# Tag Info

26

Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$. Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$. This implies that for all $\epsilon < 1/2$, we have that $$\mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0))$$ where $\mu$ is ...

11

A thin version of the usual Cantor middle third set works. The idea is just that you need to omit more than just a third of the remaining intervals as the construction proceeds, enough so as to force the Hausdorff dimension to $0$. Specifically, we construct the set in stages. At each stage, we've omitted a "middle third" from each finite interval ...

11

Negative dimension is actually much easier to talk about than complex dimension. Super vector spaces are a natural collection of objects that can have negative dimension; given a super vector space $(V_0, V_1)$ we can define its dimension to be $\dim V_0 - \dim V_1$, and this definition has many nice properties; see this blog post, for example. More ...

8

We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller. So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$. Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose ...

8

No. Just for good old $\mathbb R$, there is a homeomorphism $[0,1]\to[0,1]$ which takes the middle-thirds Cantor set (of measure $0$) to the "fat" Smith-Volterra-Cantor set (of measure $1/2$). (The excluded intervals in the two constructions, including their endpoints, correspond naturally to each other -- simply map each of them to its counterpart ...

8

I was going to make this a comment but it occurred to me there might be sufficient interest that perhaps I should not bury it in a comment. At the beginning of the paper below Erdős gives a short proof (that he attributes to Tibor Radó) making use of the Lebesgue density theorem that $E_r$ has Lebesgue measure zero, where $E$ is a closed set in ${\mathbb ... 7 You only need to consider the case$\mathfrak{h}_{s}^\ast(A) \lt \infty$, but you need to be a bit careful in choosing the outer approximations since swapping$\inf$and$\sup$certainly isn't allowed without some thinking. If you knew that you can always take the same set$E$in the$\inf$(which I will show in \eqref{eqn:ast} below) then you'd be ... 6 One notion of complex dimension that has been used extensively has to do with self-similar sets. A$t$-neighborhood (i.e. points within distance$t$) of such a set may have volume$v(t)$bounded above and below by constant multiples of$t^d$, where$d$is the dimension of the boundary and$t$is small, but such that$t^{-d} v(t)$is oscillatory and ... 5 Algebraic stacks are a far-reaching generalization of algebraic varieties. If an algebraic variety is considered as a stack, then its dimension as stack is the same as its dimension as variety. However there are many stacks that do not correspond to varieties, and some of these have negative dimension. Specifically, if$V$is a variety and$G$is an ... 5 I just checked the book and I found out that the author defines the$m$-dimensional density of a set$E\subset \mathbb{R}^n$near a point$x$for$m\le n$by $$\lim_{r\to 0} \frac{\mathcal{H}^m(E\cap B^n(x,r))}{|B^m(0,r)|}$$ Note that in the numerator, the$n$-dimensional ball appears, whereas in the denominator it's the$m$-dimensional ball! For$m=n$the ... 5 A good anotated list of textbooks on geometric measure theory can be found in this blog post. Besides comments on Federer and Mattila it has several more examples. As my personal favorite I found, while lecturing geometric measure theory, "Measure Theory and Fine Properties of Functions" by Evans and Gariepy. It is short and crisp (often you have to build ... 4 In case you are still interested in an answer, I encourage you to look at Leon Simon's notes. The idea of currents, is to serve as a far reaching generalization of oriented submanifolds. They were introduced as part of the fairly long and involved program that went into solving the Plateau's problem: such a minimization problem would require three important ... 4 As felipeh noted, the problem reduces to$p=1$by replacing$f$with$g = |f|^p$. (The case$p=\infty$should be treated separately.) Also, the annulus can be replaced by the sphere$S^{n-1}$. Indeed, define a function$h$on the unit sphere by$h(\xi) = \int_r^1 g(t\xi)\,dt$. Then$h\in L^1(S^{n-1})$, with a norm comparable to$\|g\|_{L^1(A)}$because ... 4 As Henning says, whether or not you have a Banach-Tarski paradox depends on what symmetries of the Cantor set you allow. My understanding is that you want to allow all measure preserving continuous bijections. With this understanding, the Cantor has a Banach-Tarski decomposition. Note that we can identify the Cantor set with the$2$-adic integers, ... 4 Be advised that Federer's book is not for newcomers to the subject (and it's not an online reference anyway). An accessible introduction to isoperimetric and related inequalities is The Brunn-Minkowski inequality by Gardner. It focuses on the volume of subsets of$\mathbb R^n$, thus avoiding most of the GMT machinery. A somewhat dated, but still good, ... 4 Assume we have a Jordan curve of positive area measure whose complement has connected components$U,V$as in the Jordan Curve theorem. Let$U$be the bounded component. Every point in$\partial U = \partial V $is the limit of a sequence of points in$\overline U $and the limit of a sequence of points in$V= \overline U ^c.$Thus the boundary of$\overline ...

