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Let $(X,d)$ be a complete metric space and let $f_1,\ldots,f_n$ be contractions with Lipschitz constants $q_i$. Then a unique non-empty compact set exists such that $K=\bigcup_{i=1}^n f_i(K)$.

Now the way the proof usually begins is by proving that the so-called Hutchinson operator, $F:\mathcal{K}(X)\to\mathcal{K}(X)$ is contractive, that is, for every $A, B\in\mathcal{K}(X)$

$$d_H(F(A),F(B))=d_H(\bigcup_{i=1}^n f_i(A),\bigcup_{i=1}^n f_i(B))\leq\max d_H(f_i(A),f_i(B))\leq\max q_i\cdot d_H(A,B).$$

Can someone please explain why the middle inequality holds? It's obviously true for different indices $i,j$, but I can't figure out why it's the case for the same index $i$?

Alp Uzman
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Dave the Sid
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1 Answers1

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By induction it suffices to show, for $X$ a metric space and $A_1,A_2,B_1,B_2$ four nonempty compact subsets of $X$,

$$d_{\mathcal{H}(X)}(A_1\cup A_2, B_1\cup B_2)\leq \max \{d_{\mathcal{H}(X)}(A_1, B_1),d_{\mathcal{H}(X)}(A_2, B_2)\}.$$

There are multiple ways of showing this; here is one way that uses the "ball definition" of Pompeiu-Hausdorff metric:

$$d_{\mathcal{H}(X)}(A,B)=\inf \{r\in\mathbb{R}_{\geq0}| A\subseteq [B|<r] \mbox{ and } [A|< r]\supseteq B\},$$

where $[E|<r]=\{x\in X| \exists e\in E: d(x,e)<r\}$ is the union of all open balls of radius $r$ whose center is some point in $E$.

Proof: Without loss of generality say $d_{\mathcal{H}(X)}(A_1, B_1)\geq d_{\mathcal{H}(X)}(A_2, B_2)$. Let $r>d_{\mathcal{H}(X)}(A_1, B_1)$. Then

\begin{align*} &A_1\subseteq [B_1|<r],\quad[A_1|\leq r]\supseteq B_1\\ &A_2\subseteq [B_2|<r],\quad[A_2|\leq r]\supseteq B_2. \end{align*}

Thus

$$ A_1\cup A_2\subseteq [B_1\cup B_2|<r],\quad [A_1\cup A_2|<r]\supseteq B_1\cup B_2, $$

so that

$$ \{r\in\mathbb{R}_{\geq0}| A_1\subseteq [B_1|<r] \mbox{ and } [A_1|\leq r]\supseteq B_1\} \subseteq \{r\in\mathbb{R}_{\geq0}| A_1\cup A_2\subseteq [B_1\cup B_2|<r] \mbox{ and } [A_1\cup A_2|\leq r]\supseteq B_1\cup B_2\}. $$

Taking infima concludes the proof.

Here is a relevant idea: Ultimately this can be considered as the statement that taking finite unions is Lipschitz: Fix $N\in\mathbb{Z}_{\geq0}$ and consider the $N$-ary union operation:

$$ \bigcup:\underbrace{\mathcal{H}(X)\times \mathcal{H}(X)\times \cdots \mathcal{H}(X)}_{N \mbox{ many}}\to \mathcal{H}(X), (A_1,A_2,...,A_N)\mapsto \bigcup_{i=1}^N A_i. $$

Endowing the space of tuples of subsets with the $\ell^\infty$ product metric (the distance between two tuples is the maximum of the Pompeiu-Hausdorff distances between components), gives that the Lipschitz constant (see e.g. Question on the distortion of a metric embedding and Lipschitz maps) of finitary unions is not greater than $1$.

In fact, as you mention one can reorder the sets over which one takes unions to get a possibly better estimate:

$$ d_{\mathcal{H}(X)}\left(\bigcup_{i=1}^N A_i,\bigcup_{i=1}^N B_i\right)\leq \min_{\sigma\in S_N}\max_{1\leq i\leq N} d_{\mathcal{H}(X)}(A_i,B_{\sigma(i)}), $$

though ultimately the version without the permutations seem sufficient for most "soft" (in the sense of e.g. Difference between soft analysis and hard analysis) purposes.

I should also mention that currently I am teaching a class about these kinds of things and just recently we covered this statement; though I had left your question as an exercise; see https://youtu.be/XydQoxKnodE?feature=shared&t=7753

Alp Uzman
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