22 votes
Accepted

Dimension in CW-approximation

Barratt and Milnor (An Example of Anomalous Singular Homology) proved that (for $n > 1$) the singular homology of the union of countably many $n$-spheres with one point in common and radii tending ...
Karol Szumiło's user avatar
21 votes
Accepted

Why do almost all points in the unit interval have Kolmogorov complexity 1?

I'm not an expert on Kolmogorov complexity, but this does seem like a counting argument: for any fixed $\epsilon > 0$, there are only $\sum_{i = 1}^{(1-\epsilon)n} 2^i < 2^{(1-\epsilon)n+1}$ ...
Ronnie Pavlov's user avatar
20 votes
Accepted

Two definitions of Lebesgue covering dimension

As you refer to Engelking's "Dimension theory" book, I suppose you know the following two statements, but anyway, here they are: The notions agree for separable metric spaces, by Exercise 1.7.E of ...
Matthias Wendt's user avatar
20 votes
Accepted

A set whose Hausdorff dimension gradually changes?

I assume you want a set $A\subseteq [0,1]$ such that $\dim (A\cap [0,x])=x$ for all $x$. We can define $A_1$ by taking the union of a (Borel) subset of dimension $0$ of $[0,1/2]$ with a subset of ...
Christian Remling's user avatar
14 votes
Accepted

Existence of subset with given Hausdorff dimension

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 ...
Skeeve's user avatar
  • 1,277
14 votes
Accepted

Is the complement of a zero-dimensional subset of the plane path-connected?

If the zero-dimensional set $X$ is not closed, then the answer is "no". To construct a suitable example, take any open bounded neighborhood $U\subset\mathbb R^2$ of zero, whose boundary $\partial U$ ...
Taras Banakh's user avatar
  • 40.8k
12 votes
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Hausdorff dimension of the graph of an increasing function

Theorem 1. The Hausdorff dimension of the graph $\Gamma_f$ of $f$ equals $1$. Proof. Take a partition of $[0,1]$ by intervals of length $1/n$. Since the function is increasing you can cover the graph ...
Piotr Hajlasz's user avatar
12 votes
Accepted

If $\text{dim}(X \times X) = 2\text{dim}(X)$, does $\text{dim}(X^n) = n\text{dim}(X)$?

As John Samples noted in his comment, Dranishnikov's Theory of cohomological dimension implies the positive answer to this problem for compact (even $\sigma$-compact) metrizable spaces. Namely, ...
Taras Banakh's user avatar
  • 40.8k
12 votes

Existence of subset with given Hausdorff dimension

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 ...
Piotr Hajlasz's user avatar
12 votes

Is the complement of a zero-dimensional subset of the plane path-connected?

At least if $X$ is compact, the answer is yes. Indeed, by Corollary 2 of Theorem IV 3 in: W. Hurewicz, H. Wallman, Dimension Theory. Princeton Mathematical Series, v. 4. Princeton University Press, ...
Piotr Hajlasz's user avatar
12 votes
Accepted

Unknown work of Nöbeling on topological/Hausdorff dimension

So, the sought for paper is: Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931). And here is a ``translation" (to English and to modern ...
Behnam Esmayli's user avatar
10 votes

Box dimension of the set of Pisot numbers?

The set $\mathcal P$ of all Pisot numbers is known to be closed (Salem). Its limit points $\mathcal P'$ are also known in $(1,2)$ (Talmoudi). The smallest element of $\mathcal P'$ is the golden ratio $...
Nikita Sidorov's user avatar
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 $n$, ...
Gerald Edgar's user avatar
  • 40.2k
9 votes
Accepted

Lowest Dimension for Counterexample in Topological Manifold Factorization

As asked Dusan Repovs (who is an expert in the theory of topological manifolds), and he sent me the following answer: This is indeed best possible result, since whenever a product of two spaces is a ...
Taras Banakh's user avatar
  • 40.8k
8 votes

Fractals of dimension zero

If by a fractal you mean a self-similar set and if the corresponding iterated function system (IFS) satisfies the Open Set Condition (OSC), then the answer is no. The Moran-Hutchinson formula gives ...
Nikita Sidorov's user avatar
8 votes

