Basic property of Hausdorff distance

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

$$$$\nonumber d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\}$$$$ where $$(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:

$$$$\nonumber d_H(X\cup Z,Y) \leq d_H(X,Y)+ d_H(Z,Y)$$$$

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

• What is $(X)_\varepsilon$? Also, is this taking place in some ambient metric space? Feb 27, 2021 at 20:07
• $(X)_\epsilon$ is the $\epsilon$ fattening of $X$ defined using some distance metric $d$. I have added the definition of $(X)_\epsilon$ to the question. Feb 27, 2021 at 20:58
• 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. Feb 27, 2021 at 21:36

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 $$d_{H}(X \cup Z, Y) \leq \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace \, \text{.}$$
• @JDoe2 We take an inf over $\epsilon \ge 0$. The answerer claims the inf is attained and is in fact a minimum. Feb 27, 2021 at 22:55
• 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. Feb 27, 2021 at 22:56
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 $$d_H(X\cup Z,Y)\le \max\{d_H(X,Y),d_H(Z,Y)\}$$