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 $x$.


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .