# convergenge of Hausdorff dimension for a sequence of compact sets

Recall that

1. A metric in the space of subsets of a compact set $$K$$, defined as follows. Let $$X,Y\subset K$$ and let $$D_{x,y}$$ be the set of all numbers $$\rho(x,Y)$$ and $$\rho(y,X)$$ where $$x\in X$$, $$y\in Y$$ and $$\rho$$ is a metric in $$K$$. Then the Hausdorff metric $$\operatorname{dist}(X,Y)$$ is the least upper bound of the numbers in $$D_{x,y}$$.

2. the Hausdorff dimension is: Let $$(X,d)$$ be a metric space. In what follows, for any subset $$E\subset X$$, $$\operatorname{diam}(E)$$ will denote the diameter of $$E$$.

For any $$E\subset X$$, any $$\delta \in ]0, \infty]$$ and any $$\alpha\in [0, \infty[$$ we consider the outer measure $$\mathcal{H}^\alpha_\delta (E) := \inf \left\{ \sum_{i=1}^\infty (\operatorname{[diam}\, E_i)^\alpha : E\subset \bigcup_i E_i \quad\text{and}\quad \operatorname{ diam}(E_i)< \delta\right\}.$$ The map $$\delta\mapsto \mathcal{H}^\alpha_\delta (E)$$ is monotone nonincreasing and thus we can define the Hausdorff measure $$\alpha$$-dimensional measure of $$E$$ as
$$\mathcal{H}^\alpha (E) := \lim_{\delta\downarrow 0} \mathcal{H}^\alpha_\delta (E)$$ The map $$\delta\mapsto \mathcal{H}^\alpha_\delta (E)$$ is monotone nonincreasing and thus we can define the Hausdorff α-dimensional measure of E as $$\mathcal{H}^\alpha (E) := \lim_{\delta\downarrow 0} \mathcal{H}^\alpha_\delta (E).$$

I wonder if the following is true: If $$A_n \to A$$ in the Hausdorff distance, then $$\operatorname{dim}_H A_n \to \operatorname{dim}_H A$$, where $$A$$ is a compact subset of $$[0,1]^2$$.

Unfortunately this is not correct. Let us define $$A_n:=\{\frac{k}{n}|\ k=0,1,\ldots n\}\subset [0,1].$$ Then $$A_n$$ converges in the Hausdorff sense to $$[0,1]$$. Let us proof this claim: Let $$x\in[0,1)$$. Then for all $$n\in\mathbb{N}$$ exists a $$k\in\{0,1,\ldots,n\}$$ such that $$\frac{k}{n}\leq x < \frac{k+1}{n}.$$ Hence $$\rho(x,A_n)\leq \min(|x-\frac{k}{n}|,|x-\frac{k+1}{n}|)\leq |\frac{k+1}{n}-\frac{k}{n}|=\frac{1}{n}.$$ Therefore $$dist_{\mathcal{H}}(A_n,A)\leq \frac{1}{n}\rightarrow 0\mbox{ for }n\rightarrow\infty.$$

The Hausdorffdimension of the $$A_n$$ is zero, because $$A_n$$ consists of finitely many points, while the Hausdorffdimension of $$[0,1]$$ is 1, because $$\mathcal{H}^1([0,1])=\mathcal{L}^1([0,1])=1$$ (Here $$\mathcal{L}^1$$ is the one dimensional Lebesgue mesaure).

• Your statements about convergence are a little imprecise. As sets, $A_n \to \mathbb{Q} \cap [0,1]$. Note that this convergence is with respect to the Hausdorff metric on the powerset of $\mathbb{R}$ (the Hausdorff metric is not a metric on this space, but this is the naively obvious way to interpret the given definition). This limit set is not compact, and has Hausdorff dimension $0$. On the other hand, if we understand the Hausdorff metric to be a metric on the space (equivalence classes of) compact subsets of $[0,1]$, then the result holds. Jan 8, 2021 at 14:58

The conjectured result does not hold.

For $$j=1,2,$$ define the maps $$\varphi_j : [0,1] \to [0,1]$$ by $$\varphi_1(x) = \frac{1}{3}x \qquad\text{and}\qquad \varphi_2(x) = \frac{1}{3}x + \frac{2}{3}.$$ Let $$A_0 = [0,1]$$, and recursively define $$A_n$$ by $$A_n = \varphi_1(A_{n-1}) \cup \varphi_2(A_{n-1}).$$ Thus, working out the first few sets in the sequence: \begin{align} A_0 &= [0,1] \\ A_1 &= [0,1/3] \cup [2/3,1] \\ A_2 &= [0,1/9] \cup [2/9, 1/3] \cup [2/3,7/9] \cup [8/9,1], \end{align} and so on. In general, the set $$A_n$$ will consist of $$2^n$$ closed intervals, each having length $$3^{-n}$$. As each $$A_n$$ is a finite union of compact sets, each $$A_n$$ is, itself, compact.

Moreover, it can be shown that $$A_n \to A$$ in the Hausdorff metric, where $$A$$ is the usual two-thirds Cantor set (a proof of this follows by noting that the map of sets defined by $$\Phi(E) = \varphi_1(E) \cup \varphi_2(E)$$ is a contraction mapping on the space of compact subsets of $$\mathbb{R}$$ with respect to the Hausdorff distance; the claimed result then follows from the Banach fixed point theorem; see Hutchinson's 1981 paper for details).

Now, note that $$\operatorname{dim}_H(A_n) = 1$$ for each $$n$$ (where $$\operatorname{dim}_H$$ denotes the Hausdorff dimension)—that intervals are one-dimensional is a fairly standard result. On the other hand, $$\operatorname{dim}_H(A) = \log_3(2)$$—again, I would refer you to Hutchinson's 1981 paper for a proof of this result, though, again, this is a fairly standard result. Therefore $$A_n \overset{d_H}{\longrightarrow} A \qquad\text{but}\qquad 1 = \operatorname{dim}_H(A_n) \not\to \operatorname{dim}(A) = \log_3(2).$$

Note: I am working with subsets of $$[0,1]$$. However, we can regard these sets as subsets of $$[0,1]^2$$ with no modifications. Alternatively, we might consider a similar setup with four maps $$\varphi_j : [0,1]^2 \to [0,1]^2$$ defined by \begin{align} \varphi_1(x) &= \frac{1}{3} x, & \varphi_2(x) &= \frac{1}{3} x + \begin{pmatrix} \frac{2}{3} \\ 0 \end{pmatrix}, \\ \varphi_3(x) &= \frac{1}{3} x + \begin{pmatrix} 0 \\ \frac{2}{3}\end{pmatrix}, & \varphi_4(x) &= \frac{1}{3} x + \begin{pmatrix} \frac{2}{3} \\ \frac{2}{3} \end{pmatrix}. \\ \end{align} In this case, take $$A_0 = [0,1]^2$$ and $$A_n = \bigcup_{j=1}^{4} \varphi_j(A_{n-1}).$$ The approximating sets have Hausdorff dimension $$2$$, while the limiting set is a two-dimensional Cantor dust, which has Hausdorff dimension $$\log_3(4)$$.