Basic property of Hausdorff distance

Question

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.

Answer 1

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 $$d_{H}(X \cup Z, Y) \leq \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace \, \text{.}$$

Answer 2

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 $$ d_H(X\cup Z,Y)\le \max\{d_H(X,Y),d_H(Z,Y)\} $$