In euclidean n-space, it's easy to show that given a set $S$ of radius $< r$, the $a$-neighbourhood of $S$ is a ball, for any $a \geq 2r$.



> 
> **Proof**: Let $S$ be contained in $B\_r(y)$, $y \in \mathbb{R}^n$.
>  Note that if $a \ge 2r$ then $ B\_r(y) \subset Nbd\_a(S)$.
>  Let $z\in Nbd\_a(S) \backslash B\_r(y)$.
>  Consider the triangle
>  with vertices $z$, $y$ and $s$ with $s\in S$.
>  The length of the edge $yz$ is greater than $r$ which is greater
>  than the length of the edge $ys$.
>  It follows that the angle at $z$ is less than $\pi/2$ (less than $\pi/3$, in fact),
>  which means that points on the edge $yz$ near $z$ are closer to $s$ than $z$ is,
>  which implies that these points are also in $Nbd\_a(S)$.
>  Hence $Nbd\_a(S)$ is star-shaped with respect to $y$.
> 
> 
> 


I'd like a result for a metric $PL$ manifold, of the form:



> 
> **Theorem**: For an metric $PL$ manifold $M$, there is some $\epsilon > 0$ such that every subset $S$ of radius $r$ and all $a$ with $2r \leq a \leq \epsilon$, the neighbourhood $Nbd\_a(S)$ is homeomorphic to a ball.
> 
> 
> 


Can someone provide a proof?



When is the neighbourhood of a set a ball?
If I understand the question right, the answer is no. Make a triangulated $2$-manifold with Euclidean metrics on the simplices, such that the total angle around some vertex is very small. Let $S$ consist of two points, both of which are the same small distance $D$ from that vertex and (subject to that) as far from each other as possible. If $2r$ is the distance between the points then for a pretty big range of values of $a$ the union of $a$-balls centered at these two points is topologically an annulus.


