Hausdorff metric and $\varepsilon$-thickenings Let $h(A,B)$ be the Hausdorff metric defined by:
$$
h(A,B)=\inf\{\varepsilon >0 \; | \; A \subseteq B_\varepsilon, B \subseteq A_\varepsilon \},
$$
where $A_\varepsilon$ is the $\varepsilon$-thickening of $A$ defined by $ A_\varepsilon = \{x \in \mathbb{R}^n \;|\; \exists a \in A,\; d(a,x)\le  \varepsilon\}$ (and $d$ is the Euclidian metric).
I am wondering whether $(A_{\varepsilon})_\delta=A_{\varepsilon+\delta}$.
It is fairly easy to show that:$(A_{\varepsilon})_\delta \subseteq A_{\varepsilon+\delta}$: Let $x \in (A_\varepsilon)_\delta$. Then there is a $y\in A_{\varepsilon}$ such that $d(x,y)\le \delta$. And since $y\in A_{\varepsilon}$, there is a $z \in A$ such that $d(y,z)\le \varepsilon$. By summing the two inequalities, and by the triangular inequality: $d(x,z)\in A_{\varepsilon+\delta}$.
But the other inclusion is trickier and I cannot manage to find a counter example.
Any suggestions? 

As Dave L. Renfro said in a comment, in general  the inclusion $(A_{\varepsilon})_\delta\subseteq A_{\varepsilon+\delta}$ may be strict. For example, consider $A=\{0\}$ in the metric space $\mathbb{Z}$ with the natural metric. Here $(A_{1/2})_{1/2}=\{0\}$ but $A_1 = \{-1,0,1\}$.
Equality $(A_{\varepsilon})_\delta = A_{\varepsilon+\delta}$ holds in all geodesic metric spaces [Definition: a space is geodesic if for any two points, there is a path between them whose length equals the distance between the points.] More generally, it holds in spaces $X$ with the following property: for every $x,y\in X$ and every $t\in [0,1]$ there is $z\in X$ such that 
$$d(z,x)=td(x,y)\quad\text{ and }\quad d(z,y)=(1-t)d(x,y)$$
Proof: let $p\in  A_{\varepsilon+\delta}$ and choose $r>{\varepsilon+\delta}$. There is $a\in A$ such that $d(p,a)<r$. By the assumption, there is $z\in X$ such that $d(z,a)=\varepsilon$ and $d(z,p)=r-\varepsilon$. Hence $d(p,A_\varepsilon)\le r-\varepsilon$. Since this holds for every $r>{\varepsilon+\delta}$, we have $d(p,A_\varepsilon)\le \delta$.
