uniformity for measurable subsets of $[0,1]^2$ w.r.t dimension $\alpha\in[0,2]$

Let $(X,d)$ be a metric space. If set $A\subseteq X$, let $H^{\alpha}$ be the $\alpha$-dimensional Hausdorff measure on $A$, where $\alpha\in[0,2]$ and $\text{dim}_{\text{H}}(A)$ is the Hausdorff ...
Arbuja's user avatar
  • 1
4 votes
0 answers
84 views

Counting fractals modulo "shared complements"

Previously asked at MSE: Let $\mathscr{H}$ be the space of compact nonempty subsets of $\mathbb{R}^2$ (I'm not especially wedded to dimension $2$, so feel free to tweak that if it would lead to a more ...
Noah Schweber's user avatar
1 vote
1 answer
141 views

Is there a two-dimensional unimodal function with fractal level sets

Is there an open simply connected $U\subset\mathbb{R}^2$ and a continuous non-constant function $f: U\to \mathbb{R}$, such that for all $c\in \mathbb{R}$ both sets $$ f_{<c}~=~ f^{-1}\left( (-\...
Karl Fabian's user avatar
  • 1,546
4 votes
0 answers
147 views

Naïve definition of a measure on a fractal

This question was previously posted on MSE. Let $K\subset \mathbb R^2$ be a compact fractal of Hausdorff dimension $1<d<2$. I want to define a natural measure on $K$. One option would be to use ...
Matheus Manzatto's user avatar
6 votes
0 answers
106 views

A generalized Hausdorff dimension in form of a Lower semi continuous function

Let $(X,d)$ be a compact metric space. Assume that $f:X\to \mathbb{R}$ is a positive continuous function. We say that the $f$-dimension of $(X,d)$ is equal to $0$ if for every $\epsilon>0$ ...
Ali Taghavi's user avatar
1 vote
1 answer
252 views

Hausdorff dimension of the non-differentiability set of a locally Lipschitz function

Let $f:\mathbb R^n \to \mathbb R$ and $E := \{x \in X : f \text{ not Fréchet differentiable at }x\}$. Then $E$ is Borel measurable. It is well-known that Theorem If $f$ is convex, then the Hausdorff ...
Akira's user avatar
  • 749
2 votes
1 answer
111 views

Hausdorff dimension and non-empty intersections with lines

Let $A\subseteq [0,1]^d$, $d\geq 2$, a set with Hausdorff dimension $\operatorname{dim}_{\mathcal{H}}A=s$. What is the minimum $s$ (if any) which guarantee that $A$ has non-empty intersections with a ...
RaffaeleScandone's user avatar
6 votes
1 answer
374 views

Does finite Hausdorff dimension imply finite packing dimension?

In other words, does there exist a metric space $(E,\rho)$ with finite Hausdorff dimension but infinite packing dimension? Here are my thoughts: I know that it is generally hard to relate Hausdorff ...
Peter Koepernik's user avatar
3 votes
2 answers
219 views

Hausdorff dimension of the non-differentiability set a convex function

Let $X \subset \mathbb R^d$ be open, $f : X \to \mathbb R$ and $$ E := \{x \in X : f \text{ is not Fréchet differentiable at }x\}. $$ Then we have the following result which is Theorem: If $X= \...
Akira's user avatar
  • 749
1 vote
0 answers
92 views

Lower bound estimate for the sum $\sum \text{diam}(U)^d$ over all countable covers of a cube

This question is inspired from the definition of Hausdorff measure. Let $C$ be a closed unit hypercube in $\mathbb R^d$ (side length equal to one, including boundary. The cube itself is at top ...
No One's user avatar
  • 1,545
6 votes
1 answer
461 views

Direct proof that the set of badly approximable numbers have full Hausdorff dimension without using Schmidt games

A badly approximable number is an $x$ for which there is a positive constant $c$ such that for all rational $p/q$ we have $$\left|{ x - \frac{p}{q} }\right| > \frac{c}{q^2} \ . $$ The set of badly ...
No One's user avatar
  • 1,545
4 votes
1 answer
174 views

On the set on which $|Df|$ is maximal for Lipschitz $f$

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$. That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^d$. Question: ...
Nate River's user avatar
  • 4,802
5 votes
1 answer
291 views

Hausdorff measure

Let $(X,d)$ be a metric space and let $H^\alpha$ denote the $\alpha$-dimensional Hausdorff measure on $X$, where $\alpha$ is the Hausdorff dimension of $X$. Is there any simple condition on $X$ that ...
Wreck it Ralph's user avatar
1 vote
0 answers
126 views

