Let $X$ be a compact finite dimensional Alexandrov space with curvature bounded below. 



> 
> Does there exist $\varepsilon\_0>0$ (depending on $X$) such that for any $\varepsilon \in (0,\varepsilon\_0)$ and any point $x\in X$ the open ball $B(x,\varepsilon)$ is contractible? Is similar statement true for closed balls?
> 
> 
> 


$B(x,\varepsilon)$ denotes the open ball of radius $\varepsilon$ with the center at $x$. 



Contractibility of balls in Alexandrov spaces
Formally speaking the answer is "no".


Take a 2-dimensional cone with small total angle. Then for any $\varepsilon>0$ there is a point $x$ close enuf to the tip of the cone such that $B(x,\varepsilon)$ is an annulus. 