Unknown work of Nöbeling on topological/Hausdorff dimension

It seems to be: Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931). ZBL57.0749.02. Google shows the first and sporadically the second page ...
Francois Ziegler's user avatar
8 votes
Accepted

Sequences with 0's in $\mathbb R ^\omega$

For every $n\in\mathbb N$ consider the space $$X_n=\{x\in\mathbb R^\omega:\lvert x^{-1}(0)\rvert\ge n\}.$$ Theorem. For any positive integer numbers $n<m$, the spaces $X_n$ and $X_m$ are not ...
Taras Banakh's user avatar
  • 40.8k
7 votes

Haar measure on the Grassmannian space

The Haar measure is the volume induced by the unique (up to scalar) $O(n)$ invariant Riemannian metric on $G(n,m)$, and any volume induced by a Riemannian metric has this property (since the metric ...
Ben Webster's user avatar
  • 43.9k
7 votes

Ordinal vs. cardinal dimension

A classic text is Dimension Theory by Hurevicz & Wallman (1941) Among the types of dimension for topological spaces are the "upper inductive dimension" and the "lower inductive dimension". ...
Gerald Edgar's user avatar
  • 40.2k
7 votes
Accepted

Transitive homeomorphisms of Erdős spaces

The answer to both questions is affirmative. Theorem 1. The complete Erdos space $\mathfrak E_c$ has a self-homeomorphism whose every orbit is dense in $\mathfrak E_c$. Proof. We use a known result ...
Taras Banakh's user avatar
  • 40.8k
7 votes

Dimension of a topological space equals the supremum of the dimension of open subsets in an open cover

We want to show that if $X$ has a chain $\varnothing\neq Z_0\subsetneq\dots\subsetneq Z_n$ of irreducible subsets, then some $U_i$ contains a chain of irreducible subsets of the same length. Since $...
Wojowu's user avatar
  • 27.4k
7 votes

What is special to dimension 8?

Some special properties of dimension 8, in addition to the ones you identify: Bernstein's problem holds up to dimension $n=8$. The only function of $\mathbb{R}^{n-1}$ whose graph in $\mathbb{R}^n$ ...
7 votes

What is special to dimension 8?

The following is too long to leave as a comment, so I post it as an answer. At least for the cases of the Bernstein problem for minimal graphs, Simons' theorem on stable minimal cones, etc., there is ...
7 votes

Fraction dimensional "Euclidean" space

One remark: there is no metric space $\mathbb{E}^{1/2}$ of Hausdorff dimension $\frac{1}{2}$ whose topological square $\mathbb{E}^{1/2} \times \mathbb{E}^{1/2}$ is homeomorphic to $\mathbb{E}^1$, i.e.,...
user142382's user avatar
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 ...
Gerald Edgar's user avatar
  • 40.2k
6 votes
Accepted

Can a smooth diffeomorphism of a Riemannian manifold have only positive Lyapunov exponents on a large set?

The answer is no. In fact, the following result holds: if $M$ is a compact manifold, $f\in\mathrm{Diff}^{1+\alpha}(M)$ and $\mu$ is an ergodic $f$-invariant probability measure such that all its ...
Alejandro's user avatar
  • 940
6 votes
Accepted

Is there a gap between the Hausdorff and the lower Minkowski dimensions?

I believe what you are asking is whether the Hausdorff dimension can be strictly less than the lower packing dimension (also called the lower modified box dimension). I'm pretty sure the strongest ...
Dave L Renfro's user avatar
6 votes

Fractals of dimension zero

Closed uncountable sets of Hausdorff dimension zero turn up from time to time in practice (one of my papers with Nikita Sidorov features one) but I think that there does not exist a set which is ...
Ian Morris's user avatar
  • 6,176
6 votes
Accepted

Doubling dimension vs other metric dimensions

In one direction, a rapidly branching tree will have very high doubling dimension, while having topological dimension $0$ (or $1$, if you include the edges). In another direction there is a bound, and ...
Igor Rivin's user avatar
  • 95.5k
6 votes
Accepted

Real rank 0 implies stable rank 1 on $C^\ast$-algebras?

There are "cheap" examples. If a C*-algebra $A$ has stable rank 1 then it is stably finite. On the other hand every simple, unital purely infinite C*-algebra has real rank 0. So every simple, ...
Caleb Eckhardt's user avatar

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