Hausdorff dimension of a compact Lie group [closed]

Let $G$ be a compact Lie group (for simplicity assume $G=SO(3)$). Equip $G$ with a left-invariant Riemannian metric and let $m$ be left-invariant Haar measure on $G$. Now that $G$ is a metric space ...
Adam's user avatar
  • 323
2 votes
1 answer
354 views

Fraction dimensional "Euclidean" space

The “dimension” of Euclidean space $\mathbb{E}^n$ can be explained as an algebraic property, simply as a dimension of a vector space over the field $\mathbb{R}$. It also can be understood as a ...
Kacper Kurowski's user avatar
2 votes
0 answers
169 views

Relationship between Hausdorff dimension and covering number

Let $(X,d)$ be a compact metric space and recall that the $\epsilon$-external covering number $\mathcal{N}^{\epsilon}(X)$ of $X$ is defined by: $$ \mathcal{N}^{\epsilon}(X) := \inf\left\{ N\in \mathbb{...
ABIM's user avatar
  • 4,989
7 votes
1 answer
710 views

Fubini's theorem for Hausdorff measures

$B\subset \mathbb{R}^2$ is a Borel set. Define the slices $B_x:= \{y \in \mathbb{R}: (x,y) \in B \}$. If $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$, presentations of Fubini's theorem often ...
Chertopkhanov on Malek Adel's user avatar
4 votes
1 answer
233 views

Hausdorff dimension and critical exponent of words

What is the Hausdorff dimension of the subset $S_c \subset [0,1]$ of points such that the critical exponent of their binary expansion is $c$? It's clear that $\dim_H S_{\infty}=1$, but what can be ...
Alessandro Della Corte's user avatar
3 votes
2 answers
299 views

Maximal Hausdorff dimension of the set on which derivatives do not agree

Let $f, g: [0, 1] \to \mathbb R$ be functions that are differentiable a.e. with $f’ = g’$ almost everywhere. What is the maximal Hausdorff dimension $d$ (and corresponding Hausdorff $d$-measure) of ...
Nate River's user avatar
  • 4,802
4 votes
2 answers
362 views

Hausdorff dimension of Julia set

Can anyone show me the proof "Hausdorff dimension of Julia set is strictly positive"? For purpose to prove this we might have to prove the green function of basin of attraction to infinity ...
matthew's user avatar
  • 51
4 votes
1 answer
713 views

Hausdorff dimension and surface measure

Could someone please indicate me some reference that contains the proof of the following theorem? Below $\mathcal{H}^n$ denotes the $n$-dimensional Hausdorff outer measure in $\mathbb{R}^n$. Theorem: ...
rfloc's user avatar
  • 463
5 votes
0 answers
575 views

On the Hausdorff dimension of a Cantor set

In what follows I refer to this paper by Orevkov. I am writing a paper on this, so if somebody is interested we could consider to write a joint paper. Consider a sequence $R=\{R_n\}_n$ of strictly ...
Joe's user avatar
  • 759
4 votes
1 answer
262 views

The Hausdorff codimension of singular matrices vs. the Hausdorff codimension of points with divergent trajectories

Let $G:=SL(m+n,\mathbb R)$ and $\Gamma :=SL(m+n,\mathbb Z)$ and $X:=G/\Gamma$. (1) Let $M$ denote the set of all $m \times n$ matrices with real entries. A matrix $A \in M$ is called $\textit{singular}...
No One's user avatar
  • 1,545
2 votes
0 answers
216 views

Is Kakeya conjecture open with some additional regularity condition on Kakeya map?

in the paper, ON KAKEYA MAPS WITH REGULARITY ASSUMPTIONS, the author of the paper consider the Kakeya conjecture with some regularity on Kakeya map, Kekeya conjecture[Hausdorff dimension version] If ...
katago's user avatar
  • 543
2 votes
1 answer
142 views

Natural way to thicken Brownian motion to 2D?

If we have a smooth plane curve (Hausdorff dimension 1), we can thicken it by a small amount to get a 2D set (all points within distance $\epsilon$ to the curve). What if we start with the graph of a ...
Geoffrey Irving's user avatar
2 votes
0 answers
88 views

Bounds on convergence of two orbits in the limit set of a Schottky group

Suppose we have two points in the limit set of a Schottky group, $x,y\in \Lambda(\Gamma)$. Consider the orbits of those points under a primitive subset $\Gamma'$ of $\Gamma,$ that is, one that does ...
user6419's user avatar
  • 431
3 votes
0 answers
91 views

For $\mathcal{L}^1$-a.e. $t\in R$, Hausdorff dimension of level sets of a locally Lipschitz function $f:R^n\to R$ is $n-1$?