4

The two measures are within a constant factor of each other. In K. Falconer, Fractal Geometry: Mathematical Foundations and Applications, see the material on "net measure". In the 2nd edition, that is page 36

4

A $k-$dimensional subspace $S$ of $\mathbb R^n$, with $k<n$, is the range of linear transformation of rank $k$. That means that there exists a matrix $A\in\mathbb{R}^{n\times n}$, with $\mathrm{rank}(A)=k<n$, such that $\mathrm{Ran}(A)=S$. But then $\det A=0$. The theorem of change of variables $$m(S)=\int_{\mathbb R^n}\lvert\det A\rvert\,dx=0.$$ ...

4

This is false even for curves in the plane. Consider a sequence of ellipses converging to a line segment. The length of the limit curve drops by the factor of 2. If you want an example of curves without boundary, consider a similar example in 3d space. Or consider curves which are disjoint unions of pairs of concentric circles converging to a single circle. ...

4

It is, at least partially, a chicken-and-egg problem. We'd like to define the Lebesgue integral which has nicer convergence properties when we integrate sequences of functions, and take limits of the functions and integrals -- but the definition of the Lebesgue integral requires that we already have a well-behaved measure (i.e., "area") concept to build it ...

3

The problem is poorly written. There are no nontrivial finite doubling measures on $\mathbb{R}$. Indeed, if $\mu(\mathbb{R})<\infty$, then there is an interval $I$ with $\mu(I)>(1-\epsilon)\mu(\mathbb{R})$. An adjacent interval $J$ of equal length will have $\mu(J)<\epsilon\mu(\mathbb{R})$, and $(1-\epsilon)/\epsilon$ can be arbitrarily large. If ...

3

Interesting question! The Hausdorff dimension is defined as an infimum over a subset of the positive reals. That means that it's positive and real by definition, so if you're looking for a way for a space to have negative or complex dimension, you're either going to have to generalize the Hausdorff dimension in some way or use some other definition of ...

3

I thought about allowing a negative value of $d$ in the definition given in the article in the OP. My definitions do not extend Hausdorff dimension, but considers an alternate interpretation of dimension itself as nonpositive numbers rather than nonnegative numbers. It seems to me that the definitions of Hausdorff dimension offer little leeway into ...

3

Of course most of these don't transform $A$ into a ball at once, but can be used to get there in the limit. I would add spherical symmetrization to the list; slicing $A$ by concentric spheres $S_r$, then rearranging each $A\cap S_r$ into a spherical cap (say, centered at North pole) of the same spherical measure. This is a generalization of classical ...

3

In practice, I prefer the (equivalent) following definition of upper Minkowski dimension: $$\overline{\dim}(E) = \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log(1/\delta)}.$$ $N(E, \delta)$ is the number of $\delta$-balls needed to cover the set $E$. (We need $E$ to be compact). (Throughout, we write $X \approx Y$ if there is a constant $c\geq 1$ ...

3

Obviously, the square $[0,1]$ is the union of uncountably infinitely many line segments, since $$[0,1] = \bigcup_{x\in[0,1]}(\{x\}\times[0,1])$$ A more interesting question is "can I do it with fewer line segments?" What you are referring to as "area" is mathematically formalized into measure theory, as you correctly suspect. The "standard" measure of sets ...

3

Edit: In this case, the boundary of the bounded component $U$ has positive, finite measure, but the question was asking about the closure of the open set. I'll think about this more. Your thinking is in fact correct for the reason you mention: there is a Jordan Curve in the plane with positive measure. See here. Then, by the Jordan Curve Theorem, there ...

3

Roughly speaking, the answer is "because you have a (bi)cone". Any smooth embedded manifold (in fact any $C^2$ manifold) will have the property that the appropriately defined density is precisely equal to $1$. But if you have something that is not smooth, you can have other values. If you have just a single cone $z = \sqrt{x^2 + y^2}$, near the origin ...

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