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Let $(M, d)$ be a compact metric space, where $d$ is a distance metric. If I define the Hausdorff distance, $d_H$ as:

\begin{equation}\nonumber d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\} \end{equation} where $(Z)_\epsilon$ represents the $\epsilon$-fattening of $Z$ defined as $(Z)_\epsilon=\{m \in M:\exists z \in Z \text{ such that } d(z,x)\leq \epsilon\}$. Is it true that for arbitrary compact sets $X,Y$ and $Z$ that:

\begin{equation}\nonumber d_H(X\cup Z,Y) \leq d_H(X,Y)+ d_H(Z,Y) \end{equation}

I believe this is true and has something to do with the metric subadditivity property but am unsure how to proceed proving it.

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  • $\begingroup$ What is $(X)_\varepsilon$? Also, is this taking place in some ambient metric space? $\endgroup$
    – Reveillark
    Feb 27, 2021 at 20:07
  • $\begingroup$ $(X)_\epsilon$ is the $\epsilon$ fattening of $X$ defined using some distance metric $d$. I have added the definition of $(X)_\epsilon$ to the question. $\endgroup$
    – JDoe2
    Feb 27, 2021 at 20:58
  • $\begingroup$ We should even have $d_{H}(X \cup Z, Y) \leq \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace$ -- if I am not mistaken. $\endgroup$
    – v_lentin
    Feb 27, 2021 at 21:36

2 Answers 2

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First, note that the "$\inf$" in the definition of $d_{H}$ is actually a "$\min$" since we are dealing with compact subsets. Thus, for all compact subsets $X$ and $Y$ of $M$, we have $X \subset (Y)_{d}$ and $Y \subset (X)_{d}$, where $d = d_{H}(X, Y)$.

Now, let $X$, $Y$ and $Z$ be compact subsets of $M$. Set $$d = \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace \, \text{.}$$ On the one hand, we have $$Y \subset (X)_{d_{H}(X, Y)} \subset (X)_{d} \subset (X \cup Z)_{d} \, \text{.}$$ On the other hand, we have $$X \subset (Y)_{d_{H}(X, Y)} \subset (Y)_{d} \quad \text{and} \quad Z \subset (Y)_{d_{H}(Z, Y)} \subset (Y)_{d} \, \text{,}$$ and hence $$X \cup Z \subset (Y)_{d} \, \text{.}$$ Therefore, we have $$d_{H}(X \cup Z, Y) \leq \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace \, \text{.}$$

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  • $\begingroup$ I see! Thank you yes that proof makes sense to me! Thank you! The only thing which confuses me a little is what you say about the infinimum being a minimum - what is it a minimum over in this setting? $\endgroup$
    – JDoe2
    Feb 27, 2021 at 22:35
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    $\begingroup$ @JDoe2 We take an inf over $\epsilon \ge 0$. The answerer claims the inf is attained and is in fact a minimum. $\endgroup$ Feb 27, 2021 at 22:55
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    $\begingroup$ I meant that we have $X \subset (Y)_{d_{H}(X, Y)}$, and not only $X \subset (Y)_{\epsilon}$ for all $\epsilon > d_{H}(X, Y)$. We could have done the same without this remark and with lots of $\epsilon$, but I find the proof clearer like this. $\endgroup$
    – v_lentin
    Feb 27, 2021 at 22:56
  • $\begingroup$ Oh right! I see, thank you! $\endgroup$
    – JDoe2
    Feb 27, 2021 at 23:13
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Let $r>\max\{d_H(X,Y),d_H(Z,Y)\}$.

It is clear that $(X\cup Z)_r=(X)_r\cup (Z)_r$. By assumption, $$ (X)_r\subseteq Y \wedge (Y)_r\subseteq X \wedge (Z)_r\subseteq X\wedge (X)_r\subseteq Z $$ Therefore, $$ (X\cup Z)_r\subseteq Y \wedge (Y)_r\subseteq X\cup Z $$ By definition, $d_H(X\cup Z,Y)\le r$. Since $r>\max\{d_H(X,Y),d_H(Z,Y)\}$ was arbitrary, it follows that $$ d_H(X\cup Z,Y)\le \max\{d_H(X,Y),d_H(Z,Y)\} $$

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