Let $f:R^n\to R$ be a locally Lipchitz function. Denote $H^n$ the n-dimensional Hausdorff measure. We know that for any $H^n$-measurable subset $A\subset R^n$, for $\mathcal{L}^1$-a.e. $t\in R$, $A\...
mathmetricgeometry's user avatar
2 votes
0 answers
46 views

(1, 2) stability and Hausdorff dimension

Recall that a set $E \subset \mathbb{R}^n$ is called (1,2) stable if it satisfies the following: $$ W_0^{1,2}(E) = W_0^{1,2}(E^0), $$ where $E^0$ is the interior of $E$ and $W_0^{1,2}(E)$ is defined ...
user109433's user avatar
4 votes
2 answers
1k views

A set whose Hausdorff dimension gradually changes?

Can there be a set whose Hausdorff dimension gradually changes? For instance, a set of real numbers contained in an interval, whose Hausdorff dimension is 0 at the beginning and 1 closer to the end, ...
Anixx's user avatar
  • 9,316
2 votes
0 answers
93 views

Hausdorff dimension between $(1,2)$

Is there a number $c \in (1,2)$ for which there exist some interesting geometric property/properties which hold for every set of Hausdorff dimension in $(1,c)$ and does not hold for any set of ...
truebaran's user avatar
  • 9,140
8 votes
1 answer
184 views

How do sets with unit fractional Hausdorff measure of dimension $>1$ look like?

Triggered by the recent question How can we not know the measure of the Sierpiński triangle? I would like to ask: Let $s>1$ and $s$ not be an integer. How to construct a set $A$ with $\mathfrak{H}^...
Dirk's user avatar
  • 12.3k
6 votes
1 answer
202 views

Subspaces of metric spaces having prescribed dimension

Let $(X,d)$ be a metric space having Hausdorff dimension $\alpha>0$ and let $0<\beta<\alpha$. Is there a metric subspace of $X$ having Hausdorff dimension $\beta$?
Nicola Arcozzi's user avatar
2 votes
0 answers
207 views

Examples of essentially sub-linear functions

A dimension function is an increasing, continuous function $% f:\mathbb R_{+}\rightarrow \mathbb R_{+}$ such that $f(r)\to 0$ as $r\to 0$. Say that a dimension function $f$ is essentially sub-linear ...
user48633's user avatar
1 vote
1 answer
201 views

Does fractallity depend on the Riemannian metric?

Edit: According to comment of Andre Henriques we revise the question: In this question a fractal is a metric space whose topological and Hausdorff dimensions are different. So we would like that ...
Ali Taghavi's user avatar
11 votes
0 answers
314 views

Hausdorff dimension and von Neumann dimension

There are two subjects in which non-integral dimensions appear: fractal geometry: consider the well-known Hausdorff dimension of fractals. von Neumann algebra: consider a type ${\rm II_1}$ ...
Sebastien Palcoux's user avatar
2 votes
0 answers
65 views

Universal structure of fractal spaces

In the same way that we can say manifolds are made of pieces that look like $\mathbb{R}^n$, is there any way to say that spaces with the same hausdorff dimension are made up of pieces that look the ...
QCD_IS_GOOD's user avatar
3 votes
1 answer
217 views

Hausdorff dimension and $W^{1,1}$ functions

What can be said about the Hausdorff dimension of the image of a set by a $W^{1,1}$ map? In other words, what is the relationship between $\mathrm{dim}_H f(A)$ and $\mathrm{dim}_H A$, where $f \in ...
Riku's user avatar
  • 819
7 votes
1 answer
256 views

Hausdorff dimension of the boundary of fibres of Lipschitz maps

Let $f: \mathbb{R}^m\rightarrow \mathbb{R}^{m-k}$ be a Lipschitz map. Can we get a uniform estimate on the Hausdorff dimension of the boundaries of fibres of $f$? I.e. do we have an upper bound for ...
Severin Schraven's user avatar
23 votes
3 answers
1k views

Existence of subset with given Hausdorff dimension

Let $A\subseteq \mathbb{R}$ be Lebesgue-measurable and let $0<\alpha<1$ be its Hausdorff dimension. For a given $0<\beta <\alpha$ can we find a subset $B\subset A$ with Hausdorff ...
Severin Schraven's user avatar