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Recall that

  1. the Hausdorff distance (metric) is:

    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 $[0,1]^2$.

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2 Answers 2

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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).

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  • $\begingroup$ 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. $\endgroup$
    – Xander Henderson
    Jan 8, 2021 at 14:58
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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.

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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 $\log_3(4)$.

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