Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options answers only not deleted user 454

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

2 votes

Fractal sets and dimensions

Suppose we construct sets $A,B,C$ with $A \subseteq [0,1]$, $\dim_\mathrm{H}(A) = 1/5$ and $\dim_\mathrm{P}(A) = 4/5$, $B \subseteq [2,3]$, $\dim_\mathrm{H}(B) = 1/5$ and $\dim_\mathrm{P}(B) = 2/5$, $ …
Gerald Edgar's user avatar
  • 41.1k
7 votes
Accepted

Does finite Hausdorff dimension imply finite packing dimension?

A construction used (repeatedy) in the paper Edgar, G. A., Centered densities and fractal measures, New York J. Math. 13, 33-87 (2007). ZBL1112.28004. For more information, see that paper. We constru …
Gerald Edgar's user avatar
  • 41.1k
7 votes
Accepted

Hausdorff measure

Not true for all $X$. Taking the gauge function $\phi(t) = t^{1/2}/\log|t|$, construct a Cantor set $X$ in $\mathbb R$ using Hausdorff's original method so that $H^\phi(X) = 1$. Then the Hausdorff d …
Gerald Edgar's user avatar
  • 41.1k
10 votes

Fractals of dimension zero

One example: the set of Liouville numbers has Hausdorff dimension zero. In number theory, a Liouville number is an irrational number $x$ with the property that, for every positive integer $n$, the …
Gerald Edgar's user avatar
  • 41.1k
4 votes

Construction of null sets with prescribed Hausdorff dimension and generalizations

The first reference is Hausdorff's paper, "Dimension und Äußeres Maß" He constructs a Cantor set $E \subseteq [0,1]$ with $0 < \mathcal H^h(E) < +\infty$ for any sensible $h$. So if $0 < s < 1$ this …
Gerald Edgar's user avatar
  • 41.1k
4 votes
Accepted

Hausdorff dimension of sequence space

According to Falconer[1] this is due to Besicovitch[2]. Falconer states it (generalized to $\mathbb R^n$) as Theorem 5.1, p. 65. This proves more than just $X$ and $\pi(X)$ have the same Hausdorff d …
Gerald Edgar's user avatar
  • 41.1k
3 votes

When is Hausdorff measure a Frostman measure?

K. Falconer, Fractal Geometry: Mathematical Foundations and Applications (3rd ed, 2014, Wiley). page 77 Corollary 4.12 Let $F$ be a Borel subset of $\mathbb R^n$ with $0 < \mathcal H^s(F) \le \inf …
Gerald Edgar's user avatar
  • 41.1k
2 votes

Quantitative measurement of infinite dimensionality

See my paper LINK Centered densities and fractal measures, New York Journal of Mathematics 13 (2007) 33-87 Some references are also at the end of it. In particular, Boardman, Goodey, and McClure.
Gerald Edgar's user avatar
  • 41.1k
17 votes

Fractal questions: Weierstraß-Mandelbrot

My question is whether there are nowhere differentiable continuous functions (between real numbers) whose graph are not fractals? Of course this depends on your definition of fractal. There are no …
Gerald Edgar's user avatar
  • 41.1k
7 votes

Simple definition of the Hausdorff measure using squared paper

As Charles said, this would be the box-counting dimension (a.k.a. Bouligand dimension or Minkowski dimension). Not the Hausdorff dimension. And for the measure it would usually not converge so that …
Gerald Edgar's user avatar
  • 41.1k
15 votes
Accepted

How big can the Hausdorff dimension of a function graph get?

The answer is 2. Besicovitch and Ursell, Sets of fractional dimensions (V): On dimensional numbers of some continuous curves. J. London Math. Soc. 12 (1937) 18–25. doi:10.1112/jlms/s1-12.45.18
Gerald Edgar's user avatar
  • 41.1k
6 votes
Accepted

Hausdorff dimension vs. cardinality

As stated, countable sets have Hausdorff dimension 0. So any set $S$ with $\mathrm{HD}(S)>0$ has power $\ge \aleph_1$. No need for continuum hypothesis. Without CH, though, we cannot say whether powe …
Gerald Edgar's user avatar
  • 41.